Grade 9 (Alternate) Mathematics III - Learning Modules for EASE Program of DepEd

Compilation by Ben: [email protected]

COMPILATION OF LEARNING MODULES

GRADE 9 (Alternative)

MATHEMATICS III

Effective and Alternative Secondary Education

(EASE) Third Year

Compilation by Ben: [email protected]

CONTENTS

01. Module 1 ‐ Circles 02. Module 2 ‐ Circles 03. Module 1 Properties of Quadrilaterals 03. Module 2 ‐ Triangle Trigonometry 04. Module 2 Properties of Quadrilaterals 05. Module 1 ‐ Geometric Relations 06. Module 2 ‐ Geometric Relations 07. Module 3 ‐ Geometric Relations 08. Module 1 Similarity 09. Module 2 ‐ Similarity 10. Module 3 ‐ Similarity 11. Module 1 ‐ Plane Coordinate Geometry 12. Module 2 ‐ Plane Coordinate Geometry 13. Module 3 Plane Coordinate Geometry 14. Module 1‐ Triangle Congruence 16. Module 3 ‐ Triangle Congruence 17. Module 1 Geometry of Shape and Size 18. Module 2 ‐ Geometry of Shape and Size 19. Module 3 ‐ Geometry of Shape and Size 20. Module 4‐ Geometry of Shape and Size 21. Module 5 Geometry of Shape and Size 22. Module 6 Geometry of Shape and Size 23. Module 7 ‐ Geometry of Shape and Size 24. Module 8 ‐ Geometry of Shape and Size

Module 1 Circles

What this module is about This module will discuss in detail the characteristics of a circle as well as the segments and lines associated with it. Here, you will gain deeper understanding of the angles formed in circles, how to get their measures and how they are related to one another. Furthermore, this module will also give meaning to the circle being composed of arcs and how each arc is related to the angles formed in circles. What you are expected to learn This module is written for you to

1. define a circle. 2. define and show examples of the lines and segments associated with circles. 3. describe the relationship of lines and segments that are peculiar to circles. 4. define, identify and give examples of the kinds of arcs that compose a circle. 5. identify central angle and inscribed angle. 6. discover the relationship between the measures of central angle and inscribed angle

and their intercepted arcs.

How much do you know Answer the following as indicated. S 1. Given a circle with center O. Name the following : a. the circle T b. a diameter M

c. two radii OO d. two chords which are not diameters N

e. a secant f. a tangent R

2. If a radius is perpendicular to a chord then it ________ the chord.

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3. In the given circle A, PT is a diameter, therefore MT is a ________ and 4. PTM is a _________. 5. Radius AB ⊥ CE. If CE = 8 cm, then CX = ________. 6. Using the same figure, if AX = 3 cm, What is the length of radius AC? 7. In circle O, 93=∠BOCm . What is mBC? 8. What is BACm∠ ?

9. In the figure, PR ║ ST . Using the given Find mPT and RPSm∠ . 10. A quadrilateral PQRS is inscribed in a circle. If 103=∠Pm , what is Rm∠ ? What you will do

Lesson 1

Identifying a circle, the lines, segments and angles associated with it.

A circle is defined as the set of all points that are at the same distance from a given point in the plane. The fixed given point is called the center. The circle is named after its center. Hence in the figure, given is a circle O. A The set of points on the plane containing the circle is divided into 3, (1) the circle, (2) the set of points outside the circle and (3) the set of points inside the circle. OC , OB and OA are segment whose endpoints are the center of the circle and a point on the circle. These three segments are called radii of the circle.

Radius of a circle is a segment whose endpoints are the center and a point on the circle. In the figure, AD is a segment whose endpoints are points on the circle. AD is called chord of the circle. AB is a segment whose endpoints are points on the circle and it

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passes through the center. AB is called diameter of a circle. Diameter of a circle is a chord that passes through the center. Lines on the plane containing the circle may intersect the circle at one point or at two points or not at all. Fig. 1. line a does not Fig. 2. line b intersect Fig. 3 line c intersect circle intersect circle O. circle O at point X at two points R and S. In figure 2, line b is tangent to the circle, and in figure 3, line c is a secant. Hence, we can use the following definitions. Tangent is a line that intersect a circle at one points. Secant is a line that intersect a circle at two points. Some theorems in circle show relationship between chord and radius. One of them is this theorem: Theorem: If a radius is perpendicular to a chord, then it bisects the chord.

Proof: Consider the given circle. If radius OA ⊥ BC at D, then OA bisects BC or BD = DC. One way of proving segments or angles congruent is by showing that they are corresponding parts of congruent triangles. Here, we must prove that BD and DC are corresponding sides of congruent triangles. If O and B are joined and O and C are also joined, we have ∆OBD and ∆OCD. Both of these triangles are right since BCOA ⊥ and thus ODB∠ and ODC∠ are both right angles. Since OB and OC are both radii of the same circle, hence they are congruent. And finally OBOD ≅ by reflexive property. Therefore, by

the HyL Congruency for right triangles, OCDOBD ∆≅∆ . Since the two triangles are congruent, then the remaining corresponding parts such as BD and DC are also congruent.

We have just proven the theorem here, only this time, instead of using the two column form we use the paragraph form.

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Our conclusion therefore is that a radius that is perpendicular to a chord bisects the chord. The most important considerations here were the perpendicularity and the word to bisect. Examples: 1. DEOB ⊥ at T, DT = 3x -7 , TE = x + 15 Solution: Since ⊥OB DE , then DT = TE Hence, 3x – 7 = x + 15 2x = 15 + 7 2x = 22 x = 11 Substituting the value of x, we get DT = 3(11) – 7 = 33 – 7 = 26 TE = 11 + 15 = 26 DE = DT + TE DE = 26 + 26 = 52 There are other theorems whose main idea is taken from the previously proven theorem. The next theorem serves as the converse of the first theorem and it states that: If a radius of a circle bisects a chord that is not a diameter, then it is perpendicular to the chord. If the previous theorem was proven using the HyL congruence for right triangle, the converse is proven using the reverse process, that is two angles must be proven part of congruent triangles and they are congruent and supplementary. You can prove the theorem as part of your exercise. Examples on how to use these two theorems are given below. 2. Given: AB bisects chord CD at E. CD = 6, AE = 4 Find the length of the radius of the circle. Solution: Based on the theorem, CDAB ⊥ , thus ADEACE ∆≅∆ and both are right triangles. By the Pythagorean theorem, we can solve for the length of the radius. In ∆ACE, AC 2 = AE 2 + CE 2 But CE = ½ CD so

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CE = ½ (6) CE = 3 AC 2 = AE 2 + CE 2 AC 2 = 42 + 32 AC 2 = 16 + 9 = 25 AC = 25 AC = 5 Lesson on circle is very rich with theorems and definitions, principles and postulates. Some of those theorems and definitions will be introduced as we plod along with this module. Definitions:

• Congruent circles are circles that have congruent radii. • Concentric circles are coplanar circles having the same center

Illustrations: a) b) Circle A is congruent to circle B if and These two circles are only if BYAX ≅ concentric circles Theorem: If chords of a circle or of congruent circles are equidistant from the center(s), then the chords are congruent Illustration of the theorem. Circle O ≅ circle P OX = PY Then, CDAB ≅

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Try this out A. Using the given figure, name 1. the circle 2. 2 diameters 3. 2 chords which are not diameters 4. 2 secants 5. a tangent B. Given: CDAB ⊥ at E CD is 10 cm long. How far is CD from the center if the length of the radius is

1. 13 cm 5. 12 cm 2. 7 cm 6. 10 cm 3. 14 cm 7. 5 2 cm 4. 8 cm 8. 63 cm

C. Given: CD is 20 cm long. How long is the radius of the circle if the distance of CD from the center is

1. 7 cm 3. 13 cm 2. 10 cm 4. 8 cm 5. 5 cm 7. 55 cm 6. 21 cm 8. 64 cm

D. AC is 12 cm long. How long is chord CD if its distance from the center is 1. 10 cm 5. 9 cm 2. 6 cm 6. 23 cm 3. 8 cm 7. 112 cm 4. 5 cm 8. 54 cm E. Solve the following problems. 1. MPON ⊥ ME = 7x + 5 PE = 4x – 20 Solve for ME , PE and MP

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2. In a circle are two chords whose lengths are 10 cm and 24 cm respectively. If the

radius of the circle is 13 cm, what is the maximum distance of the two chords? What is their minimum distance?

Lesson 2

Arcs and Central Angles

A part of a circle between any two points is an arc. In the figure, the set of points from A to B is an arc. A circle is in itself an arc. Arc of a circle is measured in terms of degrees. The whole arc making up the circle measures 360°.

Any arc of a circle can belong to any of these three groups. a. minor arc – an arc whose measure is between 0 and 180°. b. semicircle – an arc whose measure is exactly 180° c. major arc – an arc whose measure is between 180° and 360°

In the given figure, AB is a diameter, hence AB represents

a semicircle, AC is minor arc and ABC is a major arc. Aside from AC, another minor arc in the figure is BC. ACB also represents a semicircle. Angles in a circle are formed by radii, chords, secants and tangents. Determination of the measures of the angles formed by these lines depends upon the measure of the intercepted arcs of the given angles. Examples: In circle some angles formed by chords and radii are shown. Each of the angles intercepts an arc defined by the endpoints contained on the sides of the angle. ∠AEB intercepts AB. ∠BOC intercepts BC ∠COD intercepts CD ∠EOD intercepts ED ∠AEB intercepts AB. ∠AEB intercepts AB. ∠AEB intercepts AB.

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At this point we will discuss in detail the kinds of angles formed in a circle, their characteristics and how to get their measures from the measures of the intercepted arcs. We will start with the angle formed by two radii.

Central angle is an angle formed by two radii and the vertex is the center of the circle. In the figure, ∠AOB, ∠BOD and ∠DOC are all examples of central angles. Each of these angles has its own intercepted arc. ∠AOB intercepts AB, ∠BOD intercepts BD and ∠DOC intercepts DC. The measure of a central angle is numerically equal to its intercepted arc. In the figure, ∠BAC is a central angle and ∠BAC intercepts BC. Since mBC = 83, then m∠BAC = 83, mBDC = 277°.

In the study of geometry, every new topic or concept is always associated with study of postulates, theorems and definitions. In the study of arcs and angles in a circle, we will discuss many theorems that will help us solve problems involving the said concepts. We will start with the simplest postulate in the chapter.

Like any measure, measure of an arc is also a unique real number and as such, we

can perform the four fundamental operations on those measure. So the first postulate is the Arc Addition Postulate: The measure of an arc formed by two adjacent non-overlapping arcs is the sum of the measures of the two arcs. In the given circle, m AC = m AB + m BC Examples: 1. DG is a diameter. Find the measure of the following arcs. DG, DE, DF, GE, DGF Solution: Since DG is a diameter, then DG is a semicircle.

Therefore, m DG = 180 m DE = 180 – (60 + 70) = 180 – 130 = 50

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m DF = m DE + m EF = 50 + 60 = 110 m GE = m GF + m FE = 70 + 60 = 130

m DGF = m DG + m GF = 180 + 70 = 250 Definitions: In the same circle or in congruent circles, arcs which have the same measure are congruent. Example: 1. In the figure, m DC = 60, m BC = 60 m AB = 60 . Therefore, DC≅BC ≅ AB 2. Since every semicircle measures 180°, then all semicircles are congruent. Theorem: If two minor arcs of a circle or of congruent circles are congruent, then the corresponding chords are congruent. Examples: 1. Given: AB ≅ BC Since AB subtends AB and BC subtends BC then AB ≅ BC 2. Circle O ≅ circle M If AB ≅ XY, then XYAB ≅ Theorem: If two chords of a circle or of congruent circles are congruent, then the corresponding minor arcs are congruent. This is the converse of the previous theorem. Basically if you prove these two theorems, the steps will be just the reverse of the other. Instead of proving them, showing examples will be more beneficial to you.

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In circle A, if PQRS ≅

then RS ≅ PQ Theorem: If two central angles of a circle or of congruent circles are congruent, then the corresponding minor arcs are congruent. Example: In circle O, ∠MNO≅ ∠BOA

Therefore, MP ≅ AB Theorem: If two minor arcs of a circle or of congruent circles are congruent, then the corresponding central angles are congruent. Example: In circle A, DEBC ≅ Therefore `DAEBAC ∠≅∠ Theorem: If two central angles of a circle or of congruent circles are congruent, then the corresponding chords are congruent. Given: In circle O, AOBXOY ∠≅∠

Prove: XY AB≅ Proof

Statements Reasons 1. In circle O, AOBXOY ∠≅∠ 2. OBOX ≅ , OAOY ≅ 3. BOAXOY ∆≅∆ 4. XY AB≅

1. Given 2. Radii of the same or congruent circles are congruent 3. SAS congruency Postulate 4. Corresponding parts of congruent triangles are congruent..

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Theorem: If two chords of a circle or of congruent circles are congruent circles are congruent, then the corresponding central angles are congruent. Given: In circle A, STPR ≅ Prove: SATPAR ∠≅∠ Proof:

Statements Reasons 1. In circle A, STPR ≅ 2. ASAP ≅ ATAR ≅ 3. SATPAR ∆≅∆ 4. SATPAR ∠≅∠

1. Given 2. Radii of the same circle are

congruent.

3. SSS Congruency Postulate 4. Corresponding parts of congruent

triangles are congruent Examples: Given: AB and CD are diameters of circle E.

1. What is true about AED∠ and BEC∠ ? Why? 2. What kind of angles are they? 3. Give as many conclusions as you can

based on the previously discussed theorems. Answers:

1. BECAED ∠≅∠ . They are vertical angles and vertical angles are congruent. 2. In the circle they are central angles. Central angles are angles whose vertex is the

center of the circle. 3. a. AD ≅ BC. If two central angles of a circle or of congruent circles are congruent,

then the corresponding arcs are congruent. b. BCAD ≅

Likewise 1. BEDAEC ∠≅∠ 2. AC ≅ DB 3. DBAC ≅

Try this out A. AB is a diameter of circle O. 82=∠AOEm .

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Find the measures of: 1. AB 4. ABE 2. AE 5. BAE 3. BE

B. GE and FD are diameters of circle A. If DA = 73º, find the measures of 1. ∠DAE 5. GF 2. ∠GAF 6. DG 3. ∠EAF 7. FDE 4. ∠DAG C. Given circle A. If m BTY = 116, and m = 3n, find 1. m 5. m ∠BAT 2. n 6. m ∠TAY 3. BT 7. m ∠BAY 4. TY

D. Given circle O. AB ≅ BC. If mAB = 56, what is

AOBm∠ ? What is mABC ? Which chords are congruent?

E. A. B and C are three points on the circle. IF AC≅ AB≅ BC, what is the measure of each arc? What is true about the chords ,, ABAC and BC ? If ABC is 16 more than three times AC, find mAC, mABC.

F. P, Q and R are three points on a circle. If the ratio PQ:QR:PR = 3:4:5, find the measures of PR, QR and PS.

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G. Using the figure and the given in it, find the measures of: 1. PQ 5. POQ∠ 2. QR 6. ∠QOR 3. SR 7. ∠SOR 4. PS 8. ∠POS H. BD and EC are diameters of circle A. If 35=∠Cm , find the measures of 1. ∠B 5. ∠EAD

2. ∠E 6. BC 3. ∠D 7. CD 4. ∠BAC

Lesson 3

Arcs and Inscribed Angles

Another angle in a circle that is very important in the study of circle is the inscribed angle. Definition: An inscribed angle is an angle whose vertex lies on the circle and the sides contain chords of the circle.

Fig. 1 Fig. 2 Fig 3 Each of the angle shown above is an example of an inscribed angle. Three cases are represented here relative to the position of the sides in relation to the center of the circle.

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Case 1. the center of the circle is on one side of the inscribed angle. Case 2, the center of the circle is in the interior of the inscribed angle. Case 3, the center of the circle is on the exterior of the inscribed angle. In the study of the angles in a circle and in determining their measures, it is important to determine the intercepted arc(s) of the given angle. To understand better, let us see some examples. In the figure, the arc in the interior of the angle is the intercepted arc of the angle. The intercepted arc of BAC∠ is the minor arc AC. In the given examples of inscribed angles above the following holds:

a) In figure 1, ∠DEF is an inscribed angle ∠DEF intercepts arc DF

b) In figure 2, ∠PST is an inscribed angle,

∠PST intercepts arc PT

c) In figure 3, ∠BAC is an inscribed angle ∠BAC intercepts arc BC

Every angle whether in a circle on in any plane is associated with a unique number defined as its measure. If the measure of a central angle is equal to the measure of its intercepted arc, the next theorem will tell us how to find the measure of the inscribed angle. Theorem: Inscribed angle Theorem The measure of an inscribed angle is equal to one half the measure of its intercepted arc.

It means that in the given figure, mDFDEFm 2

1=∠

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Since there are three cases by which an inscribed angle can be drawn in a circle, then we have to prove each of those cases. Case 1 (One side of the angle is the diameter of the circle) Given: Circle O with inscribed angle DEF∠ Use the notation in the figure for clarity

Prove: ( )mDFDEFm21

=∠

Proof:

Statements Reasons 1. Circle O with inscribed angle DEF∠ 2. Draw OF to form ∆FOE 3. ∠ 1 is an exterior angle of ∆FOE 4. m∠ 1 = x + y 5. OEOF ≅ 6. ∆FOE is an isosceles triangle 7. x = y 8. m∠ 1 = x + x = 2x 9. 2x = m∠ 1, x = ½ m∠ 1 10. But ∠ 1 is a central angle 11. m∠ 1 = m DF

12. x = ( )mDFDEFm21

=∠

1. Given 2. Line determination postulate 3. Definition of exterior angle 4. Exterior angle theorem 5. Radii of the same circle are congruent

6. Definition of isosceles triangle 7. Base angles of isosceles triangle are congruent 8. Substitution (Steps 4 and 7) 9. Multiplication property of equality 10. Definition of central angle 11. Measure of a central angle equals its intercepted arc. 12. Substitution (Steps 9 and 11)

So, we have proven case 1. Let us now prove case 2 of the inscribed angle theorem. Case 2. (The center of the circles lies in the interior of the inscribed angle) Given : Circle O with inscribed PQR∠

Prove: m PQR∠ = 21 m PR

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Proof: Statements Reasons

1. Circle O with inscribed PQR∠ . Use the given notation in the figure. 2. Draw diameter QS 3. m PQR∠ = a + b 4. a = mPS2

1 b = mSR2

1 5. a + b = mPS2

1 + mSR21 = )(2

1 mSRmPS + 6. mPR = mPS + mPR

7. m `PQR∠ = 21 (mPS + mPR)

8. m `PQR∠ = 21 mPR

1. Given 2. Line determination Postulate 3. Angle Addition Postulate 4. Inscribed angle theorem (Case 1) 5. Addition Property of Equality

6. Arc Addition Postulate

7. Transitive Property of Equality 8. Transitive Property of Equality

Case 3. (The center is in the exterior of the inscribed angle) Given: BAC∠ is an inscribed angle in circle O Use the additional notation in the figure Prove: mBCBACm 2

1=∠ Proof:

Statements Reasons 1. Draw diameter AD 2. BACmDABmDACm ∠+∠=∠ 3. DABmDACmBACm ∠−∠=∠ 4. mDCDACm 2

1=∠ mDBDABm

21=∠

5. mDBmDCBACm 21

21 −=∠ = 2

1 (mDC-mDB) 6. mDC = mDB + mBC 7. mBC = mDC – mDB 8. mBCBACm 2

1=∠

1. Line determination Postulate 2. Angle Addition Postulate 3. Subtraction Property of Equality 4. Inscribed angle Theorem (Case 1) 5. Substitution 6. Arc Addition Postulate 7. Subtraction Property of Equality

8. Substitution From the proofs that were given, we can therefore conclude that wherever in the circle the inscribed angle is located, it is always true that its measure is one-half its intercepted arc.

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Examples. Use the figure at the right. 1. Given: circle O. 80=∠BODm Find: mBD, BADm∠ Solution: Since 80=∠BODm , then a. mBD = 80 b. BADm∠ = BD2

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= )80(21

= 40 2. Given: circle O. 37=∠BADm Find: mBD , BODm∠ Solution: mBDBADm 2

137 ==∠

mBD = 2(37) = 74

mBDBODm =∠

74=∠BODm Like in the study of central angles and its measure, discussing inscribed angles and its measure also involves many theorems. Each previous theorem studied is always a tool in proving the next theorem. The following theorem is one of the most useful theorem in solving problems which involve inscribed angles. Theorem: Angle in a semicircle theorem. An angle inscribed in a semicircle is a right angle. Given: Circle O. BAC is a semicircle. Prove: BAC∠ is a right angle. ( 90=∠BACm

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Proof:

Statements Reasons

1. Draw BC passing through center O. 2. ∠ABC, ∠ACB, and ∠BAC are all inscribed angles. 3. ACABCm 2

1=∠ , ABACBm 21=∠

4. mBAC = mAC + mAB 5. BAC is a semicircle 6. mBAC = 180 7. mAC + mAB = 180 8. ABACACBmABCm 2

121 +=∠+∠ = )(2

1 ABAC +9. ACBmABCm ∠+∠ = 2

1 (180) = 90 10. ACBmABCm ∠+∠ + BACm∠ = 180 ACBmABCm ∠+∠ = 90 11. BACm∠ = 90

12. BAC∠ is right angle

1. Definition of diameter 2. Definition of inscribed angles 3. Inscribed Angle Theorem 4. Arc Addition Postulate 5. Given 6. The measure of a semicircle is 180 7. Transitive Property of Equality

8. Addition Property of Equality (Step 3)

9. Substitution (Steps 7 and 8) 10. The sum of the angles of a triangle is 180. 11. Subtraction Property of Equality (Step 10 – step 9) 12. Definition of a right angle

From this point onward, you can use this very important theorem in proving or in exercises. There are other theorems on inscribed angle that are also important as the previous theorem. Of those theorems, we will prove two and the rest, you can answer as exercises. Theorem: Inscribed angles subtended by the same arc are congruent. Given: Circle O. MN subtends both ∠T and ∠P ∠T and ∠P are inscribed angles Prove: ∠T ≅ ∠P Proof:

Statements Reasons 1. In circle O, MN subtends both ∠T and ∠P. ∠T and ∠P are inscribed angles. 2. mMNTm 2

1=∠ mMNPm 2

1=∠ 3. PmTm ∠=∠ 4. PT ∠≅∠

1. Given 2. Inscribed Angle Theorem 3. Transitive Property of Equality 4. Definition of congruent angles

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The next theorem is about polygon inscribed in a circle. Definition:

A polygon inscribed in a circle is polygon whose vertices lie on the circle.

Examples: The figures below show examples of inscribed polygon. Inscribed triangle Inscribed Inscribed Inscribed Quadrilateral Pentagon Hexagon Theorem: Opposite angles of a inscribed quadrilateral are supplementary. Given: Circle A. PRST is an inscribed quadrilateral. Prove: ∠P and ∠S are supplementary ∠R and ∠T are supplementary Proof:

Statements Reasons 1. Circle A. PRST is an inscribed quadrilateral. 2. ∠m P = ½ mRST ∠m S = ½ mRPT ∠m R = ½ mPTS ∠m T = ½ mPRS 3. ∠m P + ∠m S = ½ mRST + ½ mRPT 4. ∠m P + ∠m S = ½( mRST + mRPT) 5. mRST + mRPT = 360 6. ∠m P + ∠m S = ½(360) 7. ∠m P + ∠m S = 180 8. ∠P and ∠S are supplementary 9. ∠m R + ∠m S = ½ mPTS + ½ mPRS 10. ∠m R + ∠m S = ½ (mPTS + mPRS) 11. mPTS + mPRS = 360 12. ∠m R + ∠m S = ½(360) 13. ∠m R + ∠m S = 180 14. ∠R and ∠ T are supplementary

1. Given 2. Inscribed angle theorem 3. Addition property of equality 4. Factoring 5. The arc of the whole circle is 360º 6. Substitution (Steps 4 and 5) 7. Algebraic process (step 6) 8. Definition of supplementary angles 9. Addition property of equality 10. Factoring 11. The arc of the whole circle is 360º 12. Substitution (Steps 4 and 5) 13. Algebraic process (step 6) 14. Definition of supplementary angles

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Examples: 1. Given: XY is a diameter. a. What kind of angle is ∠Z? b. If ∠m X = 35, what is ∠m Y? c. If ∠m Y = 73, what is mXZ? What is mYZ? Answers:

a. Since XY is a diameter, then XZY is a semicircle and ∠Z is inscribed in a semicircle. Therefore, ∠ Z is a right angle. b. m∠X + m∠Y = 90.

m∠Y = 90 - m∠X m∠Y = 90 – 35 m∠Y = 65 c. ∠Y intercepts XZ. m XZ = 2(75) = 150 m YZ = 180 – 150 m YZ = 30 2. MNOP is inscribed in circle E. If m M∠ = 94, what is m∠O? Answer: ∠M and ∠O are supplementary. m∠M +m∠O = 180

m∠O = 180 - m∠M = 180 – 94 = 86

3. Given: Circle O. AB is a diameter ∠m 1 = 36 and ∠m 3 = 61. Find: ∠m 2, ∠m 4, ∠m CBD,

∠m ADB, ∠m ACB, mCBD, ∠m CAD, mAD

Solution: ∠m 1 = 36, mAC = 2(36) = 72 ∠m 3 = 61, mBD = 2(61) = 122 ∠m 2 = ½ AD mAD = 180 – BD = 180 – 122 = 58 ∠m 2 = ½ (58) = 29

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∠m 4 = ½ CB mCB = 180 – mAC = 180 – 72 = 108 ∠m 4 = ½ (108) = 54 ∠m CBD = ½(m AC + mAD) = ½(72 + 58) = ½(130) = 65 ∠m ADB = 90 (Angle in a semicircle) ∠m ACB = 90 (angle in a semicircle) mCBD = mCB + mBD = 108 + 122 = 230 ∠m CAD = ½(mCBD) = ½ (230) = 115 Try this out A. Given: AB is a diameter of circle O.

79=mAC . Find: 1. ∠m AOC 2. ∠m ABC 3. ∠m COB B. Given: Circle A., XY and BE are diameters ∠m XAE = 104. Find: 4. m XE 5. m BX 6. ∠m E 7. ∠m B 8. ∠m BXY 9. ∠m YXE

A

C

O

B

B

X

A

E

Y

22

C. Using the given figure, find:

10. x 11. ∠m MNQ 12. ∠m MOQ 13. ∠m POQ 14. ∠m M 15. ∠m MON D. BD is a diameter of circle A. If m BC = 78, and m DE = 132, find:

16. m CD 23 ∠m 6 17. m BE 24. ∠m 7 18. ∠m 1 25. ∠m 8 19. ∠m 2 26. ∠m 9 20. ∠m 3 27. ∠m 10 21. ∠m 4 22. ∠m 5

E. PRST is inscribed in circle A. If ∠m T = (5x – 4)º and ∠m R = (4x + 13)º find:

28. x 29. ∠m T 30. ∠m R

F. ∆XYZ is inscribed in the circle. If XZXY ≅ , prove that ∠m X = b - a

Q M

O 3x

N P

● D

C

A B

E

3 4

1 2

9 10 5

6

3

87

P

T

S

RA ●

Y

X Za b

●

23

Let’s Summarize

1. A circle is the set of all points that are at the same distance from a given point in the

plane. 2. Some of the lines associated with circle are the following:

a. Radius b. Chord c. Diameter d. Secant e. Tangent

3. If a radius is perpendicular to a chord, then it bisects the chord. 4. If a radius of a circle bisects a chord that is not a diameter, then it is perpendicular to

the chord.

5. Congruent circles are circles that have congruent radii. Concentric circles are circles having the same center.

6. A circle is made up of arcs classified as minor arc, semicircle and major arc.

7. A central angle is an angle on the circle whose vertex is the center of the circle.

8. The measure of the central is numerically equal to its intercepted arc.

9. If two minor arcs of a circle or of congruent circle are congruent, then,

a. the corresponding central angles are congruent, b. the corresponding subtended chords are congruent

10. An inscribed angle is an angle on the circle whose vertex is a point on the circle. 11. The measure of an inscribed angle is equal to one-half its intercepted arc.

12. An angle inscribed in a semicircle is a right angle.

13. Inscribed angle subtended by the same arc are congruent.

14. The opposite angles of an inscribed quadrilateral are supplementary.

24

What have you learned

Answer as indicated.

1. If the diameter of a circle is 15 cm, what is the length of the radius? 2. A line that intersects a circle at one point is called _________. 3. If a radius bisects a chord which is not a diameter, then its is _________ to the

chord. 4. CD is a diameter of circle A. CED is a ______________. 5. CE is a _____________. 6. CDE is a ______________ 7. PTON ⊥ at E. If OE = 6 cm, and the radius

of the circle is 10 cm, what is the length of PT ?. 8. AC is a diameter of circle O. Using the given in the given figure, find a. ∠m A b. ∠m C c. m AB d. m BC 9. PT is a diameter of circle Q. Find a. ∠m PQR b. ∠m RQT

10. ∆ABC is inscribed in circle O. If the ratio of ∠m A: ∠m B: ∠m C = 2:3:5, Find a. m BC b. m AC c. m AB

A

D

C

E

●

O

TP

N

E

A

B C

●A

R Q

P

T

2x 3x

A

C

B

● O

25

Answer Key

How much do you know

1. a. circle O b. MN c. MO , ON

d. MT , MR e. MR f. MS 2. bisects 3. minor arc 4. major arc 5. 4 cm 6. 5 cm 7. 93° 8. 46.5° 9. 98°, 49° 10. 77°

Try this out Lesson 1 A. 1. circle O 2. AC , BD 3. ,AD BC

4. EC , BC 5. CF B. 1. 12 cm 5. 119 cm 2. 2 6 cm 6. 35 cm 3. 171 cm 7. 5 cm 4. 43 cm 8. 29 cm C. 1. 149 cm 5. 55 cm 2. 210 cm 6. 11 cm 3. 269 cm 7. 15 cm 4. 2 41 cm 8. 14 cm

26

D. 1. 114 cm 5. 76 cm 2. 312 cm 6. 22 cm 3. 58 cm 7. 20 cm 4. 192 cm 8. 16 Problem Solving:

1. ME = 52, PE = 52, MP = 104 2. maximum distance is 12 cm

minimum distance is 2 cm Lesson 2 A. 1. 180º 3. 98º 2. 82º 4. 278º 5. 262º B. 1. 73º 5. 73º 2. 73º 6. 107º 3. 107º 7. 253º 4. 107º C. 1. 29º 5. 29º 2. 87º 6. 87º 3. 29º 7. 116º 4. 87º D. 1. 56 2. 56 3. AB and CD E. 1. Each arc measures 120 2. AC = 86º 3. ABC = 274º F. PQ = 90 QR = 120 PR = 150 G. 1. 38º 5. 38º 2. 157º 6. 157º 3. 76º 7. 76º 4. 89º 8. 89º

27

H. 1. 35º 5. 110º 2. 35º 6. 110º 3. 35º 7. 17º 4. 110º Lesson 3 A. 1. 79 3. 101 2. 39.5 B. 4. 104 7. 52 5. 76 8. 52 6. 38 9. 38 C. 10. 36 13. 108 11. 36 14. 36 12. 72 15. 108 D. 16. 102 22. 24 17. 48 23. 39 18. 51 24. 66 19. 66 25. 24 20. 39 26. 105 21. 51 27. 75 E. 28. 19 29. 91 30. 89 F. Proof: 1. ∠X + ∠Y + a = 180 1. Sum of the measures of the angles of a triangle is 180. 2. b = m∠X + m∠Y 2. Exterior angle theorem 3. m∠Y = a 3. Angles opposite equal sides in the same triangle are congruent 4. b = m∠X + a 4. Substitution 5. m∠X = b – a 5. Subtraction Property of Equality What have you learned

1. 7.5 cm 8. a. 22.5 10. a. 72 2. tangent b. 67.5 b. 108 3. perpendicular c. 135 c. 180 4. semicircle d. 45 5. minor arc 9. a. 72 6. major arc b. 108 7. 16 cm

Module 2

Circles

What this module is about This module will discuss in detail the characteristics of tangent and secants; the relationship between tangent and radius of the circle; and how secant and tangent in a circle create other properties particularly on angles that they form. This module will also show how the measures of the angles formed by tangents and secants can be determined and other aspects on how to compute for the measures of the angles. What you are expected to learn This module is written for you to

1. Define and illustrate tangents and secants. 2. Show the relationship between a tangent and a radius of a circle. 3. Identify the angles formed by tangents and secants. 4. Determine the measures of angles formed by tangents and secants.

How much do you know If CB and CD are tangents to circle A, then

1. CB ___ CD 2. ABCB ____

3. CB and CA are tangents to circle O. If 160=∠BOAm , then =∠Cm _____.

4. If 22=∠BCOm , what is ACOm∠ ?

A

B

C

D

C

B

O

A

2

5. In the figure, if m PTA = 242, what is PALm∠ ?

6. Two secants GD and BL intersect at A. If m BG = 83 and m LD = 39, find GABm∠ .

7. In the figure, if m MX = 54, and mAX = 120, what is Nm∠ ? 8. AC and AT are tangents to the circle with C and T as the points of tangency. If ACT∆ is an equilateral triangle, find m CT. 9. AC and AT are secants. If 23=∠Am and m CT = 66, find m BM.

A

T

B L

P O

G

L

D

A B

N

X

M A

A

C

T

M

A B

M

C

T

3

10. DB is a diameter of circle O. If the ratio of DE:EB is 3:2, what is Xm∠ ? What you will do

Lesson 1

Circles, Tangents, Secants and Angles They Form

A line on the same plane with a circle may or a) may not intersect a circle. If ever a line intersects a circle, it could be at one point or at two points. The figures at the right showed these three instances. Figure a showed a line that does not intersect the circle. b) Figure b showed that line t intersects the circle at only one point. Figure c showed line l intersecting the circle at two points A and C. c) We will focus our study on figures b and c. In figure b, line t is called a tangent and point B is called the point of tangency. Therefore a tangent is a line that intersects a circle at only one point and the point of intersection is called the point of tangency. In figure c, line l intersects the circle at two points A and C. Hence, line l is called a secant. Thus a secant is a line that intersects a circle at two points. Some properties exist between tangent and circle and they will be discussed here in detail. The first theorem is given below. Theorem: Radius-Tangent Theorem. If a line is tangent to a circle, then it is perpendicular to the radius at the point of tangency.

t B

A C

l

X

E D

O

B

4

Given: line t is tangent to circle O at A. OA is a radius of the circle. Prove: t ⊥ OA Proof:

Statements Reasons 1. Let B be another point on line t. 2. B is on the exterior of circle O 3. OA < OB

4. ⊥OA t

1. The Line Postulate 2. Definition of a tangent line ( A tangent can intersect a circle at only one point . 3. The radius is the shortest segment from the center to the circle and B is on the exterior of the circle. 4. The shortest distance from a point to a line is the perpendicular segment.

Example: In the figure, if AC is tangent to circle B, then BDAC ⊥ at D. The converse of the theorem is also true. Converse: The line drawn perpendicular to the radius of a circle at its end on the circle is tangent to the circle. Illustration: If BDAC ⊥ at D, then AC is tangent to circle B. Examples: GY is tangent to circle A. 1. What kind of triangle is ∆ AGY? Give reason. 2. If 79=∠Am , what is Ym∠ ? Solutions: 1. ∆AGY is a right triangle because GY is tangent to circle A and tangent line is perpendicular to the radius of the circle. Perpendicular lines make right angles between them thus AGY∠ is a right angle making ∆AGY a right triangle.

O

A B t

B

D C A

B

D C A

A

YG

5

2. Since ∆AGY is a right triangle, then ∠m A + ∠m Y = 90 79 + ∠m Y = 90 ∠m Y = 90 – 79 = 11 A circle is composed of infinite number of points, thus it can also have an infinite number of tangents. Tangents of the same circle can intersect each other only outside the circle. At this point, we will discuss the relationship of tangents that intersect the same circle. As such, those tangents may or may not intersect each other. Our focus here are those tangents that intersect each other outside the circle. Consider the given figure: AM and AY are tangent segments from a common external point A. What relationship exists between AM and AY ? The next theorem will tell us about this relationship and other properties related to tangent segments from a common external point. Theorem: If two tangent segments are drawn to a circle from an external point then

a. the two tangent segments are congruent and b. the angle between the segments and the line joining the external point and the

center of the circle are congruent. Given: Circle A. BC and BD are two tangent segments from a common external point B. C and D are the points of tangency. Prove: a. BDBC ≅ b. DBACBA ∠≅∠ Proof:

Statements Reasons 1. Draw AC , AD , AB 2. BC and BD are two tangent segments from a common external point B. 3. BCAC ⊥ , BDAD ⊥ 4. ACB∠ and ADB∠ are right angles

1. Line determination Postulate 2. Given

3. A line tangent to a circle is perpendicular to the radius at the point of tangency 4. Definition of right angles

O • A

M

Y

A

C

D B

6

5. ∆ACB and ∆ADB are right triangles 6. ADAC ≅ 7. BDBC ≅ 8. ∆ACB ≅ ∆ADB 9. BDBC ≅ 10. DBACBA ∠≅∠

5. Definition of right triangles

6. Radii of the same circle are congruent 7. Reflexive property of Congruency 8. Hy L Congruency Postulate 9. Corresponding parts of congruent triangles10. are congruent.

Examples: a) In the figure, CB and CD are tangents to circle A at B and D.

1. If CB = 10 what is CD? 2. If 49=∠BACm , what is BCAm∠ ? 3. ,73=∠BCDm what is DCAmBCAm ∠∠ ?

Solution: 1. Since CB and CD are tangents to the same circle from the same external point, then CB ≅ CD , and therefore, CB = CD. Thus if CB = 10 then CD = 10 2. 90=∠+∠ BCAmBACm 49 + 90=∠BCAm 4990 −=∠BCAm = 41 3. )(2

1 BCDmBCAm ∠=∠ = )73(2

1 = 36.5 DCABCA ∠≅∠ 5.36=∠=∠ DCAmBCAm b) PQ , QR and PR are tangents to circle A at S, M and T respectively. If PS = 7, QM = 9 and RT = 5, what is the perimeter of ∆PQR? Solution: Using the figure and the given information, It is therefore clear that PS = PT, QS = QM and RM = RT. PQ = PS + SQ QR = QM + MR PR = PT + RT

C

B

D

A

P

Q

R

S

T

M

A

7

Perimeter of ∆PQR = PQ + QR + PR = (PS + SQ) + (QM + MR) + (PT + RT) = (PS + QM) + (QM + RT) + (PS + RT) = 2PS + 2QM + 2RT = 2(PS + QM + RT) = 2( 7 + 9 + 5) = 2 (21) = 42 Every time tangents and secants of circles are being studies, they always come with the study of angles formed between them. Coupled with recognizing the angles formed is the knowledge of how to get their measures. The next section will be devoted to studying angles formed by secants and tangents and how we can get their measures.

Angles formed by secants and tangents are classified into five categories. Each category is provided with illustration. 1. Angle formed by secant and tangent intersecting on the circle. In the figure, two angles of this type are formed, FED∠ and FEB∠ . Each of these angles intercepts an arc. FED∠ intercepts EF and FEB∠ intercepts EGF. 2. Angle formed by two tangents. In the figure, E∠ is formed by two tangents. The angle intercepts the whole circle divided into 2 arcs, minor arc FD , and major arc FGD. 3. Angle formed by a secant and a tangent that intersect at the exterior of the circle. C∠ is an angle formed by a secant and a tangent that intersect outside the circle. C∠ intercepts two arcs, DB and AD. 4. Angle formed by two secants that intersect in the interior of the circle. The figure shows four angles formed. ,MAN∠ ,NAR∠ PAR∠ and PAM∠ . Each of these angle intercepts an arc. ,MAN∠ intercepts MN, ,NAR∠ intercepts NR, PAR∠ intercepts PR and MAP∠ intercepts MP.

E

F

B D

G

F

E

D

G

C

D

A B

M N

A

R P

8

5. Angle formed by two secants intersecting outside the circle. E∠ is an angle formed by two secants intersecting outside the circle. E∠ intercepts two arcs namely QR and PR How do we get the measures of angles illustrated in the previous page? To understand the answers to this question, we will work on each theorem proving how to get the measures of each type of angle. It is therefore understood that the previous theorem can be used in the proof of the preceding theorem. Theorem: The measure of an angle formed by a secant and a tangent that intersect on the circle is one-half its intercepted arc. Given: Circle O. Secant m and tangent t intersect at E on circle O. Prove: CECEBm 2

1=∠ Proof:

Statements Reasons 1. Draw diameter ED . Join DC. 2. tDE ⊥ 3. DCE∠ is a right angle

4. DEB∠ is a right angle 5. ∆DCE is a right triangle 6. 1∠m + 902 =∠m

7. =∠+∠ BECmm 1 DEBm∠ 8. =∠+∠ BECmm 1 90 9. 21 ∠+∠ mm = BECmm ∠+∠1 10. 1∠m = 1∠m 11. 2∠m = BECm∠ 12. 2∠m = 2

1 mCE 13. BECm∠ = 2

1 mCE

1. Line determination Postulate 2. Radius-tangent theorem 3. Angle inscribed in a semicircle is a right angle. 4. Perpendicular lines form right angles 5. Definition of right triangle 6. Acute angles of a right triangle are complementary 7. Angle addition Postulate 8. Definition of complementary angles 9. Transitive Property of Equality 10. Reflexive Property of Equality 11. Subtraction Property of Equality 12. Inscribed angle Theorem 13. Substitution

Illustration: In the given figure, if mCE = 104, what is the m∠BEC? What is m CEF∠ ?

Q

P

S R

E

B E

C

D

F

O

t

9

Solution: BECm∠ = 2

1 mCE BECm∠ = 2

1 (104) = 52 m CEF∠ = ½ (mCDE) m CEF∠ = ½ (360 – 104) = ½ (256) = 128 Let’s go to the next theorem. Theorem: The measure of an angle formed by two tangents from a common external point is equal to one-half the difference of the major arc minus the minor arc. Given: Circle O. AB and AC are tangents Prove: )(2

1 BCBXCAm −=∠ Proof:

Statements Reasons 1. Draw chord BC 2. In ∆ABC, 1∠ is an exterior angle 3. 1∠m = 2∠m + Am∠ 4. Am∠ = 1∠m - 2∠m 5. 1∠m = ½ mBXC 2∠m = ½ mBC 6. Am∠ = ½ mBXC - ½ mBC 7. Am∠ = ½( mBXC – m BC)

1. Line determination Postulate 2. Definition of exterior angle 3. Exterior angle theorem 4. Subtraction Property of Equality 5. Measure of angle formed by secant and tangent intersecting on the circle is one-half the intercepted arc. 6. Substitution 7. Algebraic solution (Common monomial Factor)

Illustration: Find the Am∠ if mBC = 162. Solution: Since Am∠ = ½( mBXC –m BC) then we have to find first the measure of major arc BXC. To find it, use the whole circle which is 360o.

A

B

C

1

2

X

O

10

mBXC = 360 – mBC = 360 – 162 = 198 Then we use the theorem to find the measure of A∠

Am∠ = ½( mBXC –m BC) = ½ (198 – 162) = ½ (36) Am∠ = 18 We are now into the third type of angle. Angle formed by secant and tangent intersecting on the exterior of the circle. Theorem: The measure of an angle formed by a secant and tangent intersecting on the exterior of the circle is equal to one-half the difference of their intercepted arcs. Given: BA is a tangent of circle O BD is a secant of circle O BA and BD intersect at B Prove: )(2

1 ACADBm −=∠ Proof:

Statements Reasons 1. BA is a tangent of circle O, BD is a secant of circle O 2. Draw AD 3. 1∠ is an exterior angle of ∆ DAB 4. ADBmBmm ∠+∠=∠1 5. ADBmmBm ∠−∠=∠ 1 6. =∠1m ½m AD 7. ADBm∠ = ½ mAC 8. Bm∠ = ½ mAD – ½ mAC 9. Bm∠ = ½ (mAD– mAC)

1. Given 2. Line determination Postulate 3. Definition of exterior angle 4. Exterior angle Theorem 5. Subtraction Property of Equality 6. The measure of an angle formed by secant and tangent intersecting on the circle equals one-half its intercepted arc. 7. Inscribed angle Theorem 8. Substitution 9. Simplifying expression

Illustration: In the figure if mAD = 150, and mAC = 73, what is the measure of B∠ ?

AB

C O

D

1

11

Solution: Bm∠ = ½ (mAD– mAC) = ½ (150 – 73) = ½ (77) Bm∠ = 38.5 The next theorem will tell us how angles whose vertex is in the interior of a circle can be derived. Furthermore, this will employ the previous knowledge of vertical angles whether on a circle or just on a plane. Theorem: The measure of an angle formed by secants intersecting inside the circle equals one-half the sum of the measures of the arc intercepted by the angle and its vertical angle pair.

Given: AC and BD are secants intersecting inside circle O forming 1∠ with vertical angle pair .CED∠ (We will just work on one pair of vertical angles.) Prove: )(1 AEBmm ∠∠ = ½ (AB + DC) Proof:

Statements Reasons 1. AC and BD are secants intersecting inside circle O. 2. Draw AD 3. 1∠ is an exterior angle of ∆AED 4. 1∠m = DACm∠ + ADEm∠ 5. DACm∠ = ½ mDC ADEm∠ = ½ mAB 6. 1∠m = ½ mDC + ½ mAB 1∠m = ½ (mDC + mAB)

1. Given 2. Line determination Postulate 3. Definition of exterior angle 4. Exterior angle Theorem 5. Inscribed Angle Theorem 6. Substitution

Illustration: Using the figure, find the measure of 1∠ if mAB = 73 and mCD = 90. Solution: Using the formula in the theorem, 1∠m = ½ (mDC + mAB) = ½ ( 90 + 73) = ½ (163) = 81.5

E

A B

C

O

D

1 E

12

Let us discuss how to find the measure of the angle formed by two secants intersecting outside the circle. Theorem: The measure of the angle formed by two secants intersecting outside the circle is equal to one-half the difference of the two intercepted arcs. Given: AB and CD are two secants

intersecting outside of circle O forming BEC∠ outside the circle.

Prove: BECm∠ = ½ (AD – BC) Proof:

Statements Reasons 1. AB and CD are secants of circle O forming BEC∠ outside the circle. 2. Draw DB 3. 1∠ is an exterior angle of ∆ DBE 4. 1∠m = 2∠m + BECm∠ 5. BECm∠ = 1∠m - 2∠m 6. 1∠m = ½ mAD 2∠m = ½ mBC 7. BECm∠ = ½ mAD – ½ mBC BECm∠ = ½ (mAD – mBC)

1. Given 2. Line determination Postulate 3. Definition of exterior angle of a triangle 4. Exterior angle Theorem 5. Subtraction Property of Equality 6. Inscribed Angle Theorem 7. Substitution

Illustration: Find the measure of ∠E if mAD = 150 and mBC = 80. Solution:

Again we apply the theorem using the formula:

BECm∠ = ½ (mAD – mBC) = ½ (150 – 80 = ½ ( 70) = 35

B E

D

C

A

1

2

13

Example 1. In each of the given figure, find the measure of the unknown angle (x). 1. 2. 3. 4. 5. Solutions: 1. Given: AB = 1500 Find: xm∠ x∠ is an angle formed by a secant and a tangent whose vertex is on the circle. x∠ intercepts AB. xm∠ = ½ AB xm∠ = ½ (150) xm∠ = 75

A

B

x

1500

M

P N

O

1570

x

O

x

A

M

Y

P

●

780

x

F

D E

G

H 670

400

R

P

Q

T

S

1060

380

14

2. Given: m MP = 157 Find: xm∠ x∠ is an angle formed by two tangents from a common external point. x∠ intercepts minor arc MP and major arc MNP xm∠ = ½ ( MNP – MP) m MNP + mMP = 360 mMNP = 360 – m MP = 360 – 157 mMNP = 203 xm∠ = ½ (203 – 157) = ½ (46) = 23 3. Given: m AP = 78 AY is a diameter Find: xm∠

Since AY is a diameter, then AY is a semicircle and m AY = 180. Therefore a. m AP + m PY = 180 m PY = 180 – m AP m PY = 180 – 78 m PY = 102

b. xm∠ = ½ ( mPY – mAP) = ½ ( 102 – 78) = ½ (24) = 12 4. Given: mFD = 67, m GE = 40 Find: xm∠ x∠ is an angle formed by secants that intersect inside the circle, Hence xm∠ = ½ (mFD + mGE) = ½ (67 + 40) = ½ (107) = 53.5

15

5. Given: mSR = 38, mPQ = 106 Find: xm∠ x∠ is an angle formed by two secants whose vertex is outside the circle. Thus xm∠ = ½ (mPQ – mSR) = ½ (106 – 38) = ½ (68) = 34 Example 2: Find the unknown marked angles or arcs (x and y) in each figure: 1. 2. 3. 4.

5. 6.

M D

B C A x y

●

Q P

R

N

2450

y

y x

R S

U T

580 320 x 220 y

P

Q

S

R

300

A B C

D

E

x 350

E F

D

G

H

270 370

x

16

Solutions:

1. Given : mBMD = 210 Find: xm∠ , ym∠ mBMD + mBD = 360 (since the two arcs make the whole circle) mBD = 360 – mBMD = 360 – 210 = 150 a. xm∠ = ½ mBD = ½ (150) = 75 b. ym∠ = ½ m BMD = ½ (210) = 105 2. Given: mPNR = 245 Find: xm∠ , ym∠ mPNR + mPR = 360 (since the two arcs make a whole circle) mPR = 360 – mPNR = 360 – 245 = 115 But xm∠ = mPR (Central angle equals numerically its intercepted arc) xm∠ = 115 ym∠ = ½ (mPNR – mPR) = ½ (245 – 115) = ½ (130) = 65 3. Given: mRU = 32, mST = 58 Find: xm∠ , ym∠ xm∠ = ½ (mRU + mST) = ½ (32 + 58) = ½ (90) = 45

17

xm∠ + ym∠ = 180 Linear Pair Postulate ym∠ = 180 - xm∠ ym∠ = 180 – 45 = 135 4. Given: mQR = 30, 1∠m = 22 To check; Find : xm∠ , y 1∠m = ½ (mPS + mQR) First, find y by using m 1∠ = ½ (14 + 30 ) 1∠m = ½ ( y + 30) = ½ ( 44) 22 = ½ (y + 30) = 22 2(22) = y + 30 44 = y + 30 y = 44 – 30 y = 14

5. Given: Am∠ = 35, mCD = 110 Find: x To check: Am∠ = ½ (110 – x) Am∠ = ½ (110 – x)

35 = 2

110 x− Am∠ = ½ (110 – 40)

2(35) = 110 – x Am∠ = ½ (70) 70 = 110 – x Am∠ = 35 x = 110 – 70 x = 40 6. Given: Em∠ = 27, mDH = 37 Find: x To check: Em∠ = ½ (x – 37) Em∠ = ½ (x – 37)

27 = 237−x Em∠ = ½ (91 – 37)

54 = x – 37 Em∠ = ½(54) x = 54 + 37 Em∠ = 27 x = 91

18

Try this out. 1. BC is tangent to circle O at A. mDE = 68 mAF = 91. Find

a. mEA b. mDF c. m∠BAE d. m∠EAD e. m DAF∠ f. FACm∠

2. ∆DEF is isosceles with DFDE ≅ . ∠m 1 = 82. Find a. ∠m D b. mDE c. mEF d. ∠m 2 e. ∠m 3 3. If x = 18 and y = 23, find 1∠m . 4. If mDE = 108 and m DOC∠ = 85, find: a. mEA b. ∠m EAF c. ∠m DAC d. ∠m CAB e. ∠m 1 5. Using the given figure, find x and y.

O

F

H

E

G

2

3

D

1

B

C

A

1

y

x

D

D

E

O C

A F B

1

x

1100 600 y

D A

E

F

●

B

C

680

910

O

19

6. EC is tangent to circle O. AB is a diameter. If mDB = 47, find mAD, m ECD∠

7. A polygon is said to be circumscribed about a circle if its sides are tangent to the circle. ∆PRT is circumscribed about circle O. If PT = 10, PR = 13 and RT = 9, find AP, TC and RB. 8. PT is tangent to circle O at P. If mNP = 90, and mMXP = 186, find a. ∠m 1 b. ∠m 2 c. ∠m 3 d. ∠m 4 e. ∠m 5 f. ∠m 6 9. O is the center of the given circle. If mBD = 122 find a. ∠m 1 d. ∠m 4 b. ∠m 2 e. ∠m 5 c. ∠m 3 f. ∠m 6

A

D

C

B O

E

P

T

A B

C R

O 4

3 1

M

T P

N

X 2

5

6

3

E

D

F O

A B C

5

1 4

6

2

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Let’s Summarize

1. A tangent is a line that intersect a circle at only one point. 2. A. secant is a line that intersect a circle at two points.

3. If a line is tangent to a circle, it is perpendicular to a radius at the point of tangency.

4. If two tangents are drawn from an exterior point to a circle then

a) the two tangent segments are congruent b) the angle between the segment and the line joining the external point and the center of the circle are congruent.

5. The measure of an angle formed by a secant and a tangent intersecting on the circle

is equal to one half the measure of the intercepted arc.

6. The measure of an angle formed by two tangents from a common external point is equal to one-half the difference of the measures of the intercepted arcs.

7. The measure of an angle formed by secant and tangent intersecting outside the

circle is equal to one-half the difference of the measures of the intercepted arcs.

8. The measure of an angle formed by two secants intersecting inside the circle is equal to one-half the sum of the measure of the intercepted arc of the angle and its vertical angle pair.

9. The measure of an angle formed by two secants intersecting outside the circle is

equal to one-half the difference of the measures of the intercepted arcs.

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What have you learned 1. QP is tangent to circle O at P. If POQm∠ = 73, what is Qm∠ ? 2. DE , EF and DF are tangents to circle M. If DB =- 5, EC = 7 and AF = 4, what is the perimeter of ∆DEF? 3. PS AND PQ are secant and tangent of circle A. If mRQ = 52, what is Pm∠ ? 4. Given circle E with secants AB and CD . If mBD = 53 and mBC = 117, find BEDm∠ . 5. XY is tangent to circle at A . If mAB = 105, and AB ≅ BP, find BAPm∠ and 6. PAYm∠

O

Q P

E

D

F

B

A

C

A

S

Q

P

R

E A

D

B

C

P

B

A X Y

22

7. Given circle A with secants OQ and OP . If mRS = 32 and mQR = 2mRS, find Om∠ ? 8. AB ║ DE . BE is tangent to the circle at B and intersects DE at E. If mAB = 110 and mAD = 70, then ABFm∠ = ___________. 9. BECm∠ = __________. 10. Using the same figure, if Em∠ = 42, mBC = 60, find mAB.

A O

S P

R Q

A

F

B

E C D

23

Answer Key


1. ≅ or congruent 2. ⊥ or perpendicular 3. 40 4. 22 5. 59 6. 61 7. 33 8. 120 9. 20 10. 18

Try this out Lesson 1 1. a. 112 b. 89 c. 56 d. 34 e. 44.5 f. 45.5

2. a. 16 b. 164 c. 32 d. 82 e. 16

3. 41 4. a. 72 b. 36 c. 42.5 d. 47.5 e. 54

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5. x = 70 y = 50

6. mAD = 133 m ECD∠ = 43

7. AP = 7 TC = 3 RB = 6

8. a. 45 b. 45 c. 42 d. 93 e. 57 f. 48

9. a. 122 b. 61 c. 29 d. 29 e. 29 f. 45

What have you learned 1. 17 2. 32 3. 38 4. 58 5. 52.5 6. 75 7. 26 8. 55 9. 55 10. 84

Module 1 Properties of Quadrilaterals

What this module is about This module is about the properties of trapezoids and parallelograms. In this module, you will learn to compute problems involving the median of a trapezoid, the base angles and diagonals of an isosceles trapezoid, as well as problems involving the diagonals, angles and sides of parallelograms

What you are expected to learn This module is designed for you to

1. apply inductive/deductive skills to derive certain properties of a trapezoid.

• median of a trapezoid • base angles of an isosceles trapezoid • diagonals of an isosceles trapezoid

2. apply inductive and deductive skills to derive the properties of a parallelogram

• each diagonal divides a parallelogram into two congruent triangles

• opposite angles are congruent • non-opposite angles are supplementary • opposite sides are congruent • diagonals bisect each other.

How much do you know True or False

1. The median of a trapezoid is twice the sum of the lengths of its bases. 2. The base angles of an isosceles trapezoid are congruent. 3. The diagonals of any trapezoid are congruent. 4. Non-opposite angles of a parallelogram are complementary.

2

ABCD is a trapezoid with median EF. D 10 C 13 16 A B 5. If DC = 12 cm and AB = 23 cm, what is EF? 6. If EF = 2x + 3, DC = x + 5 and AB = 2(x + 3), what is AB?

Quadrilateral BEST is a parallelogram. 2x + 3 T S 3(x – 1) B E

7. If BE = 3(x – 1) and TS = 2x + 3, what is BE + TS? Quadrilateral ABCD is a rhombus. D C 4x + 10 5x A B

8. If m ∠A = 5x and m∠C = 4x + 10, what is m∠ADC in degrees? Quadrilateral ABCD is a parallelogram. Diagonals DB and AC intersect at E. D C E A B

3

9. Find AC if AE = x + 2 and CE = 3x - 6

Quadrilateral TEAM is a parallelogram. M A x 3x + 20 T E

10. If m∠T = x and m∠E = 3x + 20 . find m∠A? What you will do

Lesson 1

The Median of a Trapezoid

A trapezoid is a quadrilateral with exactly one pair of parallel sides. The median of a trapezoid is the segment joining the midpoints of the non-parallel sides. In the trapezoid ABCD below, EF is the median. It joins the midpoints E and F of side AD and BC respectively. D C E F A B Illustration If the length of the upper base of a trapezoid is 4 cm, and the lower base is 6 cm, what do you think is the length of the median? Do the following:

1. Using a ruler draw a segment 4 cm long. Name the segment HT. 2. Draw another segment 6 cm long parallel to segment HT. Name the

segment MA.

4

3. Connect points H and M. 4. Connect points T and A. 5. Using a ruler, carefully determine the midpoints of HM and TA. Name the

midpoints of HM and TA, G and E respectively. 6. Connect points G and E. Carefully measure the length of GE.

What did you discover? Did you discover the following?

1. GE is parallel to HT and MA.

2. GE = 21 (HT + MA)

Example 1 ABCD is a trapezoid with median EF. If DC = 8 cm and AB = 14 cm, find EF. D 8 C E F 14 A B

Solution: Step 1. Write the formula.

EF = 21 (AB + DC)

Step 2. Substitute the values of DC and AB into the formula. Solve for EF.

EF = 21 (14 + 8)

EF = 21 (22)

EF = 11

Therefore EF is 11 cm.

5

Example 2 EFGH is a trapezoid with median IJ. If HG = 12 cm and IJ =15 cm, what is EF? H 12 cm G I 15 cm J E F

Step 1. Write the formula.

IJ = 21 (EF + HG)

Step 2. Substitute the values of HG and IJ into the formula. Solve for EF.

15 = 21 (EF + 12)

15 = EF + 12 2 EF + 12 = 15(2) EF = 30 – 12 EF = 18 Therefore EF is 18 cm. Example 3 ABCD is a trapezoid with median EF. If DC = x + 5, EF = 2x + 1 and AB =4x – 10, find EF. D x + 5 C

E 2x + 1 F

A 4x – 10 B

6

Solution: Step 1. Write the formula.

EF = 21 (AB + DC)

Step 2. Substitute the values of EF, AB and DC into the formula. Solve for x.

2x + 1 = 21 (4x – 10 + x + 5)

2x + 1 = 21 ( 5x –5)

2(2x +1) = 5x – 5 4x + 2 = 5x – 5 4x – 5x = -5 – 2 - x = -7 x = 7 Step 3. Substitute 7 for x. Solve for EF EF = 2x + 1 = 2(7) + 1 = 14 + 1 = 15 Therefore EF = 15 cm. Try this out Set A ABCD is a trapezoid with median EF. D C E F A B

1. If DC = 4 and AB = 10, what is EF? 2. If DC = 7 and AB = 12, what is EF? 3. If AB = 14.5 and DC = 8.5, what is EF? 4. If AB = 10.5 and DC = 6.5, what is EF? 5. IF DC = 10 and EF = 12, what is AB? 6. If DC = 14 and EF = 16, what is AB? 7. If AB = 14.6 and EF = 10.4, what is DC?

7

8. IF AB = 22.8 and EF = 16.2, what is DC? 9. If DE = 14 what is AE? 10. If BF = 24, what is CF?

Set B EFGH is a trapezoid with median IJ. H G I J E F

1. If IJ = x, HG = 8 and EF = 12, what is x? 2. If IJ = y, HG = 14, and EF = 20, what is Y? 3. If IJ = x + 2, HG = 10 and EF = 14, what is x? 4. If IJ = y + 3, HG = 14 and EF = 18, what is y? 5. If HG = x , IJ = 16, and EF = 22, what is x? 6. If HG = Y, IJ = 24 and EF = 30, what is y? 7. If HG = x – 1, IJ = 21 and EF = 34, what is x? 8. If HG = y – 2, IJ = 20 and EF = 31, what is y? 9. If HI = x + 4 and IE = 6, what is x? 10. If GJ = 10 and FJ = x –4, what is x?

Set C KLMN is a trapezoid with median OP. N M O P K L

1. If OP = 20, NM = x + 3 and KL = x + 6, what is x? 2. If NM = x – 2, KL = x + 4 and OP = 24, what is x? 3. If OP = 22, NM = x + 4, and KL = x + 8, what is NM? 4. If OP = 24, NM = x – 3 and KL = x + 7, what is KL? 5. If NM = 12, OP = x +3, and KL = x + 10, what is x? 6. If NM = 18, OP = x – 2 and KL = x + 3, what is x? 7. If KL = 30, OP = x + 1 and NM = x – 6, what is OP?

8

D C B

8. If KL = 34, OP = x - 1, and NM = x – 7, what is NM? 9. If NM = 2x, OP = 3x, and KL = 2(x+5), what is OP? 10. If NM = 2x + 2. OP = 3x + 3 and KL = 2(x + 6), what is NM?

Lesson 2

Isosceles Trapezoid An isosceles trapezoid is a trapezoid with congruent non-parallel sides. Trapezoid DAVE below is an isosceles trapezoid. The nonparallel sides ED and VA are congruent. E V D A Do the following:

1. Using a ruler, draw isosceles trapezoid ABCD with base angles, ∠A and ∠B, on a graphing paper.

A

2. Using a protractor, find the measures of ∠A and ∠B. What do you notice? 3. Using the same protractor find the measures of ∠ D and ∠ C. What do

you notice? 4. Draw the diagonals AC and BD. Using a ruler, find their lengths. Are the

lengths equal?

9

Perhaps you discovered the following properties of an isosceles trapezoid.

1. The base angles of an isosceles trapezoid are congruent. 2. The diagonals of an isosceles trapezoid are congruent.

If you know the properties of an isosceles trapezoid, you will find the next set of exercises easy to solve. Example 1

In isosceles trapezoid ABCD, with base angles, ∠ A and ∠ B. If m∠B = 40, what is m∠A? D C

A B Solution: Step 1. Base angles of an isosceles trapezoid are congruent. ∠ A ≅∠ B m∠ A = m∠ B (The measures of congruent angles are equal) Step 2. Substitute 40 for m∠B m∠A = 40 Example 2 In isosceles trapezoid ABCD with ∠ A and ∠ B as base angles. If m∠A = x + 20 and m∠B = 2x. Find m∠A. D C A B Solution: Step 1. Base angles of an isosceles trapezoid are congruent. ∠ A ≅ ∠ B m∠ A = m∠ B (the measures two angles congruent are equal)

10

Step 2. Substitute x + 20 for m∠ A and 2x for m∠ B. x+ 20 = 2x x – 2x = -20 x = 20

Step 3. Substitute 20 for x in m∠ A = x + 20. Then solve for m∠ A. m∠ A = x + 20 = 20 + 20 = 40 Example 3 In isosceles trapezoid ABCD, AD = 10. What is BC? D C A B Solution: Step 1. Non-parallel sides of an isosceles trapezoid are congruent. BC ≅ AD BC = AD (Congruent segments have equal lengths)

Step 2. Replace BC with 10

BC = 10 Example 4. In isosceles trapezoid ABCD, AC = 4x + 4 and BD = 2x + 10. What is x? D C A B

11

Solution:

Step 1 The diagonals of an isosceles trapezoid are congruent. AC ≅ BD AC = BD ( Congruent segments have equal lengths)

Step 2. Substitute 4x + 4 for AC and 2x + 10 for BD.

4x + 4 = 2x + 10 4x – 2x = 10 – 4 2x = 6 x = 3 Try this out Set A PQRS is an isosceles trapezoid S R P Q

1. Name the two pairs of base angles. 2. Name the nonparallel sides. 3. Name the parallel sides. 4. If m∠A = 30, what is the m∠B? 5. If m∠B = 60, what is m∠A? 6. If m∠C = 110, what is m∠D? 7. If m ∠D = 105, what is m∠C? 8. If m∠A = 35, what is m∠D? 9. If m∠C = 120, what is m∠B? 10. If m∠A = 35 what is m∠C?

Set B. EFGH is an isosceles trapezoid. H G E F

12

1. If m∠E = x + 10 and m∠F = 50, what is x? 2. If m∠F = x – 15 and m∠E = 70 what is x? 3. If m∠H = x + 20 and m∠G = 100 what is x? 4. If m∠G = x – 10 and m∠H = 135 what is x? 5. If m∠E = 2x and m∠F = 46 what is x? 6. If m∠F = 3x and m∠E = 39 what is x? 7. If m∠E = 2x + 5 and m∠F = 44 what is x? 8. If m∠F = 2x – 6 and m∠E = 56 what is x? 9. If m∠H = 2(x + 4) and m∠G = 116 what is x? 10. If m∠G = 2(x –5) and m∠H = 120 what is x?

Set C. ABCD is an isosceles trapezoid. D C A B

1. If m∠A = 2x + 10 and m∠B = 3x – 20 what is m∠A? 2. If m∠A = 2x + 15 and m∠B = 4x – 11 what is m∠B? 3. If m∠D = x + 15m and m∠C = 2x –85 what is m∠D? 4. If m∠C = 3y + 12 and m∠D = 2y + 50, what is m∠C? 5. If m∠C = 4x + 70 and m∠D = 2x + 90, what is x? 6. If m∠D = 4x – 20 and m∠C = 5x – 50, what is x? 7. If AC = 60 cm, what is BD? 8. If BD = 70 cm, what is AC? 9. If AC = 4x – 6 and BD = 2x + 10, what is AC? 10. If AC = 3y + 7 and BD = 6y – 8, what is BD?

Lesson 3

Properties of a Parallelogram

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

13

Do the following:

1. On a graphing paper, draw a parallelogram similar to the one below. Name your parallelogram ABCD.

2. Draw diagonal AC. What do you notice? 3. Using a protractor, find the measures of the opposite angles of

parallelogram ABCD. Are the angles congruent? 4. Using a protractor, find the measures of each pair of non-opposite

angles. Add their measures. Are the angles supplementary? 5. Using a ruler, find the lengths of each pair of opposite sides. Are their

lengths equal? 6. Draw diagonal BD. What do you notice?

Were you careful in doing the above activity? You actually proved inductively the following properties of a parallelogram.

1. Each diagonal divides a parallelogram into two congruent triangles. 2. The opposite angles of a parallelogram are congruent. 3. The non-opposite angles of a parallelogram are supplementary. 4. The opposite sides of a parallelogram are congruent. 5. The diagonals of a parallelogram bisect each other.

These properties of a parallelogram can also be proven deductively.

1. Each diagonal divides a parallelogram into two congruent triangles.

Given: Parallelogram MATH with diagonal MT. H T Prove: ∆HTM ≅ ∆AMT

M A

D C A B

14

Proof:

Statements

Reasons

1. Parallelogram MATH with diagonal MT

2. MH // TA

3. ∠HMT ≅ ∠ATM (A)

4. MT ≅ MT (S) 5. HT // MA

6. ∠HTM ≅ ∠AMT (A) 7. ∆HTM ≅ ∆AMT

1. Given

2. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

3. If two parallel lines are cut by a transversal, then any pair of alternate interior angles are congruent.

4. Reflexive Property of Congruence

5. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.(Same as reason # 2)

6. If two parallel lines are cut by a transversal, then any pair of alternate interior angles are congruent. ( Same as reason #3)

7. ASA Congruence

2. The opposite angles of a parallelogram are congruent H T Given: Parallelogram MATH Prove: ∠H ≅ ∠ A Proof: M A

Statements Reasons 1. Parallelogram MATH 2. Draw MT 3. ∆MHT ≅ ∆TAM

4. ∠H ≅ ∠A

1. Given 2. Two points determine a line. 3. Each diagonal divides a

parallelogram into two congruent triangles. (First property)

4. Corresponding parts of congruent triangles are congruent. (CPCTC)

15

If you want to prove that ∠M ≅ ∠ T, draw diagonal HA. Then follow the above steps. 3. The non-opposite angles of a parallelogram are supplementary Given: Parallelogram MATH H T Prove: ∠H and ∠ M are supplementary M A Proof:

Statements

Reasons

1. Parallelogram MATH 2. HT // MA

3. ∠H and ∠ M are supplementary

1. Given 2. A parallelogram is a

quadrilateral in which both pairs of opposite sides are parallel

3. If two parallel lines are cut by a transversal, then the interior angles on the same side of the transversal are supplementary.

4. The opposite sides of a parallelogram are congruent Given : Parallelogram MATH H T Prove: HT ≅ MA HM ≅ TA M A Proof:

Statements

Reasons

1. Parallelogram MATH 2. Draw diagonal MT 3. ∆MHT ≅ ∆TAM

4. HT ≅ MA HM ≅ TA

1. Given 2. Two points determine a line 3. Each diagonal divides a

parallelogram into two congruent triangles.

4. Corresponding parts of congruent triangles are congruent.

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5. The diagonals of a parallelogram bisect each other.

Given: Parallelogram LOVE E V A 2 Prove: EO and LV bisect each other 3 4 1 L O Proof:

Statements Reasons 1. Parallelogram LOVE with

diagonals EO and LV 2. LE ≅ VO (S) 3. LE // VO

4. ∠1 ≅ ∠2 (A) 5. ∠3 ≅ ∠4 (A) 6. ∆LEA ≅ ∆VOA 7. EA ≅ OA LA ≅ VA 8. EO and LV bisect each other

1. Given 2. Opposite sides of a

parallelogram are congruent 3. A parallelogram is a

quadrilateral in which both pairs of opposite sides are parallel

4. If two parallel lines are cut by a transversal, then the alternate interior angles are congruent.

5. Vertical angles re congruent 6. SAA Congruence 7. Corresponding parts of

congruent triangles are congruent

8. The bisector of a segment is a point, line, segment, or plane that divides the segment into two segments (Definition of bisector of a segment)

Example 1 JACK is a parallelogram. If m∠K = 110, what is m∠A. K C J A Solution:

Step 1. Opposite angles of a parallelogram are congruent ∠A ≅ ∠K m ∠A = m∠K (Congruent angles have equal measures)

17

Step 2. Replace m∠K with 110 m∠A = 110 Example 2 ABCD is a parallelogram. If m∠ A = x + 15 and m∠C = 40, what is x? D C A B

Solution: Step 1 Opposite angles of a parallelogram are congruent ∠A ≅ ∠C m∠A = m∠C ( Congruent angles have equal measures)

Step 2. Replace m∠A with x + 15 and m∠C with 40 and solve for x. x + 15 = 40 x = 40 – 15 x = 25 Example 3 The figure below is a parallelogram. If m∠O = 2x + 10 and m∠E = x + 30, what is m∠O? E V Solution: L O Step 1. Opposite angles of a parallelogram are congruent. ∠O ≅ ∠E m∠O = m∠E (Congruent angles have equal measures)

Step 2. Substitute 2x +10 for m∠O and x +30 for m∠E. Then solve for x. 2x + 10 = x + 30 2x – x = 30 – 10 x = 20 Step 3. Substitute 20 for x in m∠O = 2x + 10 to solve for m∠O. m∠O = 2(20) + 10 m∠O = 40 + 10 m∠O = 50

18

Example 4 Quadrilateral ABCD is a parallelogram. If m∠A = 60, what is m∠B? D C A B Solution:

Step 1. Non-opposite angles of a parallelogram are supplementary ∠A and ∠B are supplementary m∠A + m∠B = 180

Step 2. Substitute 60 for m∠A and solve for m∠B 60 + m∠B = 180 m∠B = 180 – 60 m∠B = 120 Example 5 Quadrilateral ETNA is a parallelogram. If m∠E = x – 60 and m∠A = 2x, what is m∠E? A N E T

Solution: Step 1. Non-opposite angles of a parallelogram are supplementary. ∠E and ∠A are supplementary m∠E + m∠A = 180

Step 2. Substitute x – 60 for m∠E, and 2x for m∠A. x –60 + 2x = 180 x + 2x = 180 + 60 3x = 240 x = 80

Step 3. Substitute 80 for x in m∠E = x – 60. Then solve for m∠E. m∠E = x – 60 m∠E = 80 – 60 m∠E = 20

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Example 6 In the parallelogram below, m∠A = 2x and m∠C = 4x – 80. What is m∠B? D C A B Solution:

Step 1. Opposite angles of a parallelogram are congruent. ∠A ≅ ∠C m∠A = m∠C (Congruent angles have equal measures)

Step 2. Substitute 2x for m∠A and 4x – 80 for m∠C. Then solve for x.

2x = 4x – 80 2x – 4x = – 80 –2x = – 80 x = 40

Step 3. Substitute 40 for x m∠A = 2x m∠A =2(40) =80

Step 4. Non-opposite angles of a parallelogram are supplementary ∠A and ∠B are supplementary m∠A + m∠B = 180

Step 5. Substitute 80 for m∠A and solve for m∠B. 80 + m∠B = 180 m∠B = 180 – 80 m∠B = 100 Example 7 Quadrilateral EFGH is a parallelogram. If EH = 14 cm long, how long is side FG? H G E F

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Solution:

Step 1. Opposite sides of a parallelogram are congruent

FG ≅ EH FG = EH (Congruent segments have equal lengths)

Step 2. Substitute 14 for EH in the equation FG = EH FG = 14 Therefore FG is 14 cm long. Example 8 Quadrilateral GEOM is a parallelogram. If MO = 2x + 3 and GE = 4x - 15. What is MO? M O G E

Solution: Step 1. Opposite sides of a parallelogram are congruent

MO ≅ GE MO = GE (Congruent segments have equal lengths)

Step 2. Substitute 2x + 3 for MO and 4x – 15 for GE. Solve for x 2x + 3 = 4x – 15 2x – 4x = –15 – 3 –2x = –18 x = 9 Step 3. Substitute 9 for x in MO = 2x + 3 MO = 2(9) + 3 = 18 +3 = 21

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Example 9 CDEF is a parallelogram. If FD = 12 cm, what is the length of FG? F E G C D

Solution: The diagonals of a parallelogram bisect each other. CE bisects FD

Therefore: FG = 21 FD

=21 (12)

=6 cm Example 10 CDEF is a parallelogram with diagonals CE and DF intersecting at G. If FG = 3x –7 and DG = x +21, find FG.

Solution: Step 1. Draw the figure. F E G C D

Step 2. The diagonals of a parallelogram bisect each other. FG = DG

Step 3. Substitute 3x – 7 for FG and x + 21 for DG in the equation FG DG. 3x – 7 = x + 21 3x –x = 21 + 7 2x = 28 x = 14

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Step 4. Substitute 14 for x in the equation FG = 3x – 7 . (See given data) FG = 3x – 7 =3(14) – 7 =42 – 7 = 35 Try this out Set A Fill in the blanks Quadrilateral ABCD is a parallelogram.. D C 500 880 A B

1. m∠A = ________ 2. m∠ABC = _____ 3. m∠A + m∠ABC = ________ 4. m∠ABC + m∠C = ________ 5. m∠C + m∠ADC = ________ 6. ∆ ABD ≅ ________ Quadrilateral EFGH is a parallelogram. H G I E F

7. HI = __________ 8. EI = ________ 9. If EG is 16 cm, then EI = ________ 10. If HI is 7 cm, then FI = __________

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Set B

Quadrilateral ABCD is a parallelogram D C A B

1. If m∠A is 50 , what is m∠C? 2. If m∠B is 125 , what is m∠D? 3. If m∠A = x and m∠C = 30, what is x? 4. If m∠B = y and m∠D = 115, what is y? 5. If m∠A = x + 30, and m∠C = 60, what is x? 6. If m∠D = 100 and m∠B = y – 40, what is y? 7. If m∠A = 50, what is m∠D? 8. If m∠D = 130, what is m∠C? 9. If m∠B = 2x – 20 and m∠D = x + 40, what is m∠B? 10. If m∠A = 2x – 50 and m∠C = x + 10 what is m∠A?

Set C Use the figure below for exercises 1-10

Quadrilateral ABCD is a parallelogram. D C E A B

1. If AB = 17 cm, then CD = _________ 2. If AD = 25 cm, then BC = __________ 3. If AE = 5 dm, then CE = __________ 4. If BD = 36 cm, then BE = __________ 5. If AB = 2x + 10 and CD = 15, then AB = _________ 6. If AD = 4x + 15 and BC = 2x + 21 then AD = _________ 7. If AB = x + 6 and CD = 14, what is x? 8. If AD = 20 and BC = x – 5, what is x? 9. If AE = 15 and CE = x + 4, what is x? 10. If BE = 2x and DE = 6, what is x?

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Let’s summarize 1. A trapezoid is a quadrilateral with exactly one pair of parallel sides. 2. The median of a trapezoid is a segment joining the midpoints of the non-

parallel sides of a trapezoid. 3. The median of a trapezoid is parallel to its bases and half the sum of the

lengths of the bases. 4. An isosceles trapezoid is a trapezoid with congruent non-parallel sides. 5. The base angles of an isosceles trapezoid are congruent. 6. The diagonals of an isosceles trapezoid are congruent. 7. A parallelogram is a quadrilateral in which both pairs of opposite sides are

parallel. 8. Each diagonal divides a parallelogram into two congruent triangles. 9. Opposite angles of a parallelogram are congruent. 10. Non-opposite angles of a parallelogram are supplementary. 11. The diagonals of a parallelogram bisect each other.

What have you learned Multiple Choice. Choose the letter of the correct answer.

1. Non-opposite angles of a parallelogram are A. Complementary C. Adjacent B. Supplementary D. Congruent C.

2. A quadrilateral with exactly one pair of parallel sides A. Square C. Trapezoid B. Rectangle D. Rhombus

3. In the figure at the right, DC = 20 cm

And AB = 36 cm. What is EF? D C A. 16 cm

B. 56 cm E F C. 28 cm D. 46 cm A B

4. The figure below is a parallelogram. If AD = 2x - 10 and BC = x + 30, then BC =

A. 50 D C B. 60 C. 70 D. 80 A B

25

5. The figure below is a rhombus. If m ∠I = 4x and m ∠E = 2x + 60, what is m ∠I in degrees? E V

A. 100 E. 110 F. 120 G. 130 G I

6. Quadrilateral BEST is a parallelogram. If m ∠B = x + 40 and m ∠E = 2x + 20, what is m ∠B in degrees? A. 50 T S B. 60 C. 70 D. 80 B E 7. The figure below is a parallelogram. The diagonals AC and BD intersect at

E. If AE = 2x and EC = 12, what is x? A. 5 D C B. 6 C. 7 E D. 9 A B 8. Quadrilateral CDEF is a parallelogram. If m ∠C = y and m ∠E = 2y – 40,

then m ∠D is A. 80 F E B. 110 C. 140 D. 170 C D 9. Into how many congruent triangles is a parallelogram divided by one of its

diagonals? A. 1 C. 3 B. 2 D. 4 10. Base angles of an isosceles trapezoid are A. Complementary C. Congruent B Supplementary D. Adjacent

26

Answer Key How much do you know

1. False 2. True 3. False 4. False 5. 17.5 6. AB = 16 7. BE + TS = 30 8. m∠ADC = 130 9. AC = 12 10. m∠C = 40

Try this out Lesson 1 Set A

1. 7 2. 9.5 3. 11.5 4. 8.5 5. 14 6. 18 7. 6.2 8. 9.6 9. 14 10. 24

Lesson 2 Set A

1. ∠P and ∠Q; ∠R and ∠S 2. AD and BC 3. DC and AB 4. m∠B = 30 5. m∠A = 60 6. m∠D = 110 7. m∠C = 105 8. m∠D = 145 9. m∠B = 60 10. m∠C = 145

Set B 1. x = 10 2. y = 17 3. x = 10 4. y = 13 5. x = 10 6. y = 18 7. x = 9 8. y = 11 9. x = 2 10.x = 14

Set C 1. x = 15.5 2. x = 23 3. NM = 20 4. KL = 29 5. x = 16 6. x = 25 7. OP = 23 8. NM = 22 9. OP = 15 10. NM = 10

Set B 1. x = 40 2. x = 85 3. x = 80 4. x = 145 5. x = 23 6. x = 13 7. x = 19.5 8. x = 31 9. x = 54 10.x = 35

Set C 1. m∠A = 70 2. m∠B = 41 3. m∠D = 115 4. m∠C = 126 5. x = 10 6. x = 30 7. BD = 60 8. AC = 70 9. AC = 26 10. BD = 22

27

Lesson 3 Set A

1. 50 2. 130 3. 180 4. 180 5. 180 6. 42 7. FI 8. GI 9. 8 cm 10. 7 cm

What have you learned 1. B 2. C 3. C 4. C 5. C 6. D 7. B 8. C 9. B 10. C

Set B 1. m∠C = 50 2. m∠D = 125 3. x = 30 4. y = 115 5. x = 30 6. y = 140 7. m∠D = 130 8. m∠C = 50 9. m∠B = 100 10. m∠A = 70

Set C 1. CD = 17 cm 2. BC = 25 cm 3. CE = 5 dm 4. BE = 18 cm 5. AB = 15 6. AD = 27 7. x = 8 8. x = 25 9. x =11 10. x = 3

Module 2 Properties of Quadrilaterals

What this module is about This module is about the properties of the diagonals of special quadrilaterals. The special quadrilaterals are rectangles, square, and rhombus. The conditions sufficient to guarantee that a quadrilateral is a parallelogram are also discussed in this module.


1. apply inductive/deductive skills to derive the properties of the diagonals of special quadrilaterals

• rectangle • square • rhombus

2. verify sets of sufficient conditions which guarantee that a quadrilateral is a parallelogram

3. apply the conditions to prove that a quadrilateral is a parallelogram 4. solve routine and non routine problems

How much do you know True of False

1. The diagonals of a square are congruent. 2. The diagonals of a rectangle are perpendicular.

3. The diagonals of a rhombus bisect each other.

4. A square is a rhombus.

5. If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral

is a parallelogram.

2

6. If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a

parallelogram.

Quadrilateral ABCD is a rectangle. Its diagonals AC and BD intersect at E. D C E A B

7. If AC = 2(x + 10) and BD = x + 60, what is AC? 8. If AE = 4x – 5 and CE = 10 + x, what is AE?

9. Quadrilateral CDEF is a rhombus. If m∠FCE = 3x - 5 and m∠DCE = 2x, find

m∠FCD. F E

C D Quadrilateral GHIJ is a square. J I G H

10. If m∠HGI is 3(x + 5), what is x?

3

What you will do

Lesson 1

The Properties of the Diagonals of Special Quadrilaterals A diagonal of a quadrilateral is a segment which connects any two non-consecutive vertices. In the following quadrilateral, AC and BD are the diagonals. D C A B The following are the properties of the diagonals of special quadrilaterals.

1. The diagonals of a rectangle are congruent. 2. The diagonals of a square are congruent. 3. The diagonals of a square are perpendicular 4. Each diagonal of a square bisects a pair of opposite angles. 5. The diagonals of a rhombus are perpendicular. 6. Each diagonal of a rhombus bisects a pair of opposite angles

You can apply inductive skills to derive these properties of the diagonals of special

quadrilaterals. In the following activities you need a ruler, a pencil, a protractor and pieces of graphing paper. 1. Do the following activity:

a. On a graphing paper, draw a rectangle. b. Name your rectangle ABCD. c. Draw diagonals AC and BD. d. Find the lengths of AC and BD. Are their lengths equal? Are the diagonals

congruent?

Conclusion: The diagonals of a rectangle are congruent 2. Do the following activity:

a. On a graphing paper, draw a square. b. Name your square ABCD. c. Draw diagonals AC and BD.

4

d. Find the lengths of the diagonals. Are their lengths equal? Are the diagonals of the square congruent?

Conclusion: The diagonals of a square are congruent.

3. Do the following activity:

a. Construct a square on a graphing paper b. Name your square EFGH. c. Draw its diagonals EG and HF. d. Label the intersection of the diagonals, M. e. Using a protractor, find the measures of ∠HME, and ∠HMG. f. What kind of angles are the two angles? g. Are the diagonals perpendicular?

Conclusion: The diagonals of a square are perpendicular 4. Do the following activity.

a. Draw a square on a graphing paper. b. Name your square ABCD. c. Draw diagonal AC. d. What do you notice? Into how many angles are the two opposite vertex angles divided? e. What do you conclude?

Conclusion: Each diagonal of a square bisects a pair of opposite angles.

5. Do the following activity.

a. Draw a rhombus on a graphing paper. b. Name your rhombus ABCD. c. Draw the diagonals and name the point of intersection, E. d. Find the measures of ∠AED and ∠CED. e. What kind of angles are they? f. What can you say about the diagonals?

Conclusion: The diagonals of a rhombus are perpendicular. 6. Do the following activity.

a. Draw a rhombus on a graphing paper. b. Name your rhombus ABCD. c. Draw diagonal AC. d. What do you notice? Into how many angles are the two opposite vertex angles divided? e. What do you conclude?

Conclusion: Each diagonal of a rhombus bisects a pair of opposite angles.

5

These properties of the diagonals of special quadrilaterals can also be proven

deductively. Let us prove the first three properties deductively. 1. The diagonals of a rectangle are congruent. D C Given: Rectangle ABCD with diagonals AC and BD. Prove: BD ≅ AC A B Proof:

Statements

Reasons

1. Rectangle ABCD with diagonals AC and BD

2. AD ≅ BC (S)

3. ∠DAB and ∠CBA are right angles

4. ∠DAB ≅ ∠CBA (A) 5. AB ≅ AB (S)

6. ∆DAB ≅ ∆ CBA 7. BD ≅ AC

1. Given 2. Opposite sides of a

parallelogram are congruent (Remember, a rectangle is a parallelogram)

3. A rectangle is a parallelogram with four right angles

4. Any two right angles are congruent

5. Reflexive Property of Congruence

6. SAS Congruence 7. Corresponding Parts of

Congruent Triangles are Congruent

Triangles ∆DAB and ∆ CBA overlap. If you find difficulty visualizing the two

overlapping triangles, separate the figure into two triangles . D C D C

A B A B A B

2. The diagonals of a square are congruent. D C

Given: ABCD is a square with diagonals AC and BD

Prove: BD ≅ AC A B

6

Proof

Statements Reasons 1. ABCD is a square with

diagonals AC and BD 2. AD ≅ BC (S) 3. ∠DAB and ∠CBA are right angles

4. ∠DAB ≅ ∠CBA (A) 5. AB ≅ AB (S)

6. ∆ DAB ≅ ∆ CBA 7. BD ≅ AC

1. Given 2. Opposite sides of a parallelogram

are congruent (Remember, a square is a parallelogram)

3. A rectangle has four right angles (Remember that a square is a rectangle with four congruent sides and a rectangle has four right angles.)

4. Any two right angles are congruent 5. Reflexive Property of Congruence

6. SAS Congruence 7. Corresponding Parts of Congruent

Triangles are Congruent

3. The diagonals of a square are perpendicular

Given: TEAM is a square with diagonals AT and ME M A Prove: ME ⊥ AT O T E A proof can also be written in paragraph form. Proof: Side TM and side EA are congruent since they are sides of a square. A square is a rectangle with four congruent sides. MO ≅ MO by Reflexive Property of Congruence. The diagonals of a parallelogram bisect each other. Since a square is a parallelogram therefore TO ≅ AO. ∆MOT ≅ ∆MOA by SSS congruence. Since ∠MOT and ∠MOA are supplementary and congruent, then each of them is a right angle. Therefore ME ⊥ AT by the definition of perpendicular.

7

Example 1 The figure at the right is a rectangle. E V If the diagonal LV = 2x and the diagonal OE = 12 cm, find x. Solution: Step 1. The diagonals of a rectangle are congruent. L O LV ≅ OE LV = OE (Congruent segments have equal lengths) Step 2. Substitute 2x for LV and 12 for OE. Then solve for x. 2x = 12 x = 6 Answer: The value of x is 6 cm. Example 2 D C Quadrilateral ABCD at the right is a square. Find m∠CAB Solution: Step 1, Quadrilateral ABCD is a square and a square is a rectangle. A B Therefore: m∠DAB = 90. Step 2. But each diagonal of a square bisects a pair of opposite angles.

Hence: m ∠CAB = 21 m∠DAB

Step 3. Substitute 90 for m∠DAB.

m ∠CAB = 21 (90)

= 45 Answer: m ∠CAB =45 A N Example 3 EDNA is a square. If m∠END is 3(x +5), what is x? E D

8

Solution: a. m∠DCB = 90 since ABCD is a square b. Each diagonal of a square bisects a pair of opposite angles. Hence: m∠ACB = 45 3(x + 5) = 45 3x + 15 = 45 3x = 45 – 15 3x = 30 x = 10 D C Example 4 The figure at the right is a rhombus. If m ∠CAB = 30 , what is the m ∠ CAD? Step 1. Each diagonal of a rhombus bisects pair of opposite angles. A B m ∠ CAD = m ∠CAB Step 2. Substitute 30 for m ∠CAB in the above equation. m ∠ CAD = 30 Answer: The measure of ∠ CAD is 300. Example 5

DEFG is a rhombus. If m∠FGE = 5x – 8 and m∠DGE = 3x + 22, find the measure of (a) m ∠FGE (b) m∠DGE and (c) m∠FGD G F D E Solution: Step 1. Each diagonal of a rhombus bisects a pair of opposite angles. m∠FGE = m∠DGE 5x – 8 = 3x + 22 5x – 3x = 22 + 8 2x = 30 x = 15

9

Step 2. Substitute 15 for x a. m∠FGE = 5x – 8

= 5(15) – 8 = 67 b. m∠DGE = 3x + 22 =3(15) + 22 = 67 c. m∠FGD = m∠FGE + m∠DGE = 67 + 67 = 134

Answers: (a) m ∠FGE = 67 (b) m∠DGE = 67 (c) m∠FGD = 134 Example 6 BETH is a rhombus. If m∠TBE = 35, H T what is m∠HEB? M Solution: Step 1. The diagonals of a rhombus B E are perpendicular. Hence, ∠BME is a right angle and its measure is 900. m∠BME = 90 Step 2. The sum of the measures of the angles of a triangle is 1800 m∠TBE + m∠BME+ m∠HEB = 180 Step 3. Substitute 35 for m∠TBE and 90 for m∠BME in the above equation. 35 + 90 + m∠HEB = 180 125 + m∠HEB = 180 m∠HEB = 180 – 125 m∠HEB = 55. Answer: m∠HEB = 55 D C Example 7 M ABCD is a rhombus. If AM = 16 cm, what is CM? A B

10

Solution: Step 1. The diagonals of a rhombus Bisect each other. CM = AM Step 2. Substitute 16 cm for AM in the above equation. CM = 16 cm Answer: CM = 16 cm Try this out Set A . ABCD is a rectangle. D C A B True or False

1. The lengths of AC and BD are equal. 2. Diagonals AC and BD are perpendicular. 3. The diagonal AC bisects ∠DCB. 4. A rectangle is a parallelogram.

ABCD is a square D C E

5. ∠EAB ≅ ∠EBC 6. ∠DEC is a right angle.

A B

FGHI is a rhombus. I H

7. m∠FIG = m∠HIG J 8. The sum of m∠JFG and m∠JGF is 45. 9. ∆FIG ≅∆ HGI

F G 10. The diagonal FH bisects the rhombus into two congruent triangles.

11

Set B

ABCD is a rectangle D C A B Find the indicated measure

1. AC = 15 dm. Find BD 2. BD = 23 cm. Find AC

EFGH is a rhombus

Find the indicated measure. H G

3. m∠HGE = 35. Find m∠FGE. I 4. m∠HEI = 20. Find m∠FEI. 5. m∠IEF = 30. Find m∠IFE 6. m∠IHE = 58. Find m∠IEH 7. m∠IEF = 20. Find m∠IEF + m∠EIF E F 8. m∠IGH = 25. Find m∠IGH + m∠HIG

9. If ABCD is a square, D C then m∠ACB = ________ 10. If ABCD is square, then m∠ DEC =___________ E

A B Set C

ABCD is a rectangle with diagonals AC and BD. D C

1. AC = 2x + 15, BD = 3x + 10. Find AC. 2. BD = 6x + 5, AC = 5x + 14. Find BD.

A B

12

FERM is a rhombus. M R

I F E 3. If m∠IFE = x +20, m∠IEF = x + 26 ,find x. 4. If m∠IMR = 4x + 20, m∠IRM = 2x + 10, find x. 5. m∠IFE + m∠IEF = __________ 6. m∠IMR + m∠IRM = _________

BETH is a square.

H T

B E

7. If HM = x + 15, HE = 40, what is x? 8. If EM = x + 9, HE = 30, what is x? 9. If BM = x + 12, EM = 2x – 20, what is x? 10. If HM = 44 – x, TM = 4 + 3x, what is x?

Lesson 2

Conditions for a Parallelogram The following are some conditions which guarantee that a given quadrilateral is a parallelogram.

1. If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

2. If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

3. If one pair of opposite sides of a quadrilateral are both congruent and parallel, then the quadrilateral is a parallelogram.

4. If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

13

5. If the non-opposite angles of a quadrilateral are supplementary, then the quadrilateral is a parallelogram.

You can verify these sets of sufficient conditions which guarantee that a

quadrilateral is a parallelogram. In the following activities you need a pencil, a ruler, a protractor and pieces of bond paper and graphing paper.

1. If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral

is a parallelogram. Do this activity:

a. On a graphing paper, draw a quadrilateral such that both pairs of opposite sides are congruent. ( See the illustration.)

b. Are the opposite sides equidistant? Find this out by using a ruler. c. Are both pairs of sides parallel? (Remember, parallel lines are everywhere

equidistant.)

d. Can you now conclude that the quadrilateral is a parallelogram? Why?

2. If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

a. On a graphing paper, with the aids of a ruler and a protractor, construct an

quadrilateral such that both pairs of opposite angles are congruent. (See illustration)

14

b. Are the opposite sides congruent? c. Can you now conclude that the quadrilateral is a parallelogram? Why?

3. If one pair of opposite sides of a quadrilateral are both congruent and parallel, then

the quadrilateral is a parallelogram. Do this activity

a. On a graphing paper, draw a quadrilateral such that one pair of opposite sides are both congruent and parallel.

( See illustration below)

b. Are the other two opposite sides congruent? c. Can you now conclude that the quadrilateral is a parallelogram? Why?

4. If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a

parallelogram.

15

a. On a bond paper, draw segments AC and BD bisecting each other. (See the illustration below.)

D C A B

b. Connect A to B, B to C, C to D and D to A .

D C A B

c. Using a ruler, find the lengths of AB and CD. Are they equal? d. Using a ruler, find the lengths of AD and BC. Are the lengths equal? e. What kind of quadrilateral is ABCD?

5. If the non-opposite angles of a quadrilateral are supplementary, then the quadrilateral

is a parallelogram.

a. On a bond paper, draw angle A. (See the illustration below.) A

b. Draw angle ADC such that its measure is supplementary to that of angle A. ● C

A D

16

c. Draw angle DCB such that its measure is equal to that of angle A. B

● C A D

d. Find the measure of angle CBA. Is it equal to the measure of angle ADC? Are ∠A and ∠B supplementary? How about ∠B and ∠C? How about ∠D and ∠C? Why/ e. What kind of quadrilateral is ABCD?

Example 1 Determine whether the figure is a parallelogram. Identical “tick marks” indicate that the sides are congruent and identical “arrowheads” indicate the lines are parallel. Solution:

If one pair of opposite sides of a quadrilateral are both congruent and parallel, then the quadrilateral is a parallelogram.

Hence the geometric figure is a parallelogram. Example 2 Determine whether the figure is a parallelogram. Solution:

17

A pair of alternate interior angles are congruent, therefore a pair of opposite sides are parallel. These parallel sides are also congruent. As can be seen in the figure, they have the same length. Hence the figure is a parallelogram. Example 3. Find the value of x for which ABCD is a parallelogram.

A D 270 3x B C Solution: If two lines are cut by a transversal and a pair of alternate interior angles are congruent, then the lines are parallel. AD // BC since ∠ADB ≅∠CBD CD // AB if 3x = 27 x = 9 Hence the value of x should be 9. Try this out Set A True or False

1. If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral

is a parallelogram. 2. If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral

is a parallelogram. 3. If one pair of opposite sides of a quadrilateral are parallel, then the quadrilateral is a

parallelogram. 4. If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a

parallelogram. 5. If the opposite angles of a quadrilateral are supplementary , then the quadrilateral is

a parallelogram.

18

ABCD is a quadrilateral. AD = 5 cm and AB = 9 cm. y D C 5 cm x A B

9 cm 6. ABCD is a parallelogram if x = 5 cm and y = 9 cm. 7. ABCD is a parallelogram if m∠C = 60 and m∠B = 120. 8. ABCD is a parallelogram if AB // DC. 9. ABCD is a parallelogram if m∠B ≅m∠D 10. ABCD is a parallelogram if AB ≅ DC ≅ AD ≅ BC.

Set B. Determine whether each quadrilateral is a parallelogram. Identical “tick marks” indicate that the sides or angles are congruent and identical “arrowheads” indicate the lines are parallel. D C 1. A B D C 2. A B 3.

19

4. 5. 6. 10 7. 6 7 10 8. 1000 790 810 1000 300 9. 500 500 300

20

15 cm 10. 300 300 15 cm Set C. What values of x and y guarantee that each quadrilateral is a parallelogram. 1. 6. y 500 450 1350 x y 3x y 2y 2 1100 700 7. x 8 cm x y 14 cm y 126 3. 8. 6 cm x 90 x 12 cm 3y 15 cm 2x + 10 4. 9. y 4 cm 2y 24 70

x 4y 2x 600 5. 10, 5 2x – 5 y 1200 32

21

Let’s summarize

1. A diagonal of a quadrilateral is a segment which connects any two non-consecutive vertices.

2. The diagonals of a rectangle are congruent. 3. The diagonals of a square are congruent. 4. The diagonals of a square are perpendicular 5. Each diagonal of a square bisects a pair of opposite angles. 6. The diagonals of a rhombus are perpendicular. 7. Each diagonal of a rhombus bisects a pair of opposite angles 8. A square is a special type of rhombus. 9. If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral

is a parallelogram. 10. If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral

is a parallelogram. 11. If one pair of opposite sides of a quadrilateral are both congruent and parallel, then

the quadrilateral is a parallelogram. 12. If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a

parallelogram. 13. If the non-opposite angles of a quadrilateral are supplementary, then the quadrilateral

is a parallelogram. 14. A quadrilateral is a parallelogram if both pairs of opposite side are parallel


1. A parallelogram is a rhombus if A. The diagonals bisect each other B. The diagonals are perpendicular. C. Two consecutive angles are supplementary. D. The opposite sides are parallel.

2. Which of the following is sufficient to guarantee that a quadrilateral is a parallelogram?

A. The diagonals are perpendicular B. A pair of adjacent sides are congruent C. Two consecutive angles are congruent D. The diagonals bisect each other

22

3. ABCD is a rectangle. if diagonal AC = 2x + 6 and diagonal BD = 10, what is x? A. 1 C. 3 B. 2 D. 4

4. ABCD is a rhombus. D C

A B If m∠DCA = 2(x+8) and m∠BCA = 3x + 9, what is m∠DCB?

A. 40 C. 60 B. 50 D. 70

5. ABCD is a square.

D C

A B

If m∠ABD = 3(x + 10), what is x? A. 1 C. 5 B. 3 D. 7

6. ABCD is a rhombus. Diagonals AC and BD intersect each other at E.

D C E

A B If AE = 12 and CE = 3x, what is x?

A. 2 C. 6 B. 4 D. 8

23

7. ABCD is a rhombus . Diagonals AC and BD intersect at E.

D C

E

A B What is m∠AED?

A. 30 C. 60 B. 45 D. 90

8. What values of x and y guarantee that ABCD is a parallelogram. D C

y 64 x y

A. x = 64 , y = 116 C. x = 64, y = 64 B. x = 32, y = 116 D. x = 32, y = 64

9. Find the value of x for which ABCD is a parallelogram. D C

400 800 800 2x A B

A. 10 C. 30 B. 20 D. 40

10. Find the value of x for which ABCD is a parallelogram.

18 3x – 6 12 18

A. 8 C. 4 B. 6 D. 2

24

Answer Key


1. True 2. False 3. True 4. True 5. True 6. True 7. AC = 100 8. AE = 15 9. m∠FCD = 20 10. x = 10

Lesson 1 Set A

1. True 2. False 3. False 4. True 5. True 6. True 7. True 8. False 9. True 10. True

Set B 1. 15 2. 23 3. 35 4. 20 5. 60 6. 32 7. 110 8. 115 9. 45 10. 90

Set C 1. AC = 25 2. BD = 59 3. x = 22 4. x = 10 5. 90

6. 90 7. x = 5 8. x = 6 9. x = 32 10. x = 10

Lesson 2 Set A

1. True 2. True 3. False 4. True 5. False 6. True 7. True 8. True 9. True 10. False

Set B 1. Parallelogram 2. Parallelogram 3. Parallelogram 4. Parallelogram 5. Parallelogram 6. Parallelogram 7. Not a parallelogram 8. Not a parallelogram 9. Parallelogram 10. Parallelogram

Set C 1. x = 500

y = 1300 2. x = 700

y = 1100

3. x = 6 cm y = 12 cm

4. x = 15 cm y = 4 cm

5. x = 600 y = 600

6. x = 450 y = 450

7. x = 8 cm y = 7 cm

8. x = 90 units y = 42 units



What have you learned 1. B 2. D 3. B 4. C 5. C 6. B 7. D 8. A 9. B 10. B

MODULE 1 Geometric Relations What this module is about

This module is about relations of segments and angles. As you go over the exercises, you will develop your skills involving points, segments and angle pairs and solve problems on the relationships between segments and between angles. Treat the lessons with fun and take time to go back if you feel you are at loss.

What you are expected to learn

This module is designed for you to:

1. illustrate betweenness and collinearity. 2. illustrate the following:

• congruent segments

• midpoint of a segment

• congruent angles

• bisector of an angle

• complementary angles

• supplementary angles

• adjacent angles

• linear pair

• vertical angles

2

How much do you know Answer the following questions asked from the given figure: . A D I M H J P

. . . . . . . . . . -3 E O 3 6

F C

Answer the following questions:

1. What is /IP/? 2. If /IM/ = /MH/ = /IH/, then what point is between the other two?

3. If M is the midpoint of AC, what segments are congruent?

4. Name the coordinate of the midpoint of IJ

5. ∠EMF and ∠FMC are ____ angles.

6. ∠EMF form a linear pair with ______.

7. If ∠EMF is the complement of ∠FMC and on ∠FMC = 75, what is on ∠EMF?

8. If ∠AMD ≅ ∠DMC and are supplementary, what kind of angle is each?

9. What is the measure of each angle if the measure of the angles in a

supplementary pair is twice that of the other?

10. What is the measure of each angle if the two angles are both vertical and complementary?

3

What you will do

Lesson 1

Collinearity of Points

You may ask this question, “how many points are there in a line?” “ How many points are there in a plane?” Do you know the answer? Yes, it is infinite or many points. Now you take a look at BP in the figure.

• .X B C P

• Y •Z

Points B, C, P are contained in BP Points X, Y, Z are not in BP Points B, X, Y are not in BP The set of points B, C, P are collinear The set of points X, Y, Z are not collinear The set of points B, X, Y are not collinear

Collinear points is a set of points which are contained in a line

Examples:

The following set of points are collinear

1. A, O, P, L S

2. M, N, P, S A O P L M N

4

A 3. A, Q, C P . . Q 4. A, P, B, S 5. B, R, C . B .R . C S

Do you still remember the number line? A B C D E F G H I J K L M N . . . . . . . . . . . . . . . -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

The points of a line can be placed in correspondence with the real numbers in such a way that:

1. to every point of the line there corresponds exactly one real number; 2. to every real number there corresponds exactly one point of the line; 3. the distance between any two points is the absolute value of the

corresponding numbers.

The number corresponding to a given point is called the coordinate of the point.

Examples:

1. The coordinate of M is 6, the coordinate of G is 0, the coordinate of B is -5. B C . . . . .

X -1 O 1 Y

2. If the coordinate of B is x and the coordinate of C is y then, /BC/ = /X-Y/ (read as

distance BC equals the absolute value of X minus Y).

5

J B . . . . . . . . . . . -5 O 3 3) /JB/ = │-5 - +3│ = │-8│ = 8

Try this out

Which set of points are collinear: C

B .

1. A, O D D E 2. A, B, O A O 3. B, O, F 4. G, O, F 5. C, O G G . F

A B C D E F G H I J K L M . . . . . . . . . . . . . -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 Give the coordinate of each of the following points:

6. G 7. D 8. M 9. A

10. K

B. The coordinates of P and Q are listed. Find /PQ/. 1. P: O 6. P: 7 Q: 7 Q: 3

6

2. P: 12 7. P: 19 Q: 0 Q: 113 3. P: 4 8. P: 5 Q: 15 Q: - 5 4. P: 21 9. P: 56 Q: 14 Q: -18 5. P: 15 10. P: -12 Q: 6 Q: -51

M N O P Q R S C T U V W X Y Z A B C. . . . . . . . . . . . . . . . . . . . . .

-5 -4 -3 -2 -1 0 1 2 3 4 5 -4.5 0.5 ∏

Find: 1. /PS/ 6. /RX/ + /PR/ 2. /US/ 7. /CR/ + /OR/ 3. /PT/ 8. /PS/ + /RU/ 4. /NC/ 9. /SW + /WY/ 5. /NT/ 10. /NR/ + /RW/

Lesson 2

Betweenness

A B C . . .

Let A, B, C be three points. B is between A and C. If A, B and C are on one line and

/AB/ + /BC/ = /AC/. This definition of betweenness means that: 1. If B is between A and C, then A, B and C then, A, B, C are collinear and /AB/ + /BC/ = /AC/. 2. If A, B and C are collinear and /AB/ + /BC/ = /AC/ then, B is between A and C.

7

Examples: 1. If x, y, and z are collinear X Y Z

and /xy/ + /yz/ = /xz/ then, Y is between X and Z.

R 2. Given: T is between R and U

Conclusion: R, T, U are collinear and /RT/ + /TU/ = /RU/ . T .U

3. O is between S and P S O P

Find /SP/ . . . Solution: -2 0 3 /SO/ + /OP/ = /SP/ 2 + 3 = /SP/ 5 = /SP/

Try this out

A. Which point is between the other two?

1. . P . O . M

C E U

2. B O A

3.

4. A, B, and C are three points on a line with coordinates 8, 4, and 13 respectively.

8

5. R, S, and T have coordinates x, y, and z respectively x < y < z. From each of the following equations determine which point is between the other two.

6. /AB/ + /BC/ = /AC/ 7. /PO/ + /QR/ = /PR/ 8. /LM/ + /LN/ = /MN/ 9. /BC/ - /AB/ = /AC/

10. /QR/ - /PQ/ = /RP/ P Q R S T U V W X B. . . . . . . . . . .

-5 -4 -3 -2 -1 0 1 2 3 4

1. Is it true that /QS/ + /ST/ = /QT/? 2. Is it true that /PS/ + /ST/ = /PT/? 3. Is it true that /PW/ - /PR/ = /RW/? 4. Is it true that O is the coordinate of the midpoint of PX? 5. Is it true that /XP/ - /WX/ = /PW/?

A B C D E F G H I J K L M 6. – 10. . . . . . . . . . . . . . -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

6. Give the distance between points G and J. 7. Give the distance between points G and D. 8. Find the distance between G and the midpoint JL . 9. Find the coordinate of the midpoint of DK.

10. Find the midpoint of AM. C. A E I O U B . . . . . . -10 -5 0 5 10 15 Using the figure complete the statement in nos. 1 – 5.

If E is between A and B,

1. /AB/ = /AE/ + ____________________ 2. /AE/ = /AB/ - ____________________ 3. /E/ = /AB/ - ____________________ 4. /AB/ = __________________________ 5. /IB/ = ________________________ 6. What is the coordinate of the midpoint of AU?

9

7. What two segments are congruent if O is the midpoint of BE? 8. Find a segment congruent to AO with B as one endpoint. 9. O is the midpoint of a segment with I as one of the endpoints. Find the

segments. 10. AU ≅ ________.

Lesson 3

Congruent Segments and Midpoint of a Segment

Segments are congruent if and only if they have equal measures. Examples: R O 1. RS ≅ XY . 0 4 If you look at RS and XY X Y m RS = /0-4/ = 4 . . m XY = /4-8/ = 4 4 8 m RS = m XY ∴RS ≅ XY 2. AD ≅ EH A B C D E F G H I J . . . . . . . . . . -4 -3 -2 -1 0 1 2 3 4 5 3. AI ≅ BJ 4. if m MB = 25 cm. and m ST = 25 cm then MB ≅ ST 5. RM = MB = 6 then, RM ≅ MB Midpoint – is a midpoint of the segment which divides the segment into two congruent parts. Examples: 1. P is the midpoint of BC __.________.____ ___.__

B P C if and only if BP ≅ PC .

10

2. O is the midpoint of SP .S 2.5

O 2.5

3. M is between D and E .P such that MD ≅ ME , then M is called the midpoint of DE More about midpoint A A segment has exactly one midpoint If M is also midpoint of AY then .M 2 /AM/ = /AY/ 2 /MY/ = /AY/ /AM =

21 /AY/ /MY/ =

21 /AY/ Y

Any line that passes through the midpoint of a segment is called a bisector of the segment. Try this out A. D E F G H I J K L M N O P Q R S . . . . . . . . . . . . . . . . -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

1. What is the distance between D and G? 2. What is m LO? 3. What can you conclude about DG and LO? 4. What two points are at a distance 3 from L? 5. What can you say about LO and LI?

Given: MD with point A M between points M and D such that /MA/ =

21 /MD/ and /MD/ = 12 A

6. What is /MA/? 7. What is /AD/? D 8. Is /MA/ = /AD/?

11

9. Is MA ≅ AD? 10. Is A the midpoint of MD? Given ME, SO, PA, and RU with their respective measures

M . 5 cm . E S . 10 cm .O P . 10 cm .A

R . 12 cm .U Write <, > or = to compare the measures of the line segments.

1. /ME/ ________ /SO/ 2. /SO/ ________ /PA/ 3. /PA/ ________ /RU/ 4. /RU/ ________ /PA/ 5. /RU/ ________ /ME/

M N O P Q R S T U V W X Y Z A B C . . . . . . . . . . . . . . . . .

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

6. How far is it from M to P? 7. How far is it from C to Z? 8. Find /RZ/ 9. Find /CU/

10. Give two pairs of congruent segments. C. A B C D E F G H I J K . . . . . . . . . . . . . . . -10 -8 -4 -2 0 2 4 6

1. /AD/ = ________ 2. m AD _________ m DG 3. AD _________DG 4. If D is between A and G, /AD/ + /DG/ = __________. 5. D is the midpoint of ______________. 6. If G is the midpoint of EI then __________≅ _______. 7. If AD ≅ DG then_______ is the midpoint of AG 8. Suppose point B is on AC and B is not a midpoint of AC, then /AB/ + /BC/ =

___________.

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Given: Point M is the midpoint of JK and JM = 5. Complete the following:

9. /MK/ = _______ 10. /JK/ = _______ and JM ≅ ________.

Lesson 4

Congruent Angles and Angle Bisector Congruent Angles are two angles with equal measure

Angle bisector is a ray from the vertex of an angle to a point in its interior which divides the angle into two congruent parts.

Examples: C 1. m ∠AOF = 15 90 D

m ∠EOG = 15 B 45 m ∠AOF = m∠EG E . ∴∠AOF ≅ ∠EDG

A 15 2. m∠1 = 45 m∠2 = 45 F O G m∠1 = m∠2

∴∠1 ≅ ∠2

3. ∠X ≅ ∠Y X 60º Y 60º B . 4. OB is an angle bisector A . C . O

13

L 5. OB is an angle bisector of ∠LOA B 30º O 30º A 6. AC is not an angle bisector B C 40º 30º A D You can also state the definition of angle bisector this way: If D is in the interior of ∠BAC B and ∠BAD ≅ ∠DAC, then AD bisects ∠BAC and AD is called the bisector of LBAC D A C Try this out A. Complete the following statements: 1.Two angles are said to be congruent if the angles have ______ measures.

2. When two angles have equal measures, the angles are________ 3. If m∠3 = 57 and ∠4 = 57 then ∠3 and ∠4 are _____________ angles. 4. A ray from the vertex to the interior of the angle which bisects an angle is called ____.

5. If ∠LEM ≅ ∠MEN then________ L M is the angle bisector of ∠LEN

E N 6. ∠MON ≅ _________ 7. ∠SOT ≅ _________

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8. ∠POQ ≅ _________ 9. ∠MDQ ≅ _________ 10. ________ is the angle Nos. 6-10 bisector of ∠POR. M O T 30 15º N 15º 30º 45º 45º S P Q R B.

1. Name two congruent angles with R H S JK as a common side. 30º 30º

2. Give another pair of congruent angles each measuring 30º. J 3. What do you call JH? 30º 30º 4. JK is ______of angle PJQ. 5. ∠RJS____∠PJQ P K Q 6. If IT is the angle bisector name two congruent angles. Nos. 6-8 7. If m∠PIS = 90 and IT is a P T bisector what is m∠SIT? 8. If IT is a bisector of ∠PIS and m∠PIT = 33, what is m∠SIP? I S 9. If m∠O = 120 what is m∠A if ∠O ≅ ∠A? 10. If ∠O is bisected what is the measure of each part? O

A

15

Lesson 5

Complementary and Supplementary Angles

The sum of the measures of complementary angles and two angles is equal to 90. Here you can see ∠M is the complement of ∠P. ∠ is the complement of ∠P. 60º 60º M N m∠M + m∠P = 90 ∠M and ∠P are complementary angles. 30º P The sum of the measures of supplementary angles are two angles is equal to 180. Each of two supplementary angles is called a supplement of the other angle. You can see that ∠B is the supplement of ∠E. 120 60 m∠B + m∠E = 180 B E ∠B and ∠E are supplementary angles. Examples: A 1. 15º + 75º = 90º B ∠AOB and ∠BOC are 75º complementary angles. 15º O C 2. 140º + 40º = 180º R ∠SER and ∠TER are supplementary angles 140º 40º S E T 3. Given m∠LI = ∠70º, find its complement. Since the sum of the measures of complementary angles is equal

to 90, subtract the given angle from 90 to get the complement. Solution: 90 – m∠ I = complement 90 – 70 = 20

16

4. ∠X and ∠Y are supplementary. Find m∠Y if m∠X = 100. Solution: m∠X + m∠Y = 180 100 + m∠Y = 180 m∠Y = 180 - 100 m∠Y = 80 Try this out A. In the given figure, ∠AOB and ∠DOC are right angles 1 - 2. Name all pairs A B of complementary angles 1 2

3 4 3 – 4. Name all pairs of supplementary angles O

5 6 7 8 C D What is the complement of each of the angles whose measures are given: 5. 12º 8. 67º 6. 39º 9. 79º 7. 41º 10. 84.5º

B. What is the supplement of each of the angles whose measures are:

1. 11º 2. 44º 3. 121º 4. 152.5º 5. 78.6º

6. ∠A and ∠B are complementary. If ∠A = 78º then, ∠B = _______. 7. ∠C and ∠D are supplementary. If ∠C = 110.5º then, ∠D = _______. 8. An angle has a measure x. Find the measure of its complement. 9. An angle has a measure 2x. Find the measure of its supplement.

10.Two supplementary angles have measure of 2x – 15 and x + 30.

C. Find the measure of each angle. 1. If∠X and ∠Y are supplementary then, m∠X + m∠Y = _______. 2. If the m∠O + m∠P = 90 then, ∠O and ∠P are ______________.

17

3. What is the complement of ∠K whose measure is m? 4. Find the supplement of an angle whose measure is a. 5. If two congruent angles are complementary then, each angle has a measure of ____. 6. If two angle are complementary then, each is a ____of the other.

7-8. The measure of an angle is 15 greater than twice the measure of its complement.

Find the measure of each angle:

9-10. The measure of an angle is 20 less than three times the measure of its supplement. Find the measure of each angle.

Lesson 6

Adjacent Angles, Linear Pair, Vertical Angles Adjacent angles are two angles which have a common side and a common vertex but no interior points common. Examples:

B 1 A C R

2 M P Q

∠1 and∠2 are adjacent ∠S ∠AOB and∠BOC ∠MPR and ∠RPQ are adjacent∠S are adjacent ∠S M J N B L A E Y O P ∠BEL and ∠ JAY are not ∠MOP and ∠MON are not adjacent angles adjacent angles

18

A linear pair are two adjacent angles whose non common sides are opposite rays. Examples:

S C P I P S E M A O B T ∠SEP and ∠PEM ∠COA and ∠BOC ∠PIT and ∠SIT form a linear pair form a linear pair form a linear pair If you try to measure each angle forming a linear pair, you will find out that the sum of their measures is 180. So angles forming a linear pair are supplementary. Vertical Angles are two nonadjacent angles formed by two intersecting lines: Examples: A B 1 4 2 O 3 C D ∠1 and ∠3 are vertical angles ∠AOB are DOC vertical angles ∠2 and ∠4 are vertical angles ∠BOC and ∠AOD are vertical angles Use your protractor to measure each angle in the figure Find:

m∠1 = ___________ m∠2 = _________ m∠3 = ___________ m∠4 = _________ m∠AOB = _______ m∠BOC = ______ m∠DOC = _______ m∠AOD = ______ Have you found out that they have the same measure? Therefore, vertical angles are congruent.

19

Try this out A. . R U.

1 – 2 Name 2 angles adjacent to ∠1 2 . 1 3 . T O 4 Q . 5 S

3. Name an angle which form a linear pair with ∠4. 4 – 5. Name two pairs of vertical angles 6. Are ∠1 and ∠5 adjacent? 7. Are ∠4 and ∠5 adjacent? 8. What pair of ∠S are ∠3 and ∠4? 9. Are ∠1 and ∠3 vertical angles? 10. ∠SOU form a linear pair with ∠________.

B. Answer with Yes or No. 1. Are ∠3 and ∠6 vertical angles? 2 3 2. Are ∠2 and ∠3 adjacent angles? 3. Are ∠1 and ∠8 vertical angles? 4. Are ∠7 and ∠8 linear pair? 5. Are ∠4 and ∠8 linear pair? 1 6. Are ∠1 and ∠8 adjacent? 7 8 7. Are ∠1 and ∠7 linear adjacent? 4 8. Are ∠1 and ∠7 linear pair? 9. Are ∠5 and ∠6 adjacent? 10. Are ∠1 and ∠4 vertical angles? 6 5 C. True or False

1. Complementary angles are always adjacent. 2. Supplementary angles are sometimes adjacent. 3. The angles of a linear pair are always adjacent. 4. Vertical angles are sometimes adjacent. 5. If two angles are vertical then, they are either both acute or both obtuse. 6. Two adjacent right angles are supplementary. 7. Two vertical angles are always congruent. 8. If two angles form a linear pair, they are supplementary. 9. If one of the angles in a linear pair is 90, then the other angle has a measure greater

than 90. 10. If two adjacent angles are congruent and complementary, the measure of each angle

is 90.

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Let’s Summarize

Collinear points is a set of points which are contained in a line. The points of a line can be place in correspondence with the real numbers in such a way that:

• to every point of the line there corresponds exactly one real number;

• to every real number there corresponds exactly one point of the line;

• the distance between any two points is the absolute value of the difference of the corresponding numbers.

The number corresponding to a given point is called the coordinate of the point.

Let A, B, C be three points. If A, B, C are on one line and. B is between A and C then, /AB + /BC/ = /AC/ Congruent segments are segments with equal measure.

Midpoint is a point of a segment which divides the segment into 2 congruent parts. Congruent angles are angles with equal measure. Angle bisector is a ray from the vertex of an angle to a point In its interior which divides the angle into two congruent parts. Complementary angles are two angles whose measures have the sum equal to 90. Supplementary angles are two angles whose measures have the sum equal to 180. Adjacent angles are two angles which have a common side and a common vertex but no interior points in common. Linear Pair are two adjacent angles whose noncommon sides are opposite rays. Vertical angles are two nonadjacent angles formed between two intersecting lines.

21


1. What is /BD/?

2. If RA ≅ SA what do you call point A? R N

3. Name a supplement of ∠NAS.

4. What angle pair is 400 Illustrated by B E A 50º C D ∠RAN and ∠NAS? -6 -3 3 6

5. If ∠BAM ≅ ∠SAM what do you call AM? M S

6. ∠RAN is vertical to ∠_____. 7. The measure of ∠SAM = _____. 8.- 9. If the measure of an angle is twice

the measure of its complement, what is the measure of each angle? 10. The measure of ∠B is 9 more than twice the measure of ∠C. If ∠B and ∠C are supplementary angles, what is the measure of ∠B?

22

Answer Key


1. 9 6. ∠FMD 2. M is between and H 7. 15 3. AM ≅ MC 8. right 4. I 9. 60º, 120º 5. adjacent 10. 45ºTry this out

Try this out Lesson 1

A. 1. collinear 6. 0 2. not collinear 7. -3 3. collinear 8. 6 4. not collinear 9. -6 5. collinear 10. 4 B. 1. 7 6. 4 2. 12 7. 94 3. 11 8. 10 4. 7 9. 74 5. 9 10. 39 C. 1. 3 6. 7 2. 2 7. 4.5 3. 4 8. 6 4. 5 9. 5 5. 5.5 10. 7.64 Lesson 2

A. 1. O is between P and M 6. B is between A and C 2. E is between C and U 7. Q is between P and R 3. O is between B and A 8. L is between M and N 4. A is between B and C 9. A is between B and C 5. S is between R and T 10. P is between R and Q B. 1. Yes 6. 3 2. Yes 7. 3 3. Yes 8. 4 4. Yes 9. 0.5

23

5. Yes 10. G C. 1. /EB/ 6. O 2. /EB/ 7. EO ≡ OB 3. /AE/ 8. IB 4. 25 9. IU 5. 15 10. BE Lesson 3 A. 1. 3 6. 6 2. 3 7. 6 3. DG≅ LO 8. Yes 4. O and I 9. Yes 5. LO ≅ U 10. Yes B. 1. < 6. 3 2. = 7. 3 3. < 8. 8 4. > 9. 8 5. > 10. MD ≅ CZ and RZ≅ CU (answers may vary) C. 1. 6 6. EG≅ GI 2. = 7. D 3. ≅ 8. /AC/ 4. 12 9. 5 5. /AG/ (answers may vary) 10. 10, MK Lesson 4 A. 1. equal or the same 6. ∠SOR 2. congruent 7. ∠NOP 3. congruent 8. ∠QOR 4. angle bisector 9. ∠TOQ 5. EM 10. OQ B. 1. ∠PJK and ∠QJK 6. ∠PIT ≅ ∠SIT 2. ∠RJH and ∠HJS 7. 45 3. angle bisector 8. 66 4. angle bisector 9. 120 5. ≅ 10. 60 Lesson 5 A. 1. ∠3 and ∠4 6. 51º

24

2. ∠5 and ∠6 7. 49º

3. ∠1 and ∠2 8. 23º 4. ∠7 and ∠8 9. 11º 5. 78 10. 5.5º B. 1. 169º 6. 12º 2. 136º 7. 69.5º 3. 59º 8. 90 - x 4. 27.5º 9. 180 - 2x 5. 101º 10. 95, 85 C. 1. 180 6. complement 2. complementary 7. 65 3. 90 - m 8. 25 4. 180 - a 9. 130 5. 45º 10. 50 Lesson 6 A. 1. ∠2 6. Yes 2. ∠5 7. Yes 3. ∠5 and ∠ROQ 8. adjacent 4. ∠1 and ∠4 9. No 5. ∠SOT and ∠ROQ 10. ∠ROU B. 1. No 6. Yes 2. No 7. Yes 3. No 8. Yes 4. Yes 9. No 5. Yes 10. Yes C. 1. False 6. True 2. True 7. True 3. True 8. True 4. False 9. False 5. False 10. False What have you learned 1. 12 6. ∠SAM 2. Midpoint 7. 40º 3. ∠RAN 8. 60º 4. Linear pair 9. 30º 5. Angle bisector 10. 123º

Module 2 Geometric Relations

What this module is about This module will explain to you different characteristics of lines on a plane. In this module, you will discover that in a plane, lines may be parallel, perpendicular or just intersecting. This module will introduce you on the relationship among sides and angles of a triangle. You will also discover some theorems associated with perpendicularity and inequalities among sides and angles of a triangle. In addition, this module will define characteristics of lines in a plane.


This module is designed for you to

1. define and illustrate perpendicular and parallel lines 2. define and give examples of the perpendicular bisector of a segment. 3. define and illustrate exterior angles of a triangle and its relationships with other

angles of a triangle. 4. define inequalities among angles and sides of a triangle. 5. illustrate lines that serve as transversal.

How much do you know L1 Answer the following questions. 1

1. If l1 ⊥ l2 , then ∠ 1 is _________ angle. L2

2. Given: AB ⊥ BC . If ∠ABE = 2x + 15, and A ∠EBC = x, what is m∠ABE? E

B C

2

3. An angle which is adjacent and supplementary to one of the angles of a triangle is a ________ angle.

E 4. In the figure, m∠EFD + m∠EFG = ________.

D F G

5. What property of inequality supports the statement, If m > 5 and 5 > n, then m > n. W

6. If WY = 5 and YX = 8, what is the range of values of WX? Y X 7. If B is the midpoint of AX , then AB _____ BX . 8. What is the length of the hypotenuse of a right triangle whose length of the legs are

3 cm and 4 cm respectively.?

9. Coplanar lines that do not intersect are called ___________. t 10. In the figure, line t intersects lines m m and n at two points. t is called ______. n

What you will do

Lesson 1

Perpendicular Lines and Perpendicular Bisector of Segment

a In the figure, two lines a and b intersect at point X, forming four angles. Two of these angles are ∠1 and ∠2. If all these angle are right angles, 1 2 b

then a is perpendicular to b. The symbol to be x used for perpendicular is “ ⊥ “. The figure above can be represented in symbols as a ⊥ b. We will define perpendicular lines as follows: Two lines are perpendicular if and only if they intersect to form right angles. Since a ⊥ b then ∠1 and ∠2 are right angles. The two other angles in the figure are also right angles. In the definition stated, there is the phrase if and only if (iff) which means that the definition is two way. 1) If the lines are perpendicular, then the angles formed are right angles and 2) if

3

the angles are right angles, then the lines or sides of the angle are perpendicular. This goes to show that one way of proving perpendicularity is to prove that the angles formed are right angles.

For you to recognize perpendicular lines, a small square is indicated at the foot of the perpendiculars which is the intersection.

Consider the given figure. OD ⊥ XM at O. Hence there is

a symbol of small square at the intersection. D Notice that XM is divided into two segments XO and OM which has similar markings. Those X M Markings indicate that XO ≅ OM making O O

The midpoint of the segment. Therefore OD is the perpendicular bisector of XM .

The perpendicular bisector of a segment is a line, ray, segment or plane that is perpendicular to the segment at its midpoint. There are four cases by which a segment has its perpendicular bisector. t Case 1. line t is the perpendicular bisector of MN . M x N D Case 2. Ray OD is the perpendicular bisector of MN M N B O Case 3. Segment AB is the perpendicular bisector of MN . M N A Case 4 Plane P is the perpendicular bisector of XY P X Y “Any two right angles are congruent”. All right angles measure 90˚.

4

Example: ∠MOP is a right angle M R ∠RST is a right angle Conclusion: ∠MOP ≅ ∠RST O P S T Try this out In the figure, MO ⊥ OP , CO ⊥ OD 1. ∠MOP is __________ angle M C 2. m∠MOP = ________. P 3. ∠COD is __________ angle. 4. m∠COD = _________. O D Give the reason for the following statements. 5. m∠MOP = m∠COD 6. m∠MOP = m∠MOC + m∠COP 7. m∠COD = m∠COP + m∠POD 8. m∠MOC + m∠COP = m∠COP + m∠POD 9. - m∠COP = - m∠COP 10. m∠MOC = m∠POD 11. ∠MOC ≅ ∠POD In the figure, OR ⊥ bisector of SP . 12. SO ≅ ________. R If SO = 3x + 7 , OP = 5x – 19, 13. The value of X is _________. 14. The length of SO is _______. S O P 15. The length of OP is _______. 16. The length of SP is _______. AD ⊥ AG . Using the given in the figure, find D 17. x E 18. m∠DAE 3x x+3 19. m∠EAF F 20. m∠FAG x - 5 21. m∠DAF A G 21. m∠EAG In the figure, ∠STM is a right angle. S The ratio of m∠STL to m∠ LTM is 2:1. Find: L 23. m∠STL 24. m∠MTL T M

5

W 2 Y AX ⊥ AY , AW ⊥ AZ X 1 3 Z 25. What conclusion can you draw on A R W ∠ 1 and ∠ 3? 26. If m∠ROM = 2x + 22 and O m∠NOW = 3x – 12 , show that MW ⊥ RN M N 27. In the given figure, solve for y. 7y – 71 28. Given: AE bisects ∠DAS AT bisects ∠SAM T S E 2 3 1 4 D A M Prove: AE ⊥ AT Proof: Statement Reason 1. AE bisects ∠DAS 1. Given AT bisects ∠SAM 2. ∠ 1 ≅ ∠ 2, ∠ 3 ≅ ∠ 4 2. Definition of ___________ 3. m∠ 1 = m∠ 2, m∠ 3 = m∠ 4 3. Definition of congruent angles 4. m∠DAS + m∠SAM = 180 4. Definition of _____________ 5. m∠DAS = m∠ 1 + m∠ 2 5. Angle ______ Postulate m∠SAM = m∠ 3 + m∠ 4 6. m∠ 1 + m∠ 2 + m∠ 3 + m∠ 4 = 180 6. ________ Property of Equality 7. m∠ 2 + m∠ 2 + m∠ 3 + m∠ 3 = 180 7. ________ Property of Equality 8. 2 m∠ 2 + 2 m∠ 3 = 180 8. Combining like terms 9. m∠ 2 + m∠ 3 = 90 9. Multiplication Property of Equality 10. m∠EAT = m∠ 2 + m∠ 3 10. Angle Addition Postulate 11. m∠EAT = 90 11. Substitution 12. ∠EAT is a right angle 12. Definition of a __________.

13. AE ⊥ AT 13. Definition of ___________

6

Lesson 2

Exterior Angle of Triangle and Triangle Inequality Given ∆ABC. If you extend side AC B through point D, then there is a new angle formed , ∠BCD. ∠BCD is both adjacent and supplementary to one of the angles of ∆ABC. ∠BCD is called an exterior angle of ∆ABC . Two angles, ∠A and ∠B are A C D interior angles of ∆ABC which are not adjacent to ∠BCD. Therefore, ∠A and ∠B are called remote interior angles of the exterior angle. ∠BCD is adjacent to ∠BCA. So ∠BCA is called the adjacent interior angle. An exterior angle of a triangle is an angle which is adjacent and supplementary to one of the angles of a triangle. The remote interior angles are angles of a triangle which are not adjacent to the given exterior angle of the triangle. Adjacent interior angle is an interior angle which forms a linear pair with the given exterior angle. Illustration: Given ∆RST. S ∠STP is an exterior angle. ∠STP and ∠STR are adjacent and supplementary. ∠S and ∠R are the two remote interior angles. R T P ∠STR is the adjacent interior angle to ∠STP. Exterior Angle Equality Theorem The measure of an exterior angle of a triangle is equal to the sum of its two remote interior angles. In ∆CEF, ∠FED is an exterior angle F Conclusions: Since ∠C and ∠F are the two remote interior angles. Therefore, m∠ FED = m∠C + m∠F C E D Example: R If in the given triangle ∆MRE, m∠M = 31, m∠R = 34 then, m∠RED = m∠M + m∠R m∠RED = 31 + 34 m∠RED = 65 M E D and m∠REM = 180 – 65 m∠REM = 115

7

Since in the latter part of the lesson, you will be dealing with inequalities, it is necessary that you recall some properties of inequality which are of great help to you in proving statements in Geometry.

Let a, b, x and y be real numbers. Trichotomy Property. Exactly one of the following is true: a < b; a = b; or a > b Addition Property of Inequality If a > b, then a + x > b + x Subtraction Property of Inequality If a > b, then a – x > b – x and x – b > x – a Multiplication Property of Inequality If a > b and y > 0, then ay > by If a > b and y < 0, then ay < by Transitive Property of Inequality If a > b and b > c, then a > c. This properties of inequality can be applied to geometric figures since the measures of angles and segments are real numbers.

Illustrations: For segments: AB > CD if and only if AB > CD AB < CD if and only if AB < CD For angles: ∠A > ∠B if and only if m∠A > m∠B ∠A < ∠B if and only if m∠A < m∠B In Geometry, whenever possible, every statement is supported by reasons. In this lesson, you will find theorems are given but are not proven formally. Instead, illustrations are provided for easier understanding. You may opt to prove them though but you are allowed to use these theorems once you have gone through the illustrations and examples.

Theorem: The whole is greater than any of its parts. The meaning of this theorem is very clear to see.

Examples: 1. 10 = 7 + 3 Therefore: 10 > 7 or 10 > 3

8

2. Given PR with point Q in between P and R. By definition of betweenness, PR = PQ + QR P Q R Therefore: PR > PQ PR > QR A C 3. Given ∠X and C is in the interior of ∠X By the Angle Addition Postulate, B

m∠X = m∠AXC + m∠CXB X

Therefore by using the theorem, m∠X > m∠AXC m∠ x > m∠CXB Theorem: Exterior Angle Inequality Theorem. The measure of an exterior angle of a triangle is greater than the measure of either of the two remote interior angles. Illustration: In the figure, ∠BAG is an B exterior angle of ∆ABC. ∠B and ∠C are the two Remote interior angles Therefore: ∠BAG > ∠C C A G ∠BAG > ∠B Examples: 1. If m∠ BAG = 113, then m∠C < 113 and m∠B < 113 But m∠C + m∠B = 113. 2. If m∠C = 28, then m∠BAG > 28. Theorem: Triangle Inequality Theorem. In any triangle, the sum of the lengths of any two of its sides is greater than the length of its third side. C Illustration: In ∆ABC, the following side inequalities hold

a. AB + BC > AC b. AC + BC > AB c. AC + AB > BC A B

Examples: Determine if the following segments whose given measures are sides of a triangle.

1. 5 cm, 3 cm, 4 cm 2. 10 cm, 11 cm, 6 cm 3. 3 cm, 2 cm, 1 cm

9

Solution: 1. 5 + 3 > 4 3 + 4 > 5 5 + 4 > 3 Conclusion: Since the sum of any two sides is greater than the third side, then 5cm, 3cm and 4cm are measures of the sides of a triangle. 2. 10 + 11 > 6 11 + 6 > 10 10 + 6 > 11 Conclusion: Since the sum of any two sides is greater than the third side, then 10 cm, 11 cm and 6cm are measures of the sides of a triangle. 3. 3 + 2 > 1 3 + 1 > 2 2 + 1 = 3 Conclusion: Since one of the sum is not greater than the third side then, 3cm, 2cm and 1 cm are not measures of sides of a triangle.

4. If in the given triangle, the lengths of the two sides are given, what is the range of the length of the third side, PQ ?

a. 5 + 8 > PQ 13 > PQ

b. 5 + PQ > 8 PQ > 8 – 5 PQ > 3

c. 8 + PQ > 5 PQ > 5 – 8 PQ > - 3 (This cannot be since the side of triangle is always positive)

d. Combining statements a and b into a single statement will give you 13 > PQ and PQ > 3 which is written as 13 > PQ > 3.

Reversing the order, the statement can be written as 3 < PQ < 13. So the range of the third side is 3 < PQ < 13. If you have noticed, that 3 is the difference between the lengths of the two given sides, while 13 is the sum of the lengths of the two given sides.

The Pythagorean Theorem states that “In any right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two legs.”

10

Illustration: E Given: DEF∆ s a right triangle. FE is the length of the hypotenuse DE and FD are the lengths of the legs Conclusion: DE2 + FD2 = FE2 D F

Examples: 1. Given a right triangle, find x Solution: c2 = a2 + b2 x x2 = 32 + 42 3 x2 = 9 + 16 x2 = 25 x = 25 4 x = 5

2. If c = 13, a = 12, find b. Solution: c2 = a2 + b2 c 132 = 122 + b2 b 169 = 144 + b2 b2 = 169 – 144 a b2 = 25 b = 25 b = 5 3. If c = 10, b = 6, find a. c2 = a2 + b2

102 = a2 + 62 100 = a2 + 36 a2 = 100 - 36 a2 = 64 a = 64 a = 8 4. The side of a square is 3 cm. Find the length of its diagonal. Solution:

x2 = 32 + 32 x2 = 9 + 9 x2 = 18 3 x = 18

x = 3 2 3

x

11

In an isosceles right triangle like example no. 4, the legs are of equal lengths. Thus in order to find the length of the hypotenuse, you simply apply the Pythagorean formula. In an isosceles right triangle, the length of the hypotenuse is equal to the length of the leg multiplied by 2 . Try this out A. Justify each statement by stating the property, theorem, postulate, or definition that supports the statement. 1. If a < b, then 3a < 3b. 2. If m∠A = 65 and m∠D > m∠A, then m∠D > 65. 3. A is the midpoint of XY . Therefore, AX = AY. 4. If x and y are the lengths of the legs of a right triangle and z is the length of the hypotenuse, then x2 + y2 = z2. 5. ∠ 1 is an exterior angle of PQR∆ . R m∠ 1 > m∠P 1 P Q B. Supply the missing statement 6. If m∠ x = m∠ 1 + m∠ 2, then m∠ 1 _______ m∠ x 7. If AM bisects ∠DAY, then ∠DAM _____ ∠MAY Using the figure at the right, N 8. If m∠A = 31, then m∠ LEN ____ 31. 9. If m∠N = 43 and m∠A = 39, m∠ LEN = _______. A 10. ∠ LEN and ∠NEA are ___________ and ___________. L E C. Determine if the following are lengths of the sides of a triangle. 11. 5, 5, 6 12. 3, 4, 5 13. 3, 3, 2 14. 1, 1, 2 15. 6, 8, 10 16. 7.5, 6.5, 14 D. Given the lengths of the two sides of a triangle, determine the range of the length of the third side. 17. 13, 10 18. 7, 11 19. 5, 5 20. 3.25, 6.1 21. 4.73, 8.92

12

A 11 N 22. Given the lengths of the four sides of quadrilateral KANE, determine the 6

range of possible lengths of diagonals 9

AE and KN . E K 14

Y F. Given rt. XYZ∆ , compute for the missing side using the table below. X Z

XZ YZ XY 23. 7 5 24 12 13 25. 6 12

G. Find the length of the diagonal of a rectangle given the measure of a side or sides. 26. 3, 5 27. 7 28. 10 , 15 29. 3 2 30. What is the length of the hypotenuse of an isosceles right triangle if one of the leg measures 13 cm?

Lesson 3

Introduction to Parallel Lines and Transversals a Lines may be classified as: a. coplanar and intersecting like lines a and b b b. coplanar but not intersecting m like lines m and n n k c. Non-coplanar and non-intersecting like lines k and l I

13

Illustration a represents intersecting lines. The second illustration represents parallel lines and the third illustration showed skew lines.

Parallel lines are coplanar lines that do not intersect. Skew lines are non-coplanar

lines. t Given coplanar lines a and b. A third a X line t intersects lines a and b at two distinct points X and Y. We call line t a transversal. A transversal is a line that intersect s two or b more lines at two or more distinct points. Y Try this out

Determine if the following statements define intersecting, parallel or skew lines. 1. The two frames of jalousie windows 2. Electric wires near the post 3. The iron base on the railroad tracks 4. The two flyovers at Nagtahan

Write true if the statement is always true, sometimes if the statement is sometimes true

and false if it is never true. 5. Intersecting lines are coplanar. 6. Skew lines are non-coplanar. 7. Two parallel lines determine .a plane 8. Two non-intersecting lines are parallel. 9. Transversal intersect only two lines at a time. 10. Parallel lines are non coplanar.

Let’s summarize Two lines are perpendicular if and only if they intersect to form right angles.

The perpendicular bisector of a segment is a line, ray, segment or plane that is perpendicular to the segment at its midpoint.

Any two right angles are congruent.

An exterior angle of a triangle is an angle, which is adjacent and supplementary to one of the angles of a triangle. The remote interior angles are angles of the triangle which are not adjacent to the given exterior angle.

14

The measure of an exterior angle is equal to the sum of the measures of the two remote interior angles.

Properties of Inequality a. Trichotomy property of Inequality b. Addition property of Inequality c. Subtraction property of Inequality d. Multiplication property of Inequality e. Transitive property of Inequality

The whole is greater than any of its parts. Exterior angle Inequality Theorem. The measure of an exterior angle of a triangle is

greater than the measure of either of the two remote interior angles. Triangle Inequality Theorem. In any triangle, the sum of the lengths of any two of its sides is greater than the length of its third side. Pythagorean theorem. In any right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two legs. Parallel lines are coplanar lines that do not intersect. Skew lines are non-coplanar lines. Transversal is a line that intersects two or more lines at two or more distinct points.


Answer the following as indicated: R 1. ∠A is a right angle, ∠X is a right angle. ∠A _____∠X 2. Using the figure at the right, ∠RYP_____ ∠R. Y P T 3. AB ⊥ BC . BD bisects ∠B. What is m∠ 1? A D 1

2

4. What property of inequality supports the B C Following statement. If x < 7, then 5x < 35. T 5. ∆BET is a right triangle. m∠B < _________. 6. In ∆BET, if BE = 5, ET = 3, what is BT?

B E

15

A 7. Using the measures given in ∆MAN, give 4 9

the range of values or length of MN . M N 8. Prison bars are examples of ________ lines D 9. If m∠D = 23, and m∠ y = 19 , then m∠DAP = ___________. Y

P A

10. Find the perimeter of a square whose diagonal is 7 2 cm long.

16


1. right 2. 65 3. an exterior 4. 180 5. transitive property of Inequality 6. 3 < WX < 13 7. ≅ 8. 5 cm 9. parallel lines 10. transversal


1. right 2. 90 3. right 4. 90 5. Any two right angles are congruent 6. Angle Addition Postulate 7. Angle Addition Postulate 8. Transitive Property of Equality 9. Reflexive Property of Equality 10. Subtraction Property of Equality 11. Definition of Congruent Angles 12. OP 13. 13 14. 46 15. 46 16. 92 17. 23 18. 46 19. 26 20. 18 21. 72 22. 44 23. 60 24. 30 25. ∠ 1 ≅ ∠ 3 26. m∠ ROM = m∠ NOW

2x + 22 = 3x – 12 - x = - 34 x = 34

17

m∠ ROM = 2(34) + 22 m∠ ROM = 68 + 22 m∠ ROM = 90 m∠ NOW = 3(34) – 12 m∠ NOW = 102 – 12 m∠ NOW = 90 Both m∠ ROM and m∠ NOW are 90 each So ∠ ROM and ∠ NOW are right angles Therefore MW ⊥ RN

27. 7y – 71 = 90 7y = 90 + 71 7y = 161 y = 23

28. 2. angle bisector 4. supplementary angles or linear pair postulate 5. Addition 6. Addition 7. Transitive 11. Right angle 12. Perpendicular lines Lesson 2 A.

1. Multiplication Property of Inequality 2. Transitive Property of Inequality 3. Definition of Midpoint 4. Pythagorean Theorem 5. Exterior Angle Inequality Theorem 6. < 7. ≅ 8. > 9. 82 10. Adjacent and supplementary 11. Yes 12. Yes 13. Yes 14. No 15. Yes 16. No 17. 3 < third side < 23 18. 4 < third side < 18 19. 0 < third side < 10 20. 2.85 < third side < 9.35 21. 4.19 < third side < 13.65 22. a. 5 < AE < 17

b. 8 < KN < 20

18

23. 74 24. 5 25. 6 3 26. 34 27. 7 2 28. 5 29. 6 30. 13 2 cm Lesson 3

1. Parallel 2. Intersecting 3. Parallel 4. Skew 5. True 6. True 7. True 8. Sometimes 9. Sometimes 10. False


1. ≅ 2. > 3. 45 4. Multiplication Property of Inequality 5. 90 6. BT = 34 7. 5 < MN < 13 8. parallel 9. 42 10. 7(4) = 28 cm.

Module 3 Geometric Relations

What this module is about This module is about the angles formed by parallel lines (//) cut by a transversal. You will learn to determine the relation between pairs of angles formed by parallel lines cut by a transversal and solve problems involving segments and angles.

What you are expected to learn This module is designed for you to:

1. identify the angles formed by parallel lines cut by a transversal. 2. determine the relationship between pairs of angles formed by parallel

lines cut by a transversal:

• alternate interior angles • alternate exterior angles • corresponding angles • angles on the same side of the transversal

3. solve problems using the definition and properties involving relationships between segments and between angles.

How much do you know The figure below shows lines m // n with t as transversal. Name:

1. 4 pairs of corresponding angles

2. 2 pairs of alternate interior angles

3. 2 pairs of alternate exterior angles

m

n87

56

4 3

1 2

t

2

4. 2 pairs of interior angles on the same side of the transversal

5. 2 pairs of exterior angles on the same side of the transversal Using the same figure:

6. Name all numbered angles congruent to ∠ 7.


8. Name all numbered angles supplementary to ∠ 8, to∠ 7.

9. Name all numbered angles supplementary to ∠ 3, to ∠ 4.

10. Name the pairs of equal angles and supplementary angles in the figure.

Given: AB // CD AD // BC

In the figure below, AB ⊥ BD,

DF ⊥ BD,

BC // DE

11. Is m ∠ 2 = m ∠ 3? Why?

12. ∠ 2 is a complement of _____ and ∠ 3 is a complement of ____.

13. Is m ∠ 1 + m ∠ 2 = m ∠ 3 + m ∠ 4? Why?

A

B

C

D

A

B C

E D

F

1 2

3 4

3

Find the value of x in each of the following figure:

14.

15.

16.

17.

70o

55o xo

140o

110o

xo

3x

xo

4x

xo

4

In the figure, write down the pairs of parallel lines and the pairs of congruent angles.

18.

19. If m∠ 3 = 135, find the measure of each angle in the figure.

20. If m∠ 6 = 85, find the measure of each numbered angle in the figure, a // b and c // d.

BCA

D FE

G

m

n8 7

5 6

4 31 2

t

m

n

8 7

5 6

4 3 1 2

c

1110

129

1514

1613

d

5

What you will do

Lesson 1

Angles Formed by Parallel Lines Cut by a Transversal In the rectangular solid below, AB and CD are coplanar in plane x. AB and EF are coplanar in plane y. EF and HF are coplanar in plane z. Line E intersect AB and CD at two different points. Line E is a transversal of lines AB and CD. Definitions: Coplanar lines are lines that lie in one plane. Parallel lines are two lines that are coplanar and do not intersect. Transversal is a line that intersects two or more lines at different points.

Line E intersect AB and CD at 2 different points. Line E is a transversal of lines AB and CD.

A E

G C

D H

B F

x

z y

A

B E

D C

6

Examples:

1. In the figure, lines AB and CD are parallel lines cut by transversal line t. The angles formed are: Angles 1, 2, 7 and 8 are exterior angles Angles 3, 4, 5, and 6 are interior angles The pairs of corresponding angles are: ∠ 1 and ∠ 5 ∠ 2 and ∠ 6

∠ 4 and ∠ 8 ∠ 3 and ∠ 7

The pairs of alternate interior angles are: ∠ 3 and ∠ 5 ∠ 4 and ∠ 6 The pairs of alternate exterior angles are: ∠ 1 and ∠ 7 ∠ 2 and ∠ 8 The pairs of exterior angles on the same side of a transversal (SST) are: ∠ 1 and ∠ 8 ∠ 2 and ∠ 7 The pairs of interior angles on the same side of a transversal (SST) are: ∠ 4 and ∠ 5 ∠ 3 and ∠ 6

2. Given: m // n, s is the transversal. The pairs of angles formed are:

a. Corresponding angles: ∠ 3 and ∠ 9 ∠ 6 and ∠ 12 ∠ 4 and ∠ 10 ∠ 5 and ∠ 11

B

8 75 6

4 3 1 2

t

A

C D

m

14 119 10

6 5 3 4

s

n

7

b. Alternate exterior angles

∠ 3 and ∠ 11 ∠ 4 and ∠ 12

c. Alternate interior angles

∠ 6 and ∠ 10

∠ 5 and ∠ 9

d. Interior angles on the same side of the transversal (SST).

∠ 6 and ∠ 9 ∠ 5 and ∠ 10

e. Exterior angles on the same side of the transversal (SST).

∠ 3 and ∠ 12 ∠ 4 and ∠ 11

3. m // n, t is the transversal. In the figure, name and identify he pairs of angles formed:

a. Corresponding angles:

∠ 1 and ∠ 9 ∠ 8 and ∠ 12 ∠ 2 and ∠ 10 ∠ 7 and ∠ 11

b. Alternate interior angles

∠ 8 and ∠ 10 ∠ 7 and ∠ 9

12 119 10

781 2

t

m

n

8

c. Alternate exterior angles

∠ 1 and ∠ 11 ∠ 2 and ∠ 12

d. Exterior angles on the same side of the transversal (SST).

∠ 1 and ∠ 12 ∠ 2 and ∠ 11

e. Interior angles on the same side of the transversal (SST).

∠ 8 and ∠ 9 ∠ 7 and ∠ 10

Try this out

1. In the figure, lines g // h and is cut by line k. Name and identify the pairs of angles formed 2. In the figure, q // r and s // t. Name and identify the pairs of angles formed.

g

7 68 5

3 24 1

k

h

q

r

16151314

11 1210 9

s

58

67

14

23

t

9

3. In the figure, u // w. Lines x and y are transversals.

Name and identify the pairs of angles formed

4. In the figure, identify and name the pairs of parallel lines and its transversal.

5. In the figure, identify and name the pairs of parallel lines and its

transversal / transversals.

9

u7

6 8

532

41

y

w141011 13

12x

B

F

A

C DG

E

c

d

a

e

b

10

Lesson 2

Relationship Between Pairs of Angles Formed by Parallel Lines Cut by a Transversal

If two lines are cut by a transversal, then:

a. corresponding angles are congruent

b. alternate interior angles are congruent

c. alternate exterior angles are congruent

d. interior angles on the same side of a transversal are supplementary

e. exterior angles on the same side of a transversal are supplementary Examples:

1. Given: p // q, r is a transversal

Figure: What is the measure of each numbered angles if m∠ 1 = 120? Give the reason for your answer.

Answers: If ∠ 1 = 120o, ∠ 5 = 120o Corresponding angles are ≅ . If ∠ 5 = 120o, ∠ 3 = 120o Alternate interior angles are ≅ If ∠ 3 = 120o, ∠ 7 = 120o Corresponding angles are ≅ If ∠ 7 = 120o, ∠ 4 = 60o Exterior angles on the same side of a transversal are supplementary If ∠ 4 = 60o, ∠ 8 = 60o Corresponding angles are ≅ If ∠ 8 = 80o, ∠ 2 = 60o Alternate interior angles are ≅ If ∠ 2 = 60o, ∠ 6 = 60o Corresponding angles are ≅

p

q

r

1 42 3

5 86 7

11

2. Given: In the figure, if m∠ 1 = 105, determine the measures of the other numbered angles. Justify your answers.

Figure: Answers:

If ∠ 1 = 105o, ∠ 5 = 105o Corresponding angles are ≅ . If ∠ 5 = 105o, ∠ 3 = 105o Alternate interior angles are ≅ If ∠ 3 = 105o, ∠ 7 = 105o Corresponding angles are ≅ If ∠ 7 = 105o, ∠ 2 = 75o Exterior angles on the same side of a transversal are supplementary If ∠ 2 = 75o, ∠ 8 = 75o Alternate exterior angles are ≅ If ∠ 8 = 75o, ∠ 4 = 75o Corresponding angles are ≅ If ∠ 4 = 75o, ∠ 6 = 75o Alternate interior angles are ≅

3. Given: m // n, s and t are the transversals If m∠ 5 = 110 and m∠ 12 = 90, determine the measures of the other numbered ∠ s. Justify your answer. Answers: If m∠ 5 = 110, m∠ 12 + m∠ 13 = 110 Since m∠ 12 = 90, m∠ 13 = 20. Corresponding angles are ≅ m∠ 4 = 70 ∠ 4 and ∠ 5 are supplementary If m∠ 4 = 70, m∠ 11 = 70 Corresponding angles are ≅

m∠ 3 = 110 ∠ 3 and ∠ 5 are vertical angles

B

D

t

14

23

58

67

A

C

m7 6 8 5

3 2 4 1

s

n14

1011

1312

t

9

12

If m∠ 3 = 110, m∠ 9 + m∠ 10 = 110 Corresponding angles are ≅ If m∠ 9 = 90, m∠ 7 = 90 Alternate interior angles are ≅ If m∠ 12 = 90, m∠ 2 = 90 Exterior angles on the same side of a transversal are supplementary Since m∠ 9 = 90, m∠ 1 = 90 Corresponding angles are ≅ If m∠ 13 + m∠ 14 = m∠ 8, m∠ 8 = 90 Corresponding angles are ≅ Therefore, the measures of the numbered angles are:

∠ 1 = 90o ∠ 8 = 90o

∠ 2 = 90o ∠ 9 = 90o

∠ 3 = 110o ∠ 10 = 20o

∠ 4= 70o ∠ 11 = 70o

∠ 5 = 110o ∠ 12 = 90o

∠ 6 = 70o ∠ 13 = 20o

∠ 7 = 90o ∠ 14 = 70o

Try this out

1. If m∠ 6 = 85, find the measure of the other numbered angles. Justify your answers. Given: a // b, c is the transversal Figure:

ba

c14

23

58

67

13

2. If m∠ 10 = 118 and m∠ 4 = 85, find the measures of the other numbered angles. Justify your answers.

Given: f // g, r // q

3. In the figure, u // w. Lines x and y are transversals. If m∠ 8 = 62, m∠ 14 = 90, find the measures of the other numbered ∠ s. Justify your answers.

Figure:

Form an equation in x and solve the equation.

4.

f

g 16 1513 14

11 12109

r

58

67

14

23

q

u7

68

532

41

y

w141011 13

12x

9

2xo

xo

14

5.

Let’ summarize

The angles formed by parallel lines cut by a transversal are:

1. corresponding angles

2. alternate interior angles

3. alternate exterior angles

4. exterior angles on the same side of a transversal

5. interior angles on the same side of a transversal

The relationship of the angles formed by parallel lines cut by a transversal are: 1. pairs of corresponding s are ≅

2. pairs of alternate interior angles are ≅

3. pairs of alternate exterior angles are ≅

4. pairs of interior angles on the same side of a transversal are supplementary

5. pairs of exterior angles on the same side of a transversal are supplementary

120o

100o

xo

15


The figure below shows lines m // n with t as transversal. Figure: Name:

1. 4 pairs of corresponding angles.

2. 2 pairs of alternate interior angles.

3. 2 pairs of alternate exterior angles.

4. 2 pairs of interior angles on the same side of a transversal.

5. 2 pairs of exterior angles on the same side of a transversal. Using the same figure:



8. Name all numbered angles supplementary to ∠ 8, ∠ 7.

9. Name all numbered angles supplementary to ∠ 3, ∠ 4.

10. Name the pairs of equal angles and supplementary angles in the figure.

Given: AB // CD Figure: AD // BC

m

n

t

14

23

58

67

A B

CD

16

11. – 13. In the figure, AB ⊥ BD, DF ⊥ BD, BC // DE

Figure:

11. Is m∠ 2 = m∠ 3? Why?

12. ∠ 2 is a complement of _____ and ∠ 3 is a complement of _____.

13. Is m∠ 1 + m∠ 2 = m∠ 3 + m∠ 4? Why? Find the value of x in each of the following figures.

14.

15.

B

C

A

D1

EF

23

4

120o

100o

xo

50o

xo xo

17

16.

17. In the figure, Write down the pairs of parallel lines and the pairs of congruent angles.

18.

5x

xo

4x

xo

AB

CD F

E

18

19. If m∠ 1 = 135, find the measure of each angle in the figure below:

20. If m∠ 6 = 75, find the measure of each numbered angle in the figure, a // b and c // d.

m

n

t

14

23

58

67

a

b 16 1513 14

11 12109

c

5 6

8 7

1 4

2 3

d

19

Answer key How much do you know 1. ∠ 8 and ∠ 4, ∠ 7 and ∠ 3

∠ 5 and ∠ 1, ∠ 6 and ∠ 2

2. ∠ 5 and ∠ 3, ∠ 4 and ∠ 6

3. ∠ 8 and ∠ 2, ∠ 7 and ∠ 1

4. ∠ 5 and ∠ 4, ∠ 6 and ∠ 3

5. ∠ 8 and ∠ 1, ∠ 7 and ∠ 2

6. ∠ 3, ∠ 1

7. ∠ 8, ∠ 6

8. To ∠ 8: ∠ 1, ∠ 5, ∠ 7; To ∠ 7: ∠ 2, ∠ 8, ∠ 6

9. To ∠ 3: ∠ 6, ∠ 4, ∠ 2; To ∠ 4: ∠ 5, ∠ 3, ∠ 1

10. ∠A ≅ ∠C, ∠B ≅ ∠D, ∠A supplement ∠B, ∠B supplement ∠C

∠C supplement ∠D, ∠D supplement ∠A

11. They are alternate interior ∠ s

12. ∠ 1, ∠ 4

13. Yes, because they are right angles.

14. x = 55o

15. x = 110o

16. x = 45o

17. x = 36o

18. AB // DE; CD // FG; ∠ACD ≅ ∠EDC; ∠EFG ≅ ∠CDF

19. ∠ 2 = 45o; ∠ 4 = 45o; ∠ 1 = 135o; ∠ 7 = 135o; ∠ 8 = 45o; ∠ 6 = 45o; ∠ 5

= 135o

20. ∠ 4 = 85o;∠ 5 = 95o; ∠ 1 = 95o;∠ 7 = 95o;∠ 3 = 95o; ∠ 8 = 85o;∠ 2 = 85o

∠ 10 = 95o; ∠ 9 = 85o; ∠ 12 = 95o; ∠ 11 = 85o; ∠ 13 = 85o; ∠ 14 = 95o;

∠ 16 = 95o; ∠ 15 = 85o

20

Try this out Lesson 1 1. ∠ 4 and ∠ 8; ∠ 3 and ∠ 7; ∠ 1 and ∠ 5; ∠ 2 and ∠ 6 are corresponding angles;

∠ 4 and ∠ 7; ∠ 1 and ∠ 6 are exterior angles on SST; ∠ 3 and ∠ 5; ∠ 2 and ∠ 8 are alternate interior angles; ∠ 4 and ∠ 7; ∠ 1 and ∠ 6 are exterior angles on SST; ∠ 3 and ∠ 5; ∠ 2 and ∠ 8 are alternate interior angles; ∠ 3 and ∠ 8; ∠ 2 and ∠ 5 are interior angles on SST; ∠ 4 and ∠ 6; ∠ 1 and ∠ 8 are exterior angles on SST;

2. ∠ 1 and ∠ 5; ∠ 4 and ∠ 8; ∠ 2 and ∠ 6; ∠ 3 and ∠ 7 are corresponding

angles; ∠ 4 and ∠ 6; ∠ 3 and ∠ 5 are alternate interior angles; ∠ 7and ∠ 2; ∠ 8 and ∠ 1 are exterior angles on SST; ∠ 6 and ∠ 3; ∠ 5 and ∠ 4 are interior angles on SST; ∠ 5 and ∠ 13; ∠ 6 and ∠ 14; ∠ 7 and ∠ 15; ∠ 8 and ∠ 16 are corresponding angles; ∠ 8 and ∠ 15; ∠ 5 and ∠ 14 are exterior angles on SST; ∠ 7and ∠ 16; ∠ 6 and ∠ 13 are interior angles on SST; ∠ 16 and ∠ 6; ∠ 13 and ∠ 7 are alternate interior angles; ∠ 15 and ∠ 5; ∠ 14 and ∠ 8 are alternate exterior angles; ∠ 15 and ∠ 11; ∠ 14 and ∠ 10; ∠ 16 and ∠ 12; ∠ 13 and ∠ 9 are corresponding angles; ∠ 14 and ∠ 11; ∠ 13 and ∠ 12 are interior angles on SST; ∠ 15and ∠ 10; ∠ 16 and ∠ 9 are exterior angles on SST; ∠ 4 and ∠ 12; ∠ 13 and ∠ 11 are alternate interior angles; ∠ 15 and ∠ 9; ∠ 16 and ∠ 10 are alternate exterior angles; ∠ 3 and ∠ 11; ∠ 4 and ∠ 12; ∠ 2 and ∠ 10; ∠ 1 and ∠ 9 are corresponding angles; ∠ 12 and ∠ 2; ∠ 3 and ∠ 9 are alternate interior angles; ∠ 11 and ∠ 1; ∠ 4 and ∠ 10 are alternate exterior angles;

21

∠ 1and ∠ 10; ∠ 4 and ∠ 11 are exterior angles on SST; ∠ 3 and ∠ 12; ∠ 2 and ∠ 9 are interior angles on SST;

3. ∠ 1 and ∠ 9 + ∠ 10; ∠ 4 and ∠ 11 are corresponding angles;

∠ 2 and ∠ 14; ∠ 3 and ∠ 12 + ∠ 13 are corresponding angles; ∠ 4 and ∠ 14; ∠ 3 and ∠ 9 + ∠ 10 are alternate interior angles; ∠ 1 and ∠ 12 +∠ 13; ∠ 2 and ∠ 11 are alternate exterior angles; ∠ 1 and ∠ 11; ∠ 2 and ∠ 12 + ∠ 13 are exterior angles on SST; ∠ 4 and ∠ 9 +∠ 10; ∠ 3 and ∠ 14 are interior angles on SST; ∠ 6 and ∠ 10; ∠ 7 and ∠ 11; ∠ 5 and ∠ 9 + ∠ 14; ∠ 8 and ∠ 13 are corresponding angles; ∠ 7 and ∠ 9 + ∠ 14; ∠ 8 and ∠ 10 are alternate interior angles; ∠ 6 and ∠ 13; ∠ 5 and ∠ 11 + ∠ 12 are alternate exterior angles; ∠ 6and ∠ 11 + ∠ 12; ∠ 5 and ∠ 13 are exterior angles on SST; ∠ 7 and ∠ 10; ∠ 8 and ∠ 9 + ∠ 14 are interior angles on SST;

4. AB // CD with BC as transversal

CD // FE with DE as transversal

DE // FG with FE as transversal

5. a // b, e, c and d are the transversals

6. c // d, a, b and c are the transversals

Lesson 2 1. ∠ 2 = 85o; ∠ 5 = 95o; ∠ 1 = 95o; ∠ 7 = 95o; ∠ 3 = 95o; ∠ 8 = 85o;

∠ 4 = 85o

2. ∠ 12 = 118o; ∠ 9 = 62o; ∠ 11 = 62o; ∠ 16 = 118o; ∠ 13 = 62o;

∠ 14 = 118o; ∠ 15 = 62o

∠ 8 = 85o; ∠ 6 = 85o; ∠ 2 = 85o; ∠ 1 = 95o; ∠ 5 = 95o; ∠ 3 = 95o; ∠ 1 = 95o

3. ∠ 8 = 62o; ∠ 6 = 62o; ∠ 10 = 62o; ∠ 13 = 118o; ∠ 9 = 28o;

∠ 12 = 28o; ∠ 11 = 90o; ∠ 13 = 62o; ∠ 7 = 118o; ∠ 5 = 118o; ∠ 1 = 90o;

∠ 2 = 90o ; ∠ 3 = 90o; ∠ 4 = 90o

22

4. x = 60o

5. x = 140o What have you learned 1. ∠ 1 and ∠ 5, ∠ 4 and ∠ 8, ∠ 2 and ∠ 6, ∠ 3 and ∠ 7

2. ∠ 4 and ∠ 6, ∠ 3 and ∠ 5

3. ∠ 1 and ∠ 7, ∠ 2 and ∠ 8

4. ∠ 4 and ∠ 5, ∠ 3 and ∠ 6

5. ∠ 1 and ∠ 8, ∠ 2 and ∠ 7

6. ∠ 3, ∠ 5, ∠ 1

7. ∠ 8, ∠ 6, ∠ 2

8. ∠ 8: ∠ 5, ∠ 7, ∠ 1; ∠ 7: ∠ 2, ∠ 6, ∠ 8

9. ∠ 3: ∠ 6, ∠ 4, ∠ 2; ∠ 4: ∠ 5, ∠ 3, ∠ 1

10. ∠A ≅ ∠C; ∠B ≅ ∠D; ∠A supplementary ∠B ∠B supplementary ∠C ∠C supplementary ∠D ∠A supplementary ∠D

11. Yes, they are alternate interior angles

12. ∠ 1, ∠ 4

13. Yes, because their sum is equal to 90o

14. x = 65o

15. x = 140o

16. x = 30o

17. x = 36o

18. BA // CF, BC // DE, ∠ABC ≅ ∠BCF; ∠BCD ≅ ∠CDE

19. ∠ 2 = 45o, ∠ 3 = 135o, ∠ 4 = 45o, ∠ 5 = 135o, ∠ 6 = 45o, ∠ 7 = 135o, ∠ 8

= 45o

20. ∠ 4 = 75o, ∠ 1 = 105o, ∠ 5 = 105o, ∠ 7 = 105o, ∠ 3 = 105o, ∠ 8 = 75o,

∠ 2 = 75o, ∠ 9 = 75o, ∠ 10 = 105o, ∠ 12 = 105o, ∠ 11 = 75o

Module 1 Similarity

What this module is about This module is about ratio, proportion, and the Basic Proportionality Theorem and its Converse. In this module, you will learn the meanings of ratio and proportion. And as you go over the exercises, you will develop skills that you will need to solve problems especially on triangles in the next module.


1. apply the fundamental law of proportions • product of the means is equal to the product of the extremes.

2. apply the definition of proportion of segments to find unknown lengths. 3. illustrate and verify the Basic Proportionality Theorem and its Converse.


1. Express each ratio in simplest form.

a. 1812 b.

128

2. Find the value of x in each proportion.

a. x2 =

63 b. 4:8 = 8:x

3. State the means and the extremes in each proportion.

a. 3:6 = 6:12 b. 3:4 = 6: 8

4. Write each ratio as a fraction in simplest form.

a. 6 to 30 b. 16 to 48

2

5. In the figure, points W, X, Y, and Z are collinear. WX = 4, XY = 6 , YZ = 10

● ● ● ●

W X Y Z

Give each ratio in simplest form

a. xywx

b. WX to (XY + YZ)

6. In ∆ ABC, CD = 2, AD = 4, CE = 3, EB = 5

C D E A B Find the ratio: a. CE to BC b. CD to CA 7. Two consecutive angles of a parallelogram are in the ratio 1:2.

a. What is the measure of the smaller angle?

b. What is the measure of the larger angle?

8. In ∆ABC, DE //AC B

D E A C

a. Given that BD = 4, DA = 6 and BE = 5, find EC.

b. Given that BE = 4, EC = 5 and BD = 2, find DA.

3

9. In the figure ∠ADE ≅ ∠C B E

A C D

a. If AE = 2, AB = 8, AD = 3, what is AC?

b. If AD = 4, AC = 6, AE = 3, find AB.

10. In triangle ABC, DE // AB. C D E A B If DC = 6, DA = 8, CE = 2x, EB = 2x + 4

a. What is x?

b. What is CB?

What you will do

Lesson 1

Ratio and Proportion A ratio is a comparison of two numbers. The ratio of two numbers a and b where b is not equal to zero can be written in three ways: a :b, a/b and a to b. A proportion is a statement of equality between two ratios. In the

proportion a: b = c:d or ba =

dc , a, b, c, and d are called the terms of the

proportion. In a proportion, the product of the extremes is equal to the product

4

of the means. In the proportion a: b = c:d , the extremes are a and d and the means are b and c, hence, ad = bc Example 1

Express the ratio 1612 in simplest form.

Solution: Divide the numerator and the denominator by their GCF (Greatest Common Factor), 4 . 12 ÷ 4 = 3 16 ÷ 4 4

The ratio in simplest form is 43 or 3:4.

Example 2 Find the ratio and express your answer in simplest form. 6 hours to 3 days Solution: Step 1. Convert 3 days to hours 3 x 24 = 72 ( Since there are 24 hours in one day) Step 2. Write the ratio in terms of hours 6 hours to 72 hours

hourshours

726 =

121

Another solution: Step 1. Convert 6 hours to days

6 ÷ 24 = 246

= 21 (Since there are 24 hours in one day)

This may be done as follows:

6 hours x hoursday

241 =

41 of a day

5

Step 2. Write the ratio in terms of days

days

dayaof

341

= 41÷3

= 41 x

31

= 121

Example 3

a. What is AB:BC? 3 5 7 b. What is (AB + BC) CD ● ● ● ● A B C D

Solution:

a. AB = 3 BC = 5 Hence AB :BC = 3:5

b. AB = 3

BC = 5 CD = 7 Hence (AB + BC) : CD = (3 + 5 ): 7 = 8:7

Example 4

1. a. What is AX:AB? A b. What is AY :YC? 4 2 X Y 7 3.5 B C

Solution a. AX = 4 AB = AX + X B = 4 + 7 = 11 Hence AX : AB = 4:11

6

b. AY = 2 YC = 3.5 Hence AY:YC = 2:3.5

3.5 can be written as 1053 or

213 or

27 .

Therefore, YCAY =

272

But

272 can be written as 2 ÷

27 which is equal to 2 x

27 or

74

Hence YCAY =

74 or AY:YC = 4:7

Example 5 State the means and the extremes in the following statement. 3:7 = 6:14 Solution: The means are 7 and 6 and the extremes are 3 and 14. Example 6 Determine whether each pair of ratios forms a proportion.

a. 54 ,

86

b. 74 ,

148

Solution:

a. A proportion is an equality of two ratios

54 =

86

4:5 = 6:8

7

The product of the means is equal to the product of the extremes 5(6) =4(8) 30 = 32 This is a false statement Hence the two ratios do not form a proportion.

b. A proportion is an equality of two ratios c.

74 =

148

4:7 = 8:14

The product of the means is equal to the product of the extremes 7(8) = 4(14)

56 = 56 This is a true statement

Hence the two ratios form a proportion. Example 7 Find the value of x. 3 = x 10 30 Solution: Step 1. Rewrite in the ratio 3:10 = x:30 Step 2. Find the products of the means and the extremes. Then solve for x. Remember, the product of the means is equal to the product of the extremes. 10(x) = 3(30) 10x = 90 x = 9 Example 8 The measures of two complementary angles are in the ratio 1:2. Find the measure of each angle.

8

Solution: Representation: Let x = the measure one angle 90 –x = the measure of its complement Proportion:

x

x−90

= 21

x : (90-x) = 1:2

1(90 –x) = 2(x)

90-x = 2x

-x –2x = - 90

-3x = - 90

x = 30 measure of one angle 90 – x = 60 measure of its complement Example 9 The measure of two supplementary angles are in the ratio 2:3. Find the measure of each angle. Solution: Representation: Let x = the measure of one angle 180 – x = the measure of its supplement Proportion:

x

x−180

= 32

x:(180-x) = 2:3

2(180-x) = 3(x)

360 – 2x = 3x

-2x –3x = -360

9

-5x = - 360

x = 72 the measure of one angle 180 – x = 108 the measure of its supplement Example 10 In ∆ABC below, m∠C = 90. The measures of ∠ A and ∠B are in the ratio 4:5. Find the measures of ∠ A and ∠B. B C A Solution: Representation: Let x = measure of ∠ A 90-x = measure of ∠B (Because, ∠A and ∠B are acute angles) Proportion: x:90-x = 4:5 4(90-x) = 5x 360-4x = 5x -4x – 5x = -360 -9x = -360 x = 40 measure of ∠A 90-x = 50 measure of ∠B Example 11 In the figure, ∠ABD and ∠CBD form a linear pair. If the measures of ∠ABD and ∠CBD are in the ratio 7 to 3, what is the measure of ∠CBD? D ●

● ● A B C

10

Solution: Let x be the measure of ∠CBD. 180-x be the measure of ∠ABD [Remember that if two angles form a linear pair they are supplementary] Proportion: m ∠ ABD = 7 m ∠ CBD 3 Substitute 180-x for m ∠ ABD and x for m ∠ CBD

xx−180 =

37

(180-x):x = 7:3

7x = 3(180-x)

7x = 540 – 3x

7x + 3x = 540

10 x = 540

x = 54 measure of ∠CBD.

180 – x = 126 measure of ∠ABD. Example 12

In the figure, EB ⊥ EH and ET is in the interior of ∠ BEH. If the B ● T● measures of ∠BET and ∠HET are in the ratio 1:5, what is the measure of ∠HET? ● E H Solution: Let x = measure of ∠ BET 90-x = measure of ∠ HET [Remember that perpendicular rays form a right angle and the measure of a right angle is 90.] Proportion: m∠ BET = 1 m∠HET 5

11

Substitute x for m∠BET and 90-x for m∠HET

x

x−90

= 51

x : (90-x) =1:5

5x = 1(90 – x)

5x = 90 – x

5x + x = 90

x = 15 measure of ∠BET

90-x = 75 measure of ∠HET

Try this out Set A Find the ratio of each of the following. Use the colon and write your answers in simplest form.

1. 14 cm to 28 cm 2. 8 hours to 12 hours 3. 16 days to 36 days 4. 42 dm to 63 dm 5. 12 inches to 36 inches 6. 24 feet to 36 feet 7. 12 inches to 24 inches 8. 6 inches to 2 feet 9. 3 feet to 24 inches 10. 8 hours to 1 day 11. 14 days to 2 weeks 12. 3 weeks to 24 days 13. 5 minutes to 60 seconds 14. 5 centimeters to 2 decimeters 15. 4 months to 1 year 16. 3 feet to 12 inches 17. 3 hours to 72 minutes 18. 2 weeks to 8 days 19. 2 centuries to 300 years 20. 200 years to 3 centuries

Set B. Find the missing number

12

1. 42 =

10n

2. n = 2 6 3

3. 3 = 5

n 10 4. 6 = 3

10 n 5. n = 2

7 5 6. 4 = n

8 7

7. 7 = 5 n 2.5

8. 3 = 4 3.6 n

9. 2n = 5

3 2 10. 6 = 3n

5 4

11. 3n = 6 7 14

12. 4 = 2

5n 10 13. 3 = n

2 7 14. 3n = 3 5 2 15. 8 = 2

5 n

13

16. 6n = 9

8 2

17. 5n = 6 9 4

18. 5 = 12

4 n 19. 15 = n 3 4

20. 6 = 2

n 5 Set C

1. The measures of two supplementary angles are in the ratio 1:2. Find the measure of each angle. 2. The measures of two supplementary angles are in the ratio 1: 4. Find the

measure of each angle 3. The measures of two complementary angles are in the ratio 2:3. Find the

measure of each angle. 4. The measures of two complementary angles are in the ratio 3:7. Find the

measure of each angle.

5. a. What is the ratio of AB to AC? b. What is AE:ED?

A 5 6 B E 5 6 C D

6. a. What is BE:CD? D

b. What is AB :BC? E 8 4 A C

14

3 B 5 7. a. What is AB:BC? 4 5 8

b. What is AB: (BC + CD) ● ● ● ● A B C D

8. a. What is yx

b. What is x + y ● ● ● ● y + z X Y Z

Determine whether each pair of ratios forms a proportion. 9. 6 , 7

7 8

10. 8, 16 9 18

11. 6, 18 9 27

12. 7 , 42 6 36

13. 11, 22 12 24 14. 13, 7

9 6

15. 22, 4 11 2 16. 14, 12 7 6 17. The acute angles in a right triangle are in the ratio 3: 6. Find the measure

of the larger of the two angles. 18. The acute angles in a right triangle are in the ratio 2 to seven. What is

the measure of the smaller of the two angles. 19. In the figure at the right, ∠1 and ∠2 form a linear pair. If the measures of ∠1 and ∠2 are in the ratio 2 to 8. Find the measures ∠1 and ∠2. 1 2

3 6 7

15

20. In the figure below, BA ⊥ BC. BD is in the interior of ∠ABC. If the measures of ∠ABD and ∠CBD are in the ratio 7 to 2. Find the measure of ∠ABD.

A

D B C

Lesson 2

The Basic Proportionality Theorem and Its Converse The Basic Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides in distinct points, then it divides the two sides proportionally. Illustration: A

E F B C In the figure, if EF is parallel to BC and intersects AB and AC at points E and F respectively, then a. AE = AF EB FC b. AE = AF AB AC c. EB = FC AB AC

16

The Converse of the Proportionality Theorem: If a line divides the two sides of a triangle proportionally, then the line is parallel to the third line. Illustration: A E F

B C

If in ∆ABC, AE = AF, then EF // BC. EB FC

You can verify the Basic Proportionality Theorem by doing the following activity. In this activity you need a ruler, a pencil and a protractor. 1. Draw ∆ABC such AB = 5 cm and AC = 10 cm. 2. Draw point E on side AB such that it is 2 cm away from vertex B. 3. Draw a line parallel to BC passing thru point E intersecting AC at point F.

Illustration:

A

E F ● ● 2

B C

4. Verify whether line EF is really parallel to BC by measuring ∠AEF and ∠ABC. Recall: If two lines are cut by a transversal and a pair of corresponding angles are congruent, then the lines are parallel.

5. Find the lengths of AE, AF and FC.

6. Is AE = AF ? EB FC Is AE = AF ? AB AC Is EB = FC ? AB AC

17

You can verify the Converse of the Basic Proportionality Theorem by doing the following activity.

1. Draw ∆ABC such that AB = 7 cm and AC = 14 cm. 2. Draw point E on side AB such that it is 3 cm away from vertex A. What is

the length of EB? 3. Draw point F on side AC such that it is 6 cm away from vertex A. What is

the length of FC? 4. Draw a line passing through points E and F. Notice that line EF divides

sides AB ad AC proportionally.

43 =

86

Illustration: A 3 6 E ● ● F 4 8

B C

5. What can you say about line EF? Is it parallel to BC? Verify by measuring ∠AEF and ∠ABC.

Example 1 In ∆ABC, DE // BC. If AD = 6, DB = 8 and EC = 12, find AE. A 6 x D E

8 12

B C Solution: Let x = AE x = 6 12 8 x = 3 12 4 x:12 = 3:4 12(3) = 4(x) 4x = 12(3) (by Symmetric Property of Equality)

18

4x = 36 x = 9

Example 2 A In ∆ACD, BE //CD If AB = 8, BC = 4, AE = 6, find AD and ED Solution: 8 6 B E Let x = AD AB = AE 4 AC AD C D 8 = 6 8 + 4 x 8:12 = 6:x 12(6) = 8(x) 8x = 72 x = 9 AD = 9 Hence: ED =AD – AE = x – 6 = 3 Example 3 A 2x 3x D E 12 B C In ∆ABC, DE // BC, AD =2x, AB = 20, AE = 3x and EC = 12. Find AD Solution: Step 1. AD = AE AB AC

202x =

1233+xx

10x =

1233+xx

19

x(3x + 12) = 10(3x) 3x2 + 12x = 30x 3x2 + 12x – 30x = 0 3x2 – 18x = 0 3x(x – 6) = 0 by factoring 3x = 0 x – 6 = 0 Equating both factors to 0. x = 0 x = 6 Step 2. Substitute 6 for x in 2x 2x = 2(6) = 12 Hence AD = 12 Try this out

Set A In the figure, DE // BC. A D E B C True or False 1. AD = AE 6. BD = AE DB EC AD EC 2. AD = AE 7. DB = AB . AB AC EC AC 3. AD = EC 8. AE = EC DB AE AD DB

4. AD = AC 9. AD = AE AB AE EC DB

5. DB = EC 10. AD = DB

AB AC AE EC

20

B In the figure, AB // ED. D 11. CD = CA BD AE 12. BC = AC CD AE A E C 13. AE = BC EC BD 14. AE = CD AC BD

15. CD = CB AB AC

Set B In ∆ABC , DE // BC. A D E B C 1 If AD = 1, DB = 2, and EC = 4, find AE. 2. If AD = 2, DB = 4, and AE = 1, find EC. 3. If BD = 6, AD = 4, and AE = 5, find EC. 4. If BD = 5, AD = 6, and CE = 9. what is AE?

5. If AD = 2, AB = 8, and AC = 10 what is AE? In ∆ ABC, DE // BC C E A D B

21

6. If AD = 4, AB = 12 and AC = 15, what is AE? 7. If AD = 6, DB = 2 and AC = 10 what is AE? 8. If AB = 12, AE = 4 and EC = 6, what is AD? 9. If AD = 2.5, DB = 4 and EC = 6, find AE. 10. If DB = 5, AE = 3.5 and EC = 7, what is AD?

In ∆CAT, AC // GD. A G C D T

11. If CD = 10, DT = 12, AG = 8, what is GT? 12. If CT = 10, CD = 4, AT = 7, what is AG? 13. If AG = 3, TG = 6, DT = 8, what is CD? 14. If AG = 4, GT = 12, CD = 6, what is DT? 15. If AT = 12, AG = 6, CT = 16, what is CD?

Set C A D E B C

1. If AD = x + 1, DB = 2, AE = 10 and EC = 5, what is x? 2. If AD = x + 2, DB = 4, AE = 6 and EC =8, what is x? 3. If AD = 4, DB = 6, AE = x + 3 and EC = 7.5, what is x? 4. If AD = 2, DB = 6, AE = x + 2 and EC = 9, find x 5. If AD = 5, DB = x + 2, AE = 4 and EC = 4.8, what is x?

In ∆ BAC, BC // DE C E B D A

22

6. If AD = 6, AE = 9, and AC = 21, what is BD? 7. If AD = 4, BD = 5 and AE = 6, what is EC? 8. If DB = 6, AE = 6 and EC = 9 what is AD? 9. If AD = x, AB = 5, AE = 2x and EC = 4, what is AD? 10. If AD = x, AB = 5, AE = 2x and EC = 6, what is AD? In ∆ABC, EF// BC. C F A E B 11. If AE = x, EB = x + 10, AF = 4 and FC = 6, what is x? 12. If AE = 6, EB = 8, AF = y and FC = 2y –2, what is y? 13. If AE = 5, AB = x +2, AF = 10 and AC = 3x, what is x? 14. If AE = x + 4, EB = 12, AF = x + 5 and FC = 14, what is x? 15. If AE = 6, EB = 4y –1, AF = 2 and FC = 3, what is y?

Let’s summarize

1. A ratio is a quotient of two numbers . 2. A proportion is an equality of two ratios. 3. The Basic Proportionality Theorem:

If a line is parallel to one side of a triangle and intersects the other two sides in distinct points, then it divides the two sides proportionally.

4. The Converse of the Basic Proportionality Theorem: If a line divides the two sides of a triangle proportionally, then the line is parallel to the third side


1. The value of n in 4:5 = n:20 is A. 14 C. 16 B. 15 D. 17

2. The value of x in 2x = 3 6 2

A. 3.5 C. 5.5

23

B. 4.5 D. 6.5

3. Write the ratio 3018 as a fraction in lowest terms

A. 169 C.

53

B. 9

15 D. 35

4. Which of the following proportions is true?

A. 2: 3 = 4:5 C. 4: 8 = 2: 4 B. 5:7 = 3: 2 D. 4 :7 = 7: 4

5. Two consecutive angles of a parallelogram are in the ratio 2:3. What is the measure of the smaller angle?

A. 36 C. 108 B. 72 D. 90

A 6. In ∆ ABC, XY // BC.

X Y B C Given that AX = 5, XB = 6, and AY = 8, what is YC?

A. 9 C. 7.6 B. 7 D. 9.6

7. In the figure below, What is AB:(BC + CD)? 3 7 2

● ● ● ● A B C D

A. 73 C.

143

B. 9

10 D. 31

8. Which of the following pairs of ratios forms a proportion?

A. 53 ,

95 C.

76 ,

98

24

B. 75 ,

1410 , D.

65 ,

56

9. In ∆ ABC, DE // AB. C

D E

A B

If DC = x, AD = 4, CE = x + 1 and EB = 8, what is x? A. 4 C. 2 B. 3 D. 1

10. In the figure, DE // AC. B

D E A C If BE =2, BC = 6 and BD = 3, what is BA?

A. 8 C. 5 B. 9 D. 12

11. Write the ratio 2 hours : 1 day as a fraction in lowest terms A. 2:1 C. 1:12 B. 1:2 D. 12:1

12. Which of the following proportions is false? A. 2:7 =8:28 C. 4:7 = 8:14 B. 3:5 = 9:15 D. 5:6 = 15:24

13. In ∆ABC, DE // BC. Which of the following proportions is false? B

D

A E C

25

A. AD = AE C. EC = DB DB EC AC AB

B. AB = AC D. AD = EC AD AE AB AC

14. in the figure. Points A, B, and C are collinear. The ratio of ∠ABD to ∠CBD is 5:1 . What is m∠CBD?

D ●

● ● ● A B C

A. 150 C. 50 B. 30 D. 100

15. ∆ ABC is a right triangle. If the ratio of ∠A to ∠B is 2:3, what is m∠ B?

B C A

A. 18 C. 20 B. 27 D. 36

26


1. a. 32

b. 32

2. a. x = 4

b. x = 16

3. The means are 6 and 6 The extremes are 3 and 12

4. a. 51

b. 31

5. a. 32

b. 41

6. a. 83

b. 31

7. a. 60

b. 120

8. a. 7.5 b. 2.5

9. a. AC = 12 b. AB = 4.5

10. a. x = 4 b. CB = 28

27


1. 1:2 2. 2:3 3. 4:9 4. 2:3 5. 1:3 6. 2:3 7. 1:2 8. 1:4 9. 3:2 10. 1:3 11. 1:1 12. 7:8 13. 5:1 14. 1:4 15. 1:3 16. 3:1 17. 5:2 18. 7:4 19. 2:3 20. 2:3

Lesson 2 Set A

1. True 2. True 3. False 4. False 5. True 6. False 7. True 8. True 9. False 10. True 11. False 12. False 13. False 14. False 15. False

Set B 1. n = 5 2. n = 4 3. n = 6 4. n = 5 5. n = 2.8 6. n = 3.5 7. n = 3.5 8. n = 4.8 9. n = 3.75 10. n = 1.6 11. n = 1 12. n = 4 13. n = 10.5 14. n = 2.5 15. n = 1.25 16. n = n = 6 17. n = 2.7 18. n = 9.6 19. n = 20 20. n = 40

Set C 1. 60 and 120 2. 36 and 144 3. 36 and 54 4. 27 and 63 5. a.1:2 b. 1:1 6. a. 1:2 b. 3:5 7. a. 4:5 b. 4:13 8. a. 1:2 b. 9:13 9. No 10. Yes 11. Yes 12. Yes 13. Yes 14. No 15. Yes 16. Yes 17.60

Set B 1. 2 2. 2 3. 7.5 4. 10.8 5. 2.5 6. 5 7. 7.5 8. 4.8 9. 3.75 10. 2.5 11. 9.6 12. 2.8 13. 4 14. 18 15. 8

Set C 1. 3 2. 1 3. 2 4. 1 5. 4 6. 8 7. 7.5 8. 4 9. 3 10. 2 11. 20 12. 3 13. 4 14. 2 15. 2.5

28


1. C 2. B 3. C 4. C 5. B 6. D 7. D 8. B 9. D 10. B 11. C 12. D 13. D 14. B 15. D

Module 17

Similar Triangles

What this module is about? This module is about similar triangles, definition and similarity theorems. As you go over the exercises you will develop your skills in determining if two triangles are similar and finding the length of a side or measure of an angle of a triangle. What are you expected to learn? This module is designed for you to:

1. apply the definition of similar triangles in: a. determining if two triangles are similar b. finding the length of a side or measure of an angle of a

triangle 2. verify the Similarity Theorems:

a. AAA Similarity b. SAS Similarity c. SSS Similarity

3. apply the properties of similar triangles and the proportionality theorems to calculate lengths of certain line segments.

How much do you know? A. For each pair of triangles, indicate whether the two triangles are similar or not. If they are similar, state the similarity theorem or definition that supports your answer. 1.

2

450 3

4

450 6

2. 3. 4. 5. B. 6 – 10. The length of the sides of a triangle are 14, 8 and 6. Find the length of the two sides of a similar triangle if the length of the shortest side is 12.

3

3

1

2

3

3

2

2

800

500

500

6 4

3 2 4

12

Lesson 1 Definition of Similar Triangles

Similar Triangles

Two triangles are similar if corresponding angles are congruent and corresponding sides are proportional. Examples: 1. If in ∆ABC and ∆XYZ

∠A≅ ∠X ∠B≅ ∠Y ∠C≅ ∠Z

XYAB =

YZBC =

XZAC

then ∆ABC ∼ ∆XYZ (Triangle ABC is similar to triangle XYZ) 2. ∆BMP ∼ ∆SEC Find the value of x and y

Since the corresponding sides are proportional SEBM =

ECMP =

SCBP

x3 =

84 =

y5

x3 =

21 =

y5

x3 =

21 ;

21 =

y5

x = 6 y = 10

A

B

C X

Y

Z

B P

M

3 4

5 S

E

C

x

y

8

Exercises: A. Given: MOL∆ ∼ REY∆

Fill the blanks.

1) M∠ ≅ ______

2) O∠ ≅ ______

3) _____ ≅ Y∠

4) RYML =

?MO

5) RYML =

?LO

6) 32 =

15?

7) 64 =

?2

B. State whether the proportion is correct for the indicated similar

triangles 1. RST∆ ∼ XYZ∆ 4. DEF∆ ~ HIS∆

XYRS =

YZST

HIDE =

IJEF

2. ABC∆ ~ DEF∆ 5. KLM∆ ~ PQR∆

DEAB =

EFBC

PRKM =

QRLM

3. RST∆ ~ LMK∆ 6. XYZ∆ ~ UVW∆

LMRT =

MKST

UVXY =

UWXZ

L O

M

2

4

5

Y

R

E

3

6

215

Complete the proportions 7. ABC∆ ~ DEF∆ 9. RST∆ ~ XYZ∆

?AB =

?BC =

?AC

?XY =

?XZ =

?YZ

8. KLM∆ ~ RST∆ 10. MNO∆ ~ VWX∆

KL? =

LM? =

KM?

?VX =

?VW =

?WX

C. Find the missing length 1. ABC∆ ~ XYZ∆ find b and a

2. SIM∆ ~ PON∆

SI = 6cm. IM = 4cm. SM= 8cm,

Find the lengths of the sides of PON∆ if the ratio of the lengths of

the corresponding sides is 1:3.

A

B

C

3 a

b X Z

Y

5 10

15

S 6cm.

I

4cm.

M

8cm

P O

N

Lesson 2

AAA Similarity Theorem If in two triangles the corresponding angles are congruent, then the two triangles are similar. If BOS∆ ↔ VIC∆

B∠ ≅ V∠ , O∠ ≅ I∠ , S∠ ≅ C∠ then BOS∆ ~ VIC∆

AA Similarity If two angles of one triangle are congruent to the corresponding two angles of another triangle, the triangles are similar. Given: A∠ ≅ O∠

J∠ ≅ B∠ Then: JAM∆ ~ BON∆ Examples: Are the triangles similar by AA Similarity? 1.

B

O

S V

I

C

J

A

M

O

B N

330 470

1000 330

470 1000

2. 3. 4. 5.

410

390

600 300

950

950

600

600 600

600

SAS Similarity Theorem If in two triangles two pairs of corresponding sides are proportional and the included angles are congruent, then the triangles are similar. If ICE∆ ↔ BOX∆ OBCI =

OXCE

and C∠ ≅ O∠ then ICE∆ ~ BOX∆ Examples: AOB∆ ~ DOC∆ by SAS since

DOAO =

COBO and

AOB∠ ≅ DOC∠ Can you explain why is RAT∆ ~ RAM∆ by SAS ?

I

C

E

B

O

X

B

A

O

D

C 9

6 12

8

R

A T

M 3 3

4

SSS Similarity Theorem If the two triangles three corresponding sides are proportional, then the triangles are similar. If SUN∆ ↔ BLK∆

BLSU =

LKUN =

BKSN

then SUN∆ ~ BLK∆ Example: 1.

OBSN =

PENA=

DESA

64 =

96 =

128

since the corresponding sides are proportional then

NSAPOE ∆∆ ~ by SSS Similarity.

2. Explain: Any two congruent triangles are similar.

S N

U

B K

L

P E

O

6 12

9

N A

S

8 4

6

Exercises: A. Tell whether the two triangles are similar. Cite the Similarity

Postulate or Theorem to justify your answer. (Identical marks indicate ≅ parts.)

1. 2. 3.

4.

6 8

10

3

5

4

6

8

3

4

3

6

6

3

9 6 8

10

12

15

5. 6. 7. 8.

6 6

6 6

5

5

3

3

2

8

8 4

9.

PSQS =

STRS

10. B. Determine whether each pair of the following triangles are

similar by SAS, AAA, SSS or not at all. 1.

Q

R

S

P

T

750

12 14 750

6 7

2. 3. 4. 5.

700

700

850

26

25

850 12.5

13

60 60

9

16

6 14 21

24

C. Answer the following: 1. Two isosceles triangles have an angle of 500. Does it follows that

the triangles are similar? 2. Two angles of BEL∆ have measures of 20 and 50. Two angles of

JAY∆ have measures of 30 and 100. IS JAYBEL ∆∆ ~ ? 3. Is it possible for two triangles to be similar if two angles of one

have measures 50 and 75, where as two angles of the other have measures 55 and 70?

4. Two angles of have measures 40 and 80, where as the two angles

of the other have measures 60 and 80, are the two triangles similar?

5 – 6. The lengths of the sides of a triangle are 12 and 15. If the

length of the shortest side of a similar triangle is 12, find the lengths of the other two sides.

7- 8. In the figure, if AE = 8 AB=4, BC=10, ED=3 Find BD and DC. 9 – 10. Explain: Any two equilateral triangle are similar.

500 500 A

B

D

C

E

Lesson 3

Application of the Properties of Similar Triangles to Calculate Lengths of certain line segments. Examples:

DEFABC ∆∆ ~ . Find the missing measure. 1. AB = 36, BC = 24, DE = 48 Bm∠ = 110, Em∠ = 110; EF =_______.

DEAB =

EFBC

4836 =

x24

36x = 1152 x = 32 EF = 32 2. AB = 38, BC =24, AC=30, DE=12, EF=16, DF= _______.

DEAB =

DFAC

1218 =

x30

23 =

x30

3x = 60 x = 20 DF = 20 3. AC = 15, DE=12, DF=20, =∠Am 35=∠Dm , AB =__________

DEAB =

DFAC

12x =

2015

12x =

43

4x = 36 x = 9 AB = 9

4. The lengths of the sides of a triangle are 14, 8 and 6. Find the perimeter of a similar triangle if the length of its shortest side is 12.

Let x = the length of one side of a triangle x = the length of the longest side of the triangle

126 =

x8

21 =

x8

x = 16

126 =

y14

21 =

y14

y = 28

P = 12 + 16 + 28 P = 56

5. On a level ground, a 5ft. person and a flagpole cast shadows of 10 feet and 60 feet respectively. What is the height of the flagpole?

Let x = the height of the flagpole

x5 =

6010

x5 =

61

x = 30 feet the height of the flagpole

Exercises 1. A yardstick casts a shadow of 24in. at the same time an electric

post cast a shadow of 20ft. 8 in. What is the height of the electric post?

2. Two triangles are similar. The lengths of the sides of one triangle are 5, 12 and 13. Find the lengths of the missing sides of the other triangle if its longest side is 39.

3. The perimeter of a triangle is 32cm. and the ratio of the sides is 3:6:7. Find the length of each side of the triangle.

4. A tall building at Makati casts a shadow of 12m. at the same time

a 7m. light pole cast a shadow of 3m. Find the height of the building.

5. If the shadow of the tree is 20m. long and the shadow of the person, who is 190cm. tall, is 250cm. long. How tall is the tree?

Let’s Summarize 1. Two triangles are similar if their vertices can be paired so that

corresponding angles are congruent and the lengths of corresponding sides are proportional.

2. The AAA Similarity. If the corresponding of two triangles are

congruent, then the two triangles are similar. 3. The AA Similarity. If two pairs of corresponding angles of two

triangles are congruent, then the two triangles are similar. 4. The SSS Similarity. If the lengths of corresponding sides of two

triangles are proportional, then the two triangles are similar. 5. The SAS Similarity. If one pair of corresponding angles of two

triangles are congruent and the lengths of the corresponding sides that include these angles are proportional, then the two triangles are similar.

What have you learned? A. Given the figures and the information below can you conclude that RPSKMN ∆∆ ~ ? If so what Similarity Theorem? 1. 80=∠Mm 80=∠Pm 2. 80=∠Mm , 100=∠+∠ SmRm 3. RS = 26 KN = 40

B. Find the missing side

4. 5. 6 – 7. If the shadow of the tree is 14cm. long and the shadow of the

person who is 1.8m. tall is 4m long, how tall is the tree? 8 – 10. A pole 3m. high has a shadow 5m long when the shadow of a

nearby building is 110m. long. How tall is the building?

K

24 M

36

N

R 16

P

24

S

24 x 5

6

16

20

3 4

5 9 12

X =?

Key

How much do you know?

1. yes, SAS 2. No 3. yes, SAS 4. yes, AAA 5. yes, SSS 6. 16,28

Lesson 1 Exercises:

A.1. R∠ B. 1. Correct C. 1. a=6 2. E∠ 2. correct b=1 3. L∠ 3. Not correct 4. RE 4. correct 2. NP=24 5. YE 5. correct NO=12 6. 5 6. correct PO = 18 7. 3 7. DE, EF, DF 8. RS, ST, RT

9. RS, RT, ST 10. MO, MN, NO

Lesson 2

A.1. Similar, SSS B.1. Similar, SAS C.1. yes 2. Similar, SAS 2. Similar, AA 2. no 3. Similar, SAS 3. Similar, SAS, AA 3. no 4. Similar, SSS 4. Similar, AA 4. yes 5. Similar, AAA 5. Similar, SSS 5 6. Similar, SSS 6 7. Similar, SAS 7. BD = 5.7’ 8. not similar 8. DC = 7.5 9. Similar, SAS 9 10. Similar, SSS 10

18, 22.5

Equilateral ∆ is equiangular by AAA. Any 2 equilateral ∆ are similar.

Lesson 3

Exercises

1. 372in. 2. 15, 36 3. 6cm., 12cm., 14cm. 4. 28m. 5. 152cm.

What have your learned?

1. Similar, SAS 2. Similar, SAS 3. not similar 4. 4 5. 15 6-7. 6.3m 8 - 10. 66m.

Fun with Math

Activities

Related to

Similar Figure

List all the triangles in the figure that are similar to MAN∆ .

C

B

A

E

D

D

O

L M N

Polymonies are made up of a number of squares connected

by common sides. Thirteen sticks were used to make this one with

four squares. Investigate the numbers of sticks needed to make

others.

How many rectangles are there in this

diagram?

Make a rectangular grid of squares. Cut it along the grid

lines to make two identical pieces.

Module 3 Similarity

What this module is about This module is about similarities on right triangles. As you go over the exercises you will develop skills in applying similarity on right triangles and solve for the missing lengths of sides using the famous Pythagorean theorem.


1. Apply AA similarity on Right triangles

2. In a right triangle, the altitude to the hypotenuse separates the triangle into two triangles each similar to the given triangle and similar to each

other.

3. On a right triangle, the altitude to the hypotenuse is the geometric mean of the segments in which it divides, each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent

to it.

4. Pythagorean Theorem and its application to special right triangles

How much do you know Use the figure to answer each of the following:

1. What is the hypotenuse of rt. ∆ABC?

C

BAD

2

2. If ∠C is the right angle of ∆ABC and CD ⊥ AB then ∆ABC ∼ ∆BDC ∼ ___.

3. Complete the proportion: BDCD

AD ?=

4. In rt. ∆PRO, ∠R is a right angle OR = 24 and PO = 26, find PR: 5. In a 30°-60°-90° triangle the length of the hypotenuse is 14. Find the

length of the longer leg.

6. In a 30°-60°-90° triangle, the length of the hypotenuse is 1121 . Find the

length of the shorter leg.

7. In a 45° - 45° - 90° triangle, the length of the hypotenuse is 16. Find the length of a leg.

8. Find the length of the altitude of an equilateral triangle if the length of a

side is 6.

9. Find the length of the diagonal of a square if the length o a side is 10 cm.

10. ∆BAC is a right triangle ∠C is right angle CD ⊥ AB. Find CD if AD = 14, DB = 6

O

P R

24 26

C

BD

A14 6

3

What will you do

Lesson 1

Similarity on Right Triangle

Let us recall the AA Similarity Theorem. Given a correspondence between the vertices of two triangles. If two pairs of corresponding angles are congruent, then the triangles are similar. From the theorem, if ABC ↔ RST and ∠A = ∠R, ∠B = ∠S then ∆ABC ∼ ∆RST. We can apply this theorem to prove another theorem, this time in a right triangle.

Theorem: In a right triangle, the altitude to the hypotenuse separates the triangle into to two triangles each similar to the given triangle and similar to each other.

Given: Right ∆ABC with altitude CP Prove: ∆ACP ∼ ∆CBP ∼ ∆ABC To prove this theorem, we apply the AA Similarity Theorem

APB

C

S

T R

B

A C

4

Examples: If you are given ∆PRT a right triangle and RM an altitude to the hypotenuse then we can have three pairs of similar triangles. ∆RMP ∼ ∆TRP

∆TMR ∼ ∆TRP

∆RMP ∼ ∆TMR Try this out

A. Use the figure to answer each of the following:

1. Name the right triangle of ∆ABC

2. What is the altitude to the hypotenuse of ∆ABC?

3. Name the hypotenuse of ∆ABC

4. Two segments of the hypotenuse Are AD and ____.

5. The hypotenuse of ∆BCD if CD ⊥ AB is ____.

6. Name the right angle of ∆ACD

7. Name the hypotenuse of right ∆BCD

8. ∆ADC ∼ _____

9. ∆ABC ∼ _____

10. ∆ABC ∼ _____

B. Name the pairs of right triangles that are similar.

1.

2.

3.

R

PM

T

D

C

BA

S

R TO

5

4.

5.

6.

C. Use the figure at the right.

1. Name all the right triangles.

2. In ∆ABC, name the altitude to the hypotenuse.

3. Name the hypotenuse in ∆ADC.

4. Name the hypotenuse of ∆ACB.

One of the segments shown is an altitude to the hypotenuse of a right triangle. Name the segment.

5. 6. 7. Name the three pairs of similar triangles: 8.

9.

10.

A

D

C B

B

C A D

M

N

OP

M

S

RO

I D

J

K G F

H

E

6

Lesson 2

Geometric Mean in Similar Right Triangles

The previous theorem states that: In a right triangle, the altitude to the hypotenuse divides the triangle into similar triangles, each similar to the given triangle. If : ∆ACB is right with ∠C, the right angle CD is the altitude to the hypotenuse AB. Then : ∆ADC ∼ ∆ACB ∆CDB ∼ ∆ACB ∆ADC ∼ ∆CDB Corollary: 1. In a right triangle, the altitude to the hypotenuse is the geometric mean of the segments into which divides the hypotenuse In the figure:

DBCD

CDAD

=

Corollary 2: In a right triangle, each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to it. In the figure:

ADAC

ACAB

=

ADBC

BCAB

=

C

BA D

C

B D

A

C

A

B

D

7

Examples: 1. How long is the altitude of a right triangle that separates the hypotenuse into lengths 4 and 20?

20

4 aa=

a2 = 80

a = 80

a = 516 ⋅

a = 54

2. Use the figure at the right to solve for x and y.

8

2 xx=

x2 = 16

x = 16

x = 4 Solve for y

8

6 yy=

y2 = 48

y = 48

y = 316 ⋅

y = 4 3

20 4

a

2

y x

8

8

Try this out A. Supply the missing parts:

1. ?RT

RTRW

=

2. RSTSWS

=?

3. RSTS

TS=

?

4. ?RT

RTRS

=

Give the indicated proportions.

5. The altitude is the geometric mean

6. The horizontal leg is the geometric mean

7. The vertical leg is the geometric mean Find: 8. BS

9. RS

10. ST

B. Solve for x and y:

1.

x = y =

F

S

O P

R

B

T S

8

2

x

4

7

y

T

S WR

9

x y

2.

x = y =

3. x = y =

4. x = y =

5. x = y = C. Given: Right ∆POM OR ⊥ PM,

x y

5

10

6

O

M R

P

10

x y

4 10

y

10 20

x

10

Find the missing parts:

1. PR = 5, RM = 10, OR =

2. OR = 6, RM = 9, PR =

3. PR = 4, PM =12, PO =

4. RM = 8, PM =12, OM =

5. PO = 9, PR = 3, PM =

6. PR = 6, RM = 8, PO =

7. PR = 4, RM = 12, OM =

8. PR = 4, PO = 6, RM =

9. PR = 8, OR = 12, RM =

10. PM =15, OM = 12, RM =

Lesson 3

The Pythagorean Theorem In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. In the figure: ∆BCA is right with leg lengths, a and b and hypotenuse length, c. The Pythagorean Theorem in symbol: c2 = a2 + b2 Pythagorean Theorem is named after Pythagoras, a Greek Mathematician of the sixth century BC. This theorem can be used to find a missing side length in a right triangle. Examples:

1. In the figure c = 13, b = 12 Find a: c2 = a2 + b2 132 = a2 + 122

B

C A

a c

b

a = ? c = 13

b = 12

11

a2 =132 - 122 a2 =169 – 144 a2 = 25 a = 25 a = 5

2. Find c, if a = 16 and b = 12 c2 = a2 + b2 c2 = 162 + 122 c2 = 256 + 144 c = 400 c = 20

3. c2 = a2 + b2 82 = 42 + b2 b2 = 82 – 42 b2 = 64 –16 b2 = 48 b = 48 b = 316 ⋅ b = 4 3

Try this out:

A. State whether the equation is correct or not 1.

a2 + b2 = c2

2. r2 = s2 + t2

a = 16

b = 12

c = ?

a = 4

b = ?

c = 8

a c

b

r t

s

12

3. e2 = f2 – d2 4. a2 = c2 – b2 5. k2 = l2 – m2 6. p2 = r2 + q2 7. x2 = 32 + 42

e d

f

a c

b

l k

m

p

q

r

3 x

4

13

8. x2 = 102 – 72 9. x2 = 42 + 62 10. x2 = 72 – 52

B. Write the equation you would use to find the value of x. 1. 2.

10 7

x

x 4

6

x 7

5

x 4

3

x 6

5

14

3.

4. Classify each statement as true or false

5. 32 + 42 = 52

6. 102 – 62 = 82

7. 12 + 12 = 22

8. 22 + 22 = 42

9. 72 – 52 = 52

10. 92 + 122 = 152 C. Given the lengths of two sides of a right triangle. Find the length of the third

side

1. a = 6, b = 8, c = 2. a = 5, b = 12, c =

3. a = 12, c = 15, b = 4. b = 4, c = 5, b = 5. a = 24, c = 26, b = 6. b = 16, c = 20

x

7 10

x 5

6

a c

b

B

C A

15

7. 8. 9. 10

Lesson 4

Special Right Triangle

Isosceles Right Triangle or 45° – 45° – 90° Theorem: In a 45° – 45° – 90° triangle, the length of the hypotenuse is equal to the length of a leg times 2 . In the figure: If ∆ABC, a 45° – 45° – 90° triangle when AC = BC = s then AB = s 2 . 30° - 60° - 90° Theorem: In a 30° - 60° - 90° triangle, the length of the hypotenuse is twice the length of the shorter leg, and the length of the longer leg is 3 times the length of the shorter leg. In the figure: If ∆PRT where ∠R is a right angle and ∠T = 30°, Then:

a. PT = 2PR b. RT = PR 3

a

b

c

7

24

?

4

6

?

7

9

?

6 3

?

12

s s

C

B A 45° 45°

P

R T

60°

30°

16

Examples: 1.Find the length of the hypotenuse of an isosceles right triangle with a leg 7 2 cm long.

Hypotenuse = leg 2⋅ .

= 7 2 2⋅ = 7 · 2 = 14

2. Find the length of each leg of a 45° - 45° - 90° triangle with a hypotenuse 12 cm long.

Leg = 2

hypotenuse

= 2

12 = 22

212

⋅ = 2

212 = 6 2 cm

3. Find the length of the longer leg and the length of the hypotenuse. Longer leg = shorter leg · 3 = 30 · 3 = 30 3 m

hypotenuse = shorter leg · 2 = 30 · 2 = 60 m

Try this out A. Use the figure to answer the following:

1. The hypotenuse of a rt. ∆ABC is ___________.

2. The shorter leg of rt. ∆ABC is ___________.

7 2

45° 45°

45°

45°

12 cm

30°

30 m

30°

30°

60°

60°

C

B A D

17

3. The shorter leg of rt. ∆ADC is ___________.

4. The longest side of rt. ∆ADC is _________.

5. The altitude to the hypotenuse of ∆ACD is ________.

6. The longer leg of rt. ∆ACB is _________.

7. The longer leg of rt. ∆ADC is ________.

8. When CD = 2 then ____ = 4.

9. When CB = 6 then _____ = 6 3

10. When CB = 6 then _____ = 3 B. Find the value of x in each of the following: 1. 2. 3. 4. 5. 6. 7. 8. 9.

10.

x

45°

45°

10

30°x

60°24

30°x

60°30 x

45°

45°

12

x 45°

45°

7

30°

x60°

26

30°x

60°

18

30°

60°

x16

30°

60°

x

10

60°

30°

x 6

18

30°60°

xy

10

C. Find the missing lengths, x and y.

1. 2. 45° 45°

3. 4.

5. 6. 7. 8.

60° 30° 9. 10.

45°

y x

5

45° 45°

x y

3

y

x 7

30°

10 3 x

y 30°

x

y 5

45°

45°y x

23

45° 45°

x y

23

45°

x

y 1.5

60°

30°.5 3

xy

19

Beyond the Pythagorean Theorem In symbol c2 = a2 + b2, where c is the hypotenuse and a and b are the legs of a right triangle. Figure shows acute triangles

Figure shows obtuse triangles

Figure shown right triangle

20

Activity: This activity will help you extend your understanding of the relationship of the sides of a triangle. Materials: Strips of paper cut in measured lengths of 2, 3, 4, 5, 6 and 8 units. Procedure:

1. Form triangles with strips indicated by the number triplets below.

2. Draw the triangle formed for each number triple.

3. Fill out the table:

Number triplets What kind of triangle Compute c2 Compute a2 + b2

1. 3 4 5 Right 52 = 25 32 + 42 = 25 2. 2 3 4 3. 2 4 5 4. 5 4 8 5. 6 5 8 6. 4 5 6 7. 2 3 3 8. 3 3 4

After the computation, the completed table will look like this

Number triplets Kind of triangle c2 a2 + b2 Comparison of c with

(a2 + b2 ) 1. 3 4 5 Right 25 25 Equal to 2. 2 3 4 Obtuse 16 13 Greater than 3. 2 4 5 Obtuse 25 20 Greater than 4. 5 4 8 Obtuse 64 41 Greater than 5. 6 5 8 Obtuse 64 61 Greater than 6. 4 5 6 Acute 36 41 Smaller than 7. 2 3 3 Acute 9 13 Smaller than 8. 3 3 4 Acute 16 18 Smaller than

1. What kind of ∆ did you get from triplet no. 1?

2. In triplet no. 1, what is the relation between c2 and (a2 + b2)?

3. Which triplets showed obtuse triangle?

21

4. For each obtuse triangle compare the result from c2 and (a2 + b2).

5. For acute triangles how will you compare the result of c2 and (a2 + b2) Fill in the blanks with <, =, >:

6. In a right triangle, c2 ____ a2 + b2

7. In an obtuse triangle, c2 ______ a2 + b2

8. In an acute triangle, c2 ____a2 + b2

Let’s Summarize

Theorem: In a right triangle, the altitude to the hypotenuse separates the triangle into two triangles each similar to the given triangle and similar to each other.

Corollary 1: In a right triangle, the altitude to the hypotenuse is the

geometric mean of the segments into which it divides the hypotenuse.

Corollary 2: In a right triangle, each leg is the geometric mean of the

hypotenuse and the segment of the hypotenuse adjacent to it.

Pythagorean Theorem: The square of the length of the hypotenuse is

equal to the sum of the squares of the legs.

45°-45°-90° Theorem: In a 45°-45°-90° triangle, the length of the hypotenuse is equal to the length of a leg times

2 .

30°- 60° - 90° Theorem: In a 30°- 60°- 90° triangle, the length of the hypotenuse is twice the length of the shorter leg, and the length of the leg is 3 times the length of the shorter leg.

22

What have you Learned Fill in the blanks:

1. The _______ to the hypotenuse of a right triangle forms two triangles each similar to the given triangle & to each other.

2. The lengths of the ________ to the hypotenuse is the geometric mean of

the lengths of the segments of the hypotenuse.

3. In the figure

?MA

MAAB

=

for nos. 3 & 4 4. If BP = 8

AB = 4 Find PM ___

5. If in a right triangle the lengths of the legs are 8 and 15, the length of the

hypotenuse is _______ 6. Find the length of an altitude of an equilateral triangle if the length of a

side is 10.

7. In a 30° – 60° – 90° triangle, the length of the hypotenuse is 8. Find the length of the shorter leg.

8. - 9. ∆ACB is an isosceles right triangle. CD bisects ∠C, the right angle. Find AB and CB.

10. What is the height of the Flag Pole?

M

PB

A

C

A BD

3

2 m

8 m

23

Answer key


1. AB 6. 5 43

2. ∆ACB 7. 8 2 3. CD 8. 3 3 4. 10 9. 10 2 5. 7 3 10. 2 21

Lesson 1: A. B.

1. ∠C or ∠ACB 1. ∆ROS ∼ ∆RST 2. CD 2. ∆TOS ∼ ∆RST 3. AB 3. ∆ROS ∼ ∆TOS 4. BD or DB 5. BC 4. ∆MST ∼ ∆MOR 6. ∠ADC 5. ∆RSO ∼ ∆MOR 7. BC 6. ∆MST ∼ ∆RSO 8. ∆BDC 9. ∆ADC 10. ∆BDC

C.

1. ∆ADC, ∆BDC, ∆ACB 2. CD 3. AC 4. AB 5. BD 6. GH 7. OK 8. ∆MNR ∼ ∆MPO 9. ∆ONP ∼ ∆MPO 10. ∆MNP ∼ ∆ONP Lesson 2

A. B. C. 1. WS 1. x = 2 11 1. 5 2 2. TS y = 2 7 2. 4

24

3. WS 2. x = 10 3 3. 4 3 4. RW y = 30 4. 4 6

5. SPOS

OSSF

= 3. x = 2 14 5. 27

y = 2 35 6. 4 3

6. PFPO

POSP

= 4. x = 20 7. 8 3

y = 5 5 8. 9

7. PFOF

OFFS

= 5. x = 3216

6100 or 9. 18

y = 10 10. 9.6 8. 4

9. 2 5 10. 4 5 Lesson 3 A. B. C. 1. correct 1. x2 = 32 + 42 1. 10 2. correct 2. x2 = 62 – 52 2. 13 3. not 3. x2 = 102 – 72 3. 9 4. correct 4. x2 = 62 – 52 4. 3 5. correct 5. true 5. 10 6. not 6. true 6. 12 7. correct 7. false 7. 25 8. correct 8. false 8. 2 13 9. not 9. false 9. 130 10. not 10. true 10. 6 Lesson 4 A. B. C. x y

1. AB 1. 12 1. 5 3 5

2. BD 2. 10 3 2. 23

223

3. CD 3. 8 3. 3 3

4. AC 4. 18 3 4. 2

23 2

23

25

5. CD 5. 13 5. 2

25 2

25

6. AC 6. 7 2 6. 2

25.1 2

25.1

7. AD 7. 12 3 7. 14 7 3

8. AC 8. 12 2 8. 5 5 3

9. AC 9. 15 3 9. 5 3 15

10. DB 10. 3

310 10. 5 10


1. altitude

2. altitude

3. AP

4. 4 2

5. 17

6. 5 3

7. 4

8. 3 2

9. 3

10. 2 3

Module 1 Plane Coordinate Geometry

What this module is about

This module will explain to you the relationship among lines on the plane. This will

also tell you about intersecting line, non-intersecting lines, characteristics of parallel lines and perpendicular lines. Furthermore, this will tell you how to determine the intersection of two lines if there is any.

What you are expected to learn This module will help you

1. Determine the point of intersection of two lines 2. Compute for the coordinates of the intersection of two lines. 3. Determine without graphing if the given lines are parallel, perpendicular or neither. 4. Define algebraically parallel and perpendicular lines.

How much do you know Given each pair of lines, determine if they are a) intersecting but not perpendicular, b) perpendicular and c) parallel 1. y = 3x – 7 y = 3x + 1 2. 2x + 3y = 5 3x – 2y = 8 3. x + 2y = 9 4x + 3y = 1 4. 3x – 2y = 3 3x + 5y = 14 5. 2x – 4y = 1 4x – 8y = 7

2

6. The slopes of parallel lines are ___________. 7. What is the slope of the line parallel to 2x + 4y – 3 = 0? 8. What is the slope of the line perpendicular to x – 6y + 5 = 0? 9. Write equation of the line parallel to 5x + 8y = 7 and passing through (2, 4). 10. At what point do the lines 3x – y = 7 and 5x + y = 9 intersect? What you will do

Lesson 1

Intersection of Two Lines Lines on the plane can either intersect or not. If two lines intersect, then there is a common point between them. That common point is where the two lines intersect. In a plane, this point has two coordinates, the x and y coordinates. The coordinates of the intersection can be solved algebraically. In second year algebra, you are taught how to solve systems of linear equations. That knowledge will help you a lot in understanding this lesson. Since every equation of the line assumes the general form ax + by + c = 0, to get the intersection of two lines, you solve for the value of x and y common to both equations.

To solve for the value of x and y algebraically, there are methods that can be used. The following steps can help you in solving systems of linear equations.

1. Eliminate one variable by using a. addition or subtraction b. finding the value of one variable in terms of the other variable

2. Solve for the value of the other variable 3. Substitute the computed value of one variable on either of the two equations to

solve for the value of the remaining variable. Examples: 1. Find the point of intersection of the lines whose equations are 3x – y =10 and 5x + y =14. Solution: To find the intersection of the two lines, solve the system by eliminating one variable through addition. Add the similar terms of the first equation to that of the second equation. Equation 1 3x – y = 10 Equation 2 5x + y = 14 Add 8x + 0 = 24 8x = 24 x = 3

3

Then replace x with 3 in either Equation 1 0r Equation 2 Equation 1 3x – y = 10 3(3) – y = 10 9 - y = 10 - y = -1 y = 1 Therefore the point of intersection of the two lines is (3, -1) 2. Find the point of intersection of the two lines whose equations are x + 2y = 4 and 3x + 5y = -21. Solution:

Another method of eliminating one variable is through substitution. You get the value of one variable in terms of the other variable. Equation 1 x + 2y = 4 Equation 2 3x – 5y = -21 Using Equation 1, solve for x in terms of y x + 2y = 4 x = 4 – 2y Using Equation 2, replace x with 4 – 2y Equation 2 3x – 5y = -21 3(4 – 2y) - 5y = -21 12 – 6y – 5y = - 21 - 11y = - 21 – 12 - 11y = - 33 y = 3 Then replace y with 3 on either Equation 1 or Equation 2 to solve for the value of x. Equation 1 x + 2y = 4 x+ 2(3) = 4 x + 6 = 4 x = -2 So the intersection of the two lines is the point whose coordinates are (-2, 3). 3. At what point do the lines 3x + 2y = 12 and 4x – 3y = -1 intersect?

4

Solution.

To eliminate a variable in this example, first find equivalent equations with equal but opposite coefficient in x or in y by multiplying one or both equations by a number factor. Equation 1 3x + 2y = 12 Equation 2 4x – 3y = -1 3(3x + 2y) = 12 9x + 6y = 36 2(4x – 3y) = -1 8x – 6y = -2 Add Equation 1 and 2 17x + 0 = 34 7x = 34 x = 2 Then replace x with 2 in either Equation 1 or 2 Equation 1 3x + 2y = 12 3(2) + 2y = 12

6 + 2y = 12 2y = 6 y = 3

Therefore, the point of intersection is (2, 3).

Example 4. Find the point of intersection of the lines y = 2x – 2 and y – 2x = 1. Solution: Using any of the previous methods, solve the system of equations. Equation 1 y = 2x – 2 Equation 2 y – 2x = 1 Using Equation 1, solve for y in terms of x y = 2x - 2 Replace y by 2x - 2 in the second equation. (2x – 2) – 2x = 1 2x - 2 – 2x = 1 -2 = 1 , this is a false statement. Solving the system led to a false statement. Therefore the system is inconsistent and has no solution . So the lines do not intersect. There is a way of determining whether the two lines intersect or not. The following theorem can be used to determine if two lines intersect or not.

5

Theorem: If two non-vertical lines intersect, then their slopes are not equal. This theorem can be verified as follows. Y B (x2,y2) C(x3,y3) A (x1, y1) X Let A, B and C be three distinct points on a plane. By line determination postulate, line AB and line AC can be constructed with A as a common point or point of intersection.

Get the slopes of the two lines. For line AB, the slope

m1 = 12

12

xxyy

−−

For line AC, the slope

m2 = 13

13

xxyy

−−

Since the numerators and denominators in the two fractions are different, then the two slopes are not equal. Thus 21 mm ≠ The converse of the theorem, “ If two non-vertical lines have different slopes, then the lines are intersecting,” is also true. 5. Without graphing, show that 5x – y = 7 and 3x + 2y = 4 are intersecting lines. Solution: Transform each equation to slope-intercept form. Equation 1 5x – y = 7 - y = -5x + 7 y = 5x - 7 slope (m1) = 5

6

Equation 2 3x + 2y = 4 2y = - 3x + 4

y = 223

+− x

slope (m2) = 23

−

Since 21 mm ≠ , then the two lines are intersecting. Try this out A. Without graphing, show that the following pairs of lines are intersecting. 1. y = 3x – 4 y = x + 7

2. y = 421

−x

y = 623

+x

3. x + 4y = 3 2x + y = 7 4. 2x – y = 8 2x + y = 7 5. 3x – 5y = 4 x + y = 4 B. Find the intersection of the following pairs of lines. 1. 3x – 4y = 1 3x + y = -4 2. x + 5y = 13 2x – y = -7 3. x – 6y = 11 3x + 3y = 12 4. 2x – 5y = -7 3x + 2y = 18 5. 3x – 7y = -15 2x + 6y = -10

7

Lesson 2

Parallel and Perpendicular Lines

In the previous lesson, there was an example in which the solution of the system of equations led to a false statement. This showed that the two lines do not intersect. In a plane, if two lines do not intersect, then they are parallel. Let us consider the given in example 4. Equation 1 y = 2x - 2 Equation 2 y – 2x = 1 If we graph the two lines on the same set of axes, they will look like this.

The graphs showed that the two lines are parallel. Without graphing, you can also determine if the lines are parallel or if they intersect. The following theorem can be used to prove that two lines are parallel.

Theorem: If two non vertical lines are parallel, then the slopes are equal. The converse of this theorem is also true. Converse theorem: If two lines which are not coincident have the same slope, then the lines are parallel.

8

Consider the given lines above. To prove that they are parallel, their slopes must be equal, transform the equations into y-form or slope-intercept form.

Equation 1 y = 2x - 2

The equation is already in y - form. slope(m1) = -2 Equation 2 y – 2x = 1 Again, transform to slop-intercept form y = -2x + 1

slope (m2) = -2 You can see that m1 = m2 . Therefore, the two lines are parallel. The slope of horizontal line is zero (0), thus all horizontal lines are parallel to each other. Slope of vertical line is undefined and so all vertical lines are also parallel to each other. In a plane, some lines are parallel, and others are intersecting. Intersecting lines may be perpendicular or not. The next theorem will help you determine if the pair of lines is perpendicular or not. Theorem: If two non vertical lines are perpendicular, then their slopes are negative reciprocals. Examples: 1. Show that the lines whose equations are 3x + y = 4 and x – 3y = 7 are perpendicular. Solution: Equation 1 3x + y = 4

Transform to slope-intercept form

y = -3x + 4 slope (m1) = -3 Equation 2 x – 3y = 7 Transform to slope-intercept form - 3y = -x + 7

9

y = 3

731

−+

−− x

y = 37

31

−x

slope (m2) = 31

Comparing their slopes, it is obvious that they are the negative reciprocal of each other. Hence, we can conclude that the two lines are perpendicular. 2. Determine which pairs of lines are parallel, perpendicular or just intersecting.

a. Line 1 5x – 2y = 7 b. Line 2 2x + 5y = 4 c. Line 3 4x + 10y = 6 d. Line 4 3x + y = 5 e. Line 5 x – 3y = 6 f. Line 6 2x + 6y =1

Solution: Transform each equation to slope-intercept form

a. 5x – 2y = 7 - 2y = - 5x + 7

y = 2

725

−+

−− x

y = 27

25

−x

m1 = 25

b. 2x + 5y = 4 5y = -2x + 4

y = 54

52

+− x

m2 = 52

−

c. 4x + 10 y = 6 10 y = -4x + 6

10

y = 106

104

+− x

y = 53

52

+− x

m3 = 52

−

d. 3x + y = 5 y = -3x + 5 m4 = -3 e. x – 3y = 6 - 3y = -x + 6

y = 3

63 −+

−− x

y = 231

−x

m5 = 31

f. 2x + 6y = 1 6y = -2x + 1

y = 61

62

+− x

y = 61

31

+− x

m6 = 31

−

By comparing the computed slopes, you can determine the relationship of a line with the other lines. Since the slopes of line 1 and line 2 are negative reciprocals, then line 1 is perpendicular to line 2. Line 1 and line 3 are also perpendicular to each other since their slopes are negative reciprocals. The slopes of line 2 and line 3 are equal so line 2 is parallel to line 3. The slopes of line 4 and line 5 are also negative reciprocals so these lines are also perpendicular. On the other hand, other pairs of lines like line 4 and line 6, line 5 and line 6 are just intersecting line since their slopes are not equal. The summary of the relationship between each pair of lines is given below.

11

Pair of Lines Relationship Pair of Lines Relationship Line 1 and line 2 Perpendicular Line 2 and line 6 Just intersecting Line 1 and line 3 Perpendicular Line 3 and line 4 Just intersecting Line 1 and line 4 Just intersecting Line 3 and line 5 Just intersecting Line 1 and line 5 Just intersecting Line 3 and line 6 Just intersecting Line 1 and line 6 Just intersecting Line 4 and line 5 Perpendicular Line 2 and line 3 Parallel Line 4 and line 6 Just intersecting Line 2 and line 4 Just intersecting Line 5 and line 6 Just intersecting Line 2 and line 5 Just intersecting

3. Prove that the quadrilateral whose vertices are given is a parallelogram.

A(-3, 3), B(-2, -2), C(8, 3), D(7, 8) Solution:

Parallelogram is defined as quadrilateral whose opposite sides are parallel. To prove that ABCD is a parallelogram, show that opposite sides have equal slopes. Slope of AB:

mAB = )3(2

32−−−−−

= 15−

mAB = -5 Slope of CD:

mCD = 8738

−−

= 1

5−

mCD = -5 Since the slopes of side AB and side CD are equal, then AB║CD. Slope of BC:

mBC = )2(8)2(3

−−−−

= 2823

++

= 105

mBC = 21

12

Slope of AD:

mAD = )3(7

38−−−

= 105

mAD = 21

Since the slope of side BC is equal to the slope of side AD then BC║AD.

Therefore we can conclude that ABCD is a parallelogram since the opposite sides are parallel. 4. Prove that the following points are vertices of a right triangle. A(-3, 5), B(6, 1) and C(-7, -4). Solution:

Since one angle of a right triangle is right, then two sides must be perpendicular and you have to find the slopes of the three sides to prove this. Slope of AB:

mAB = )3(6

51−−

−

mAB = 94−

Slope of BC:

mBC = 6714

−−−−

= 135

−−

mBC = 135

Slope of AC:

mAC = )3(7

54−−−−−

= 49

−−

mAC = 49

13

From the computed slopes of the three sides AB, BC and AC, it is clear that AB is perpendicular to AC since their slopes are negative reciprocals. Hence we can conclude that BAC∠ is a right angle. It only proved that ∆ABC is a right triangle. If we sketch the graph on the Cartesian plane, it will look like this. Y (-3, 5)

5. Write an equation in standard form of the line passing through

a. point (1, -2) and parallel to y = 3x + 2 b. point (3, 4) and perpendicular to y = 2x – 7.

Solution: a. Since parallel lines have equal slopes, then the slope of the required line is 3. Using the point slope form, determine the equation of the line. y – y1 = m(x – x1) y – (-2) = 3(x – 1) y +2 = 3x -3 -3x + y +2+3 = 0 -3x + y + 5 = 0 or 3x – y - 5 = 0, this is the required equation of the line

b. Perpendicular lines have slopes which are negative reciprocal of one another.

Since the slope of the given line is 2, then the slope of the required line is 21

− . Using

the point-slope form of the line, determine the equation of the required line. y – y1 = m(x – x1)

y – 4 = )3(21

−− x

X

(-7,-4)

(6,1)

14

2(y – 4) = -1(x – 3) 2y – 8 = -x + 3 x + 2y – 8 – 3 = 0 x + 2y – 11 = 0 , the required equation of the line. 6. A right triangle has its right angle at (-4, 1) and the equation of one of its legs is 2x – 3y + 11 = 0. Find the equation of the other leg. Solution:

The two legs of a right triangle are perpendicular at the vertex of the right angle. Their slopes are the negative reciprocal of each other. Since the equation of one of its legs is 2x – 3y + 11 = 0, then we have to compute for the slope first. Get the slope of the given leg. 2x – 3y + 11 = 0 Transform the equation to slope intercept form. - 3y = - 2x - 11

y = 3

1132

−−

+−− x

y = 3

1132

+x

Since the slope of the given line is 32 , then the slope of the required line is

23

− .

Using the point - slope form, determine the equation of the line: y – y1 = m(x – x1)

y – 1 = [ ])4(23

−−− x

2(y – 1) = -3(x + 4) 2y – 2 = -3x – 12 3x + 2y – 2 + 12 = 0 The equation of the other leg is 3x + 2y + 10 = 0 Try this out A. Determine if the pair of lines are parallel, perpendicular or neither. 1. y = 2x + 5

y = 721

−− x

15

2. y = 4x – 1 y = 4x + 3

3. y = 23

25

+x

y = 352

+− x

4. y = 3x – 7 y = x + 8 5. 2y = x + 4 y = 2x – 5 6. x + y = 5 x – y = 3 7. 2x + y = 9 x – 2y = - 4 8. 2x + y = 8 4x + 2y = -3 9. 3x + 2y – 1 = 0 2x – 3y – 7 = 0 10. 2x + 7y + 6 = 0 2x + 7y – 11 = 0 B. Prove that the following points are vertices of a parallelogram or not. Justify. 1. A(3, 7), B(4, 3), C(-2, 1), D(-3, 5) 2. L(4, 2), O(8, 3), V(6, -5), E(2, -6) 3. P(-3, -2), L(1, -1), A(1, -8), N(-3, -6) C. Verify if the following are vertices of a right triangle. Explain your answer. 1. E(1, 6), A(6, 2), r(1, 2) 2. A(1, 4), C(-3, -1), E(7, -5) 3. P(3, -1), E(6, -3), N(3, -7) D. Find the equation of the line in standard form given the following conditions. 1. Passing through (-1, -1) and parallel to y = 4x + 1. 2. Passing through (0, 4) and parallel to y = -3x – 2. 3. Passing through (3, 0) and parallel to x – 3y = 7.

16

4. Passing through (1, -5) and parallel to 2x + 5y – 4 = 0. 5. Passing through (2, 1) and perpendicular to x – y = 3. 6. Passing through (3, -4) and perpendicular to y = 3x + 1. 7. Passing through (-2, -4) and perpendicular to 5x + 2y = 4. E. Solve the following problems. 1. The line containing points (-6, k) and (4, 3) is perpendicular to the line containing points (9, 1) and (13, k+1). Find k. 2. A right triangle has its angle at (5, 7) and the equation of one of its legs is 2x – y – 3 = 0. Find the equation of the other leg. 3. Write the equations of three lines parallel to 2x + y = 4. 4. Write the equations of three lines perpendicular to 3x + y – 4 = 0 5. Show that the lines 6x + 7y – 6 = 0 and 7x – 6y + 9 = 0 are the legs of a right triangle. Let’s summarize

1. If two lines intersect, there is a common point between them which is the intersection of the two lines.

2. The coordinates of the point of intersection of two lines can be determined

algebraically by using specific method according to the given problem.

3. If two non vertical lines intersect, their slopes are not equal.

4. If two non vertical lines are parallel, then their slopes are equal.

5. If two non vertical lines are perpendicular, then their slopes are negative reciprocals.

What have you learned Nos. 1. – 5. Determine if the following pairs of lines are parallel, perpendicular or neither. 1. y = 2x + 3 y = 2x – 8 2. y = x – 5 y = -x + 5 3. y = 3x + 7

17

y = 2x – 1 4. 2x + y = 4 x – 2y = 5 5. 3x + 7y – 6 = 0 3x + 7y + 1 = 0 6. What is the slope of the line parallel to 5x + 4y – 3 = 0? 7. Find the slope of the line perpendicular to 4x – 2y – 7 = 0. 8. Find the equation of the line passing through (1, 3) and parallel to y = 4x + 7. 9. Find the equation of the line perpendicular to 3x + y – 7 = 0 and passing through (4, 3). 10. At what point do lines 3x – 4y = 8 and 3x + y = -2 intersect?

18


1. parallel 2. perpendicular 3. intersecting but not perpendicular 4. intersecting but not perpendicular 5. parallel 6. equal

7. 21

−

8. -6 9. 5x + y – 14 = 0 10. (2, -1)

Lesson 1 1. Equation 1 y = 3x – 4 m1 = 3 Equation 2 y = x + 7 m2 = 1 Therefore m1 ≠ m2 , so the lines are intersecting.

2. Equation 1 y = 421

−x

m1 = 21

Equation 2 y = 623

+x

m2 = 23

Since m1 ≠ m2 , so the lines are intersecting. 3. Equation 1 x + 4y = 3 4y = -x + 3

y = 43

41

+− x

m1 = 41

−

Equation 2 2x + y = 7 y = -2x + 7 m2 = -2

19

Since m1 ≠ m2 , so the lines are intersecting. 4. Equation 1 2x – y = 8 - y = - 2x + 8 y = 2x – 8 m1 = 2 Equation 2 x + 3y = 1 3y = - x + 1

y = 31

31

+− x

m2 = 31

−

Since m1 ≠ m2 , so the lines are intersecting. 5. Equation 1 3x – 5y = 4 – 5y = – 3x + 4

y = 5

453

−+

−− x

y = 54

53

−x

m1 = 53

Equation 2 x + y = 4 y = -x + 4 m2 = -1 Since m1 ≠ m2 , so the lines are intersecting. B. 1. (-1, -1) 2. (-2, 3) 3. (5, -1) 4. (4, 3) 5. (-5, 0) Lesson 2 A. 1. Equation 1 y = 2x + 5 m1 = 2

Equation 2 y = 721

−− x

m2 = 21

−

20

The two slopes are the negative reciprocals of each other, hence the two lines are perpendicular. 2. Equation 1 y = 4x – 1 m1 = 4 Equation 2 y = 4x + 3 m2 = 4 The two slopes are equal, therefore the two lines are parallel.

3. Equation 1 y = 23

25

+x

m1 = 25

Equation 2 y = 352

+− x

m2 = 52

−

Since the two slopes are negative reciprocals, then the two lines are perpendicular. 4. Equation 1 y = 3x – 7 m1 = 3 Equation 2 y = x + 8 m2 = 1 Since m1 ≠ m2 , and are also not the negative reciprocals, then the lines are neither parallel nor perpendicular. 5. Equation 1 2y = x + 4

y = 24

2+x

y = 221

+x

m1 = 21

Equation 2 y = 2x – 5 m2 = 2 Since m1 ≠ m2 , and are not negative reciprocals, then the lines are neither parallel nor perpendicular.

21

6. Equation 1 x + y = 5 y = -x + 5 m1 = -1 Equation 2 x – y = 3 -y = -x + 3 y = x – 3 m2 = 1 Since the negative reciprocal of 1 is -1, then the two lines are perpendicular. 7. Equation 1 2x + y = 9 y = - 2x + 9 m1 = -2 Equation 2 x – 2y = - 4 - 2y = -x – 4

y = 24

2 −−

+−− x

y = 221

+x

m2 = 21

Since m1 and m2 , are negative reciprocals, then the lines are perpendicular. 8. Equation 1 2x + y = 8 y = -2x + 8 m1 = -2 Equation 2 4x + 2y = -3 2y = -4x – 3

y = −− x24

23

m2 = -2 Since m1 = m2 , then the two lines are parallel. 9. Equation 1 3x + 2y - 1 = 0 2y = -3x + 1

y = +− x23

21

m1 = 23

−

Equation 2 2x – 3y + 7 = 0

22

-3y = -2x -7

y = 37

32

−−

+−− x

y = 37

32

+x

m2 = 32

Since m1 and m2 , are negative reciprocals, then the lines are perpendicular. 10. Equation 1 2x + 7y + 6 = 0 7y = -2x – 6

y = −− x72

76

m1 = 72

−

Equation 2 2x + 7y – 11 = 0 7y = -2x + 11

y = +− x72

711

y = 711

72

+− x

m2 = 72

−

Since m1 = m2 , then the two lines are parallel. B. To prove that the given quadrilateral is a parallelogram, show that the slopes of opposite sides are equal, otherwise, the quadrilateral is not a parallelogram. Step 1. Get the slopes of ,AB BC , CD and AD

mAB = 14

3473 −=

−−

= -4

mBC = 42

31−−−

= 62

−−

= 31

23

mCD = )2(3

15−−−

−

= 23

4+−

= 41

4−=

−

mAD = 33

75−−−

= 62

−−

= 31

In this quadrilateral, the opposite sides are AB and CD. Likewise, BC and AD are also opposite sides. Note that the computed slopes of the opposite sides are equal, thus the opposite sides are parallel. Therefore by definition, ABCD is a parallelogram. 2. Compute for the slopes of the opposite sides.

mLO = 4823

−−

= 41

mOV = 8635

−−−

= 28

−−

= 4

mVE = 62

)5(6−−−−

= 6256

−+−

= 41

mLE = 4226

−−−

= 28

−−

= 4

24

Since the two pairs of opposite sides of LOVE are parallel, then LOVE is a parallelogram. 3. Compute for the slopes of the sides of quadrilateral PLAN.

mPL = )3(1)2(1

−−−−

= 3121

++−

= 41

mLA = 11

)1(8−−−−

= 07− , undefined

mAN = 13

)8(6−−−−−

= 4

86−+−

= 4

2−

= 21

−

mPN = )3(3)2(6

−−−−−−

= 3326

+−+−

= 04− , undefined

From the computed values, LA and PN have equal slopes, hence they are parallel. On the other hand, the slopes of PL and AN are not equal, so they are not parallel. Therefore, PLAN is not a parallelogram. C. To verify if the vertices were that of a right triangle, show that the slopes of a pair of adjacent sides are negative reciprocals. 1. Compute for the slopes of the sides EA, AR and ER.

25

mEA = 1662

−−

= 54−

mAR = 522

−−

= 5

0−

= 0

mER = 1162

−−

= 04− , undefined

From the computed values, it is clear that the slopes of AR and ER are negative reciprocals, hence the two sides are perpendicular. Therefore, ∆EAR is a right triangle. 2. Compute for the slope of each side of the triangle.

mAC = )1(3

41−−−

−−

= 13

5+−

−

= 25

−−

= 25

mCE = 73

)5(1−−−−−

= 10

51−+−

= 104

−

= 52

−

mAE = 71)5(4

−−−−

26

= 854

−+

= 89

−

Since the slopes of two sides, AC and CE are negative reciprocals, then AC and CE are perpendicular. Hence ∆ACE is a right triangle because two of the sides form a right angle in between them. 3. Find the slopes of PE, EP and PN.

mPE = 36

)1(3−−−−

= 3

13+−

= 32

−

mEN = 36

)7(3−−−−

= 3

73+−

= 34

mPN = 33

)7(1−−−−

= 0

71+−

= 06 , undefined

Based on the computation, no two slopes are negative reciprocals, hence no sides are perpendicular. Therefore, ∆PEN is not a right triangle. D. 1. 4x – y + 7 = 0 2. 3x + y – 4 = 0 3. x – 3y – 3 = 0 4. 2x + 5y + 8 = 0 5. x + y – 3 = 0 6. x + 3y + 15 = 0 7. 2x – 5y – 16 = 0

27

E. 1. k = 8 2. Equation 1 2x – y -3 = 0 - y = -2x + 3 y = 2x – 3 m1 = 2

m2 = 21

−

Using the point slope form y – y1 = m(x – x1)

y – 7 = 21

− (x – 5 )

2(y – 7) = -1 (x – 5) 2y – 14 = -x + 5 x + 2y – 19 = 0 , the required equation 3. Some of the possible equations of the line parallel to 2x + y = 4 can be of this form. 2x + y = c, where c is any real number.

4. All equation of the line whose slope is 31 .

5. Equation 1 6x + 7y – 6 = 0 Rewrite to slope-intercept form 7y = - 6x + 6

y = 76

76

+− x

m1 = 76

−

Equation 2 7x – 6y + 9 = 0 - 6y = - 7x - 9 6y = 7x + 9

y = 69

67

+x

y = 23

67

+x

m2 = 67

Their slopes are negative reciprocals.

28


1. Parallel 2. Perpendicular 3. Neither parallel nor perpendicular 4. Perpendicular 5. Parallel

6. 45

−

7. 21

−

8. 4x – y – 1 = 0 9. x – 3y + 5 = 0 10. (0, -2)


What this module is about This module will discuss how the distance between two points can be derived by applying the Pythagorean theorem. By using the derivation of distance, this module will also define and discuss the midpoint formula. Furthermore, this module will also define and verify figures and their characteristics on the coordinate plane using the coordinate proof. This will also enhance your knowledge of distances between two points and how to get the lengths of segments and sides of polygons. What you are expected to learn

This module is written for you to

1. Derive the distance formula using the Pythagorean theorem. 2. Apply the distance formula in finding lengths of segments. 3. Verify congruence of segments by applying the distance formula. 4. Derive the midpoint formula. 5. Verify the midpoint of a segment using the distance formula. 6. Solve problems that are application of the distance and midpoint formula. How much do you know Answer the following questions as indicated.

1. If the coordinate of point X is - 2 , and the coordinate of point Y is 3, what is the length of XY ?

Find the distance between the following given pairs of points. 2. M(5, 5) and N(9, 8) 3. R(0, 6) and S(8, 0) 4. P(4, -3) and Q(4, 4)

Find the coordinate of the midpoint X of the segments whose endpoints are: 5. (3, 0), (7, 6) 6. (-2, -3), (-6, 9) 7. (8, 1), (-5, 5)

2

8. Find the perimeter of a triangle whose vertices are at the given points A(2, 3), B(5, 7), C(0, 1).

9. If A is the midpoint of MN , determine the coordinates of A if the coordinates of the endpoints are M(-4, -3) and N(5, 7)

10. The coordinates of the vertices of a quadrilateral are (6, 0), (2, 3), (3, -4) and (-1, -1). What kind of quadrilateral is formed when you connect the vertices?

What you will do

Lesson 1

The Distance Formula

When you refer to the distance between any two points on the plane, either horizontally, vertically or any other positions, then what you mean is getting the length of the segment joining the two points.

Illustrations: Horizontal Distance A B

-3 -2 -1 0 1 2 3 4 5 6

The distance between A and B, which is denoted by AB is 4 AB = │5 – 0│= │5│= 5.

Vertical Distance 3 E The distance between E and F, which is denoted by EF is EF = │3 – (-1)│ = │4│ = 4. 2 Distance between two points or length of a segment is always a unique positive real number. That is why it is necessary to use the symbol ││ for 1 absolute value. If x1 and x2 are the coordinates of A and B respectively on a 0 horizontal number line, then the distance between A and B is denoted by -1 F If y1 and y2 are the coordinates of E and F respectively on a -2 vertical number line, then the distance between E and F is denoted by

AB = │x2 – x1│

EF = │y2 – y1│

3

Examples: Find the length of the following segments:

a. CD Y b. XY c. RS

X Solution: a. Points C and D have the same y-coordinate. Therefore, distance CD is denoted by CD = │8 – 1│ = │7│ = 7

b. Points X and Y have the same x-coordinate. So distance XY can computed as XY = │2 –(-3)│ = │2 + 3│= 5

c. Points R and S have the same y-coordinate. So distance RS is RS = │-7 - 0│ = │-7│= 7

Suppose the given segment on a coordinate plane is neither horizontal nor vertical. How are you going to find the distance? Let M (x1, y1) and N(x2, y2) be two points on a Cartesian coordinate plane. Let there be another point A where an imaginary horizontal segment through M intersects an imaginary segment through N. Y N(x2,y2) M(x1,y1) A(x2,y1) X

D (8,4) C (1,4)

X (6,2)

Y (6,-3) R (-7,-4) S (0,-4)

0

-5

5

5 5

5

-5

-5

-5

4

Observe that right triangle MAN is formed on the coordinate plane. The distance between M and N is equal to the length of the hypotenuse MN of the right triangle. By the Pythagorean theorem, (MN)2 = (MA)2 + (NA)2 But in the earlier discussion, you were given the some formula how to get the distance on the horizontal number line and the vertical number line. You can just call them horizontal distance and vertical distance. Therefore if you substitute the previous formula to the above formula you will get (MA)2 = │x2 – x1│ and (NA)2 = │y2 – y1│ Putting together the formula above and the Pythagorean theorem, you will have (MN)2 = │x2 - x1│2 + │y2 – y1│2. Simplifying, you will get the distance formula. The distance between two points M(x1, y1) and N(x2, y2) is given by the formula,

This we can do since the absolute value of a number is non negative, so is the square of a number is also non negative. Examples: 1. Use the distance formula to find the length of the given segments in the coordinate plane. a. AB Y b. LP c. RS d. TU X

MN = 212

212 )()( yyxx −+−

-5

-5

-5

-5

5

5

5

5

0

A

B

S

T

U

R

P

L

5

Solutions: In each of the following segments, determine first the coordinates of the endpoints from the graphs. a. AB ; A(2, 7), B(5, 3) x1 = 2 x2 = 5 y1 = 7 y2 = 3 AB = 2

122

12 )()( yyxx −+−

= 22 )73()25( −+−

= 22 )4()3( −+

= 169 + = 25 AB = 5 b. LP ; L(-1, 5) , P(3, 0) x1 = -1 x2 = 3 y1 = 5 y2 = 0 LP = 2

122

12 )()( yyxx −+−

= 22 )50()]1(3[ −+−−

= 22 )5()4( −+

= 2516 + LP = 41 c. RS ; R(-3, 5) , S(-5, -4) x1 = -3 x2 = -5 y1 = 5 y2 = -4 RS = 2

122

12 )()( yyxx −+−

= 22 )54()]3(5[ −−+−−−

= 22 )9()35( −++−

= 22 )9()2( −+−

= 814 + RS = 85 d. TU ; T(-2, -2), U(5, -6) x1 = -2 x2 = 5 y1 = -2 y2 = -6

6

TU = 212

212 )()( yyxx −+−

= 22 )]2(6[)]2(5[ −−−+−−

= 22 )26()25( +−++

= 22 )4()7( −+

= 1649 + TU = 65 2. Draw a triangle with vertices A(1, 5), B(3, 1), C(-3, 3). Show that ∆ABC is isosceles. Solution: a. Plot the given points (vertices) on a Cartesian coordinate plan. Y X b. To show that ∆ABC is isosceles, find the length of the sides. For the triangle to be isosceles, at least two of the sides must have the same length. 1. Find AC; A(1, 5), C(-3, 3) x1 = 1 x2 = -3 y1 = 5 y2 = 3 AC = 2

122

12 )()( yyxx −+−

= 22 )53()13( −+−−

= 22 )2()4( −+−

= 416 + = 20 AC = 2 5

2. Find BC; B(3, 1) , C(-3, 3) x1 = 3 x2 = -3

y1 = 1 y2 =3

4 2

4

-2

-4

2

-2

4

6

-2

0

6

4

2

-2 2

-4

C

B

A

7

BC = 2

122

12 )()( yyxx −+−

= )13()33( 2 −+−−

= 22 )2()6( +−

= 436 + = 40 BC = 2 10 3. Find AB; A(1, 5) B(3, 1 x1 = 1 x2 = 3 y1 = 5 y2 = 1 AB = 2

122

12 )()( yyxx −+−

= 22 )51()13( −+−

= 22 )4()2( −+

= 164 + = 20 AB = 2 5 Since the length of AC equals the length of AB, then AC AB≅ . Therefore ∆ABC is an isosceles triangle. 3. Find the perimeter of a quadrilateral whose vertices are P(-2,2), Q(5, 2), R(4, -3) and S(-3, -3). What kind of quadrilateral is PQRS? Solution: a. Find PQ; the y-coordinate of P and Q is the same PQ = 2)]2(5[ −−

= 2)25( +

= 2)7( PQ = 7 b. Find QR QR = 22 )23()54( −−+−

= 22 )5()1( −+−

= 251+ QR = 26

8

c. Find RS; the y-coordinate of R and S is the same RS = 2)43( −−

= 2)7(−

= 49 RS = 7 d. Find PS PS = 22 )23()]2(3[ −−+−−−

= 22 )5()1( −+−

= 251+ = 26 The perimeter of quadrilateral PQRS = PQ + QR + RS + PS = 7 + 26 + 7 + 26 = 14 + 2 26 Based on the computed lengths of the sides, PQ = RS and QR = PS which means that the opposite sides of the quadrilateral are congruent. Therefore, PQRS is a parallelogram. 4. Show by the distance formula that the following points R(3, 5), S(0, -1) and T(1,1) are collinear. Solution: To illustrate that the given points R, S and T are collinear, you have to show that the sum of the lengths of the two short segments is equal to the length of the longer segment. a. Find RS RS = 22 )51()30( −−+−

= 22 )6()3( −+−

= 369 + = 45 RS = 3 5 b. Find ST ST = 22 )]1(1[)01( −−+−

= 22 )2()1( +

= 41+ ST = 5

9

c. Find RT RT = 22 )51()31( −+−

= 22 )4()2( −+−

= 164 + = 20 RT = 2 5 Now add ST and RT. ST + RT = 5 + 2 5 = 3 5 which means that ST + RT = RS. This conclusion satisfies the definition of betweenness and so R, S and T are collinear. To verify further, plot the three points on the Cartesian coordinate plane. 5. The endpoints of the base of an isosceles triangle are A( 1, 2) and B(4, -1). Find the y-coordinate of the third vertex if its x-coordinate is 6. Solution:

Let C(6, y) be the coordinates of the third vertex.

Since ∆ABC is isosceles, and AB is the base, then AC = BC. AC = 22 )2()16( −+− y

BC = 22 )]1([)46( −−+− y But AC = BC, therefore to solve for y, equate the values of AC and BC. 22 )2()16( −+− y = 22 )]1([)46( −−+− y Y

22 )2(5 −+ y = 22 )1(2 ++ y

Squaring both sides 25 + y2 – 4y + 4 = 4 + y2 +2y +1 5 5 – 6y = -24 y = 4

So the third vertex is C(6, 4). The triangle is shown in the figure at the right. -5 -5

X

-2 5 10

5 10

A (1,2)

C (6,y)

B (4,-1)

10

6. A point A(x, 1) is 29 units from B(8, 3). Find x. Solution: AB = 22 )13()8( −+− x = 29

22 21664 ++− xx = 29

64 – 16x + x2 + 4 = 29 Squaring both sides of the equation

x2 - 6x + 39 = 0

(x – 13) (x - 3) = 0 x – 13 = 0 x – 3 = 0 x = 13 x = 3 There are two values of x. Therefore the two points are (13, 1) and (3, 1) 7. Three of the vertices of a square are points A(2, 4), B(-2, 4), C(-2, 0). Find the fourth vertex D(x, y). Solution: Plot the points on the coordinate plane. Y X Try this out A. Find the distance between the following pairs of points. 1. (0, 4 ), ( 0, 6 ) 2. ( 2, -1), ( 7, -1 )

B(-2,4)

C(-2,0)

A(2,4)

D(x, y)

11

3. ( 4, -3 ), ( -7, -3 ) 4. ( 1, 5 ), ( 3, 8 ) 5. ( -4, -7 ), ( 0, 5 ) 6. ( 2, 8 ), (-5, -1) 7. ( -5, 4 ), ( -3, -3 ) 8. ( 6, 2 ), ( 5, -2 ) 9. ( -1, 6 ), ( 5, -1 ) 10. ( -4, -5 ), ( 6, 0 ) B. Find the perimeter of the polygons with vertices at the given points. 11. ( 1, 2 ), ( 4, 6 ), ( 7, 2 ) 12. ( -1, 7 ), ( -1, 1 ), ( -9, 1) 13. ( 2, -2 ), ( -1, -5 ), ( -3, -1) 14. ( 2, -6 ), ( 2. -9 ), ( -3, -6 ), (-3, 9) 15. ( 4, -1 ), ( 7, -2 ), ( 5, -6), (2, -5) 16. ( -2, 4 ), ( 0, 6 ), ( 2, 4 ), (0, 0) 17. ( -5, -4 ), ( -3, -6 ), ( -5, -9 ), ( -8, 8 ), ( -8, -5 ) C. Given the distance(d) between two points D and F and the coordinates of one of the endpoints. Find the coordinates of the other endpoint if either x or y coordinate is given. 18. d = 13, D(-4, 1), F( x, -4) 19. d = 7, D( 5, 0), F( 1, y) 20. d = 5, D(-4, y), F(0, -3) 21. d = 5 , D(x, 4), F(3, 5) D. Solve the following problems: 22. Draw a triangle with vertices ( 6, 3 ), ( 2, 7 ), ( 10, 7 ). What kind of triangle is it?

23. Use the distance formula to show that (3, 0), (0, 4), ( 6, -4) are collinear.

24. The distance from (5, 7) to (x, 2) is 34 . Find all possible values for x.

25. Find the fourth vertex S of a rectangle whose three vertices are p(-3, 2), Q(-3, 7) and R(2, 7).

26. The point (5, y) is 17 units from (6, 2). Find y.

27. A line segment 5 units long has one of its ends at (3, 1). The y-coordinate of the other end is 5. Find its x-coordinate.

12

Lesson 2

The Midpoint Formula The midpoint of a segment is a point that divides a segment into two (2) congruent segments. Illustrations: 1. C is the midpoint of AB . Then A B AC BC≅ . So AC = CB. X 2. If M is the midpoint of XY then M XM MY≅ , thus XM = MY. T S 3. S is the midpoint of XY Y Therefore, RS ST≅ , and RS = ST. R From the illustrations given, you can say that the midpoint of a segment should lie between the endpoints of the segments and the three points must be collinear. How do you get the midpoint of the segment on the coordinate plane? Consider the coordinate plane below and the segments on it.

C

X(-2,3) Y(6,3)

Q(-4,-3)

P(-4,2) M

R

Y

X

5

-5 -5

-5

-5

5 5

5

0

13

If M and R are the midpoints of XY and PQ respectively, how do your determine the coordinates of M and R?

To get the midpoint of XY , you have to consider that the segment is horizontal, thus the y-coordinate is the same. Since point M lies between the two endpoints and in the middle, the x-coordinate of M is the average of the x-coordinates of the two endpoints of the segment and the y-coordinate is 3. So the coordinates of M is

M(xm, ym) =

+− 3,

262

=

3,

24

= (2, 3)

To get the coordinates of R, consider PQ . Since the segment is vertical, the two endpoints have the same x-coordinate which is -4. To get the y-coordinate, get the average of the y-coordinates of the two endpoints. That is

R(xm, ym ) =

−+−

2)3(2,4

=

−−

21,4

Therefore, to get the coordinates of the midpoint of horizontal segment the formula

below is used. Since the two endpoints have the same y – coordinate the midpoint (M) is

M

+ yxx ,

221

For vertical segments, since the two endpoints have the same x-coordinate, then the

formula is

M

+

2, 21 yyx

How do you get the coordinates of the segment on the coordinate plane which is

neither horizontal nor vertical? AB illustrated below is neither horizontal nor vertical.

14

The endpoints of AB are A(2, 1) and B(4, 7). Let M be the midpoint of AB . To

determine the coordinates of M, draw horizontal segment passing through A and a vertical segment passing through B. The two segments intersect at a point whose coordinates are (4, 1). Get the coordinates of the midpoint H of the horizontal segment.

H

+ 1,

242 or H(3, 1)

Then get the coordinates of the midpoint V of the vertical segment .

V

+

271,4 or V(4, 4).

The points H and V suggest that the midpoint of M are (3, 4). To check if M is really the midpoint of AB , we have to show that AM = MB.

AM = 22 )14()23( −+−

= 22 31 + = 91+ = 10 MB = 22 )47()34( −+−

= 22 31 + = 91+ = 10 Since AM = MB = 10 , then M is really the midpoint of AB . For segments on the coordinate plane which are neither horizontal nor vertical, the formula for finding its midpoint M is given below.

H

(4,1)

B(4,7)

A(2,1)

M V

X

Y

0

15

The Midpoint Formula:

If A(x1, y1) and B(x2, y2) are any two points in a coordinate plane, then the midpoint M of AB has coordinates

Examples: 1. Find the coordinates of the midpoint M of the segments whose endpoints are a. (3, 5), (7, 1) b. (0, 1), (-4, 3) c. (-3, -6), (2, -11) d. (4, -1), (-7, 3) Solutions: Using the midpoint formula, a. (3, 5), (7, 1) x1 = 3, x2 = 7 y1 = 5, y2 = 1

M

++

251,

237 = M

26,

210

= M(5,3) b. (0, 1), (-4, 3) x1 = 0 , x2 = -4 y1 = 1 , y2 = 3

M

++−

213'

204

M(-2, 2) c. (-3, 6), (2, 11) x1 = -3, x2 = 2 y1 = 6 , y2 = 11

M

−+−+−

2)11(6,

223

M

−−

217,

21

M

++

2,

22121 yyxx

16

d. (4, 1), (-7, 3) x1 = 4 , x2 = -7 y1 = 1 , y2 = 3

M

+−+

213,

2)7(4

M

−=

− 2,

23

24,

23

2. M(-1, -3) is the midpoint of ST , If the coordinates of S are (-3, 2), find the coordinates of T. Solution: Step 1. Let T have coordinates (x, y). By the midpoint formula, the midpoint of ST is

+−+

22,

2)3( yx =

+−

22,

23 yx

Step 2. We are given that the coordinates of M is (-1, -3) and M is the midpoint of ST . Therefore,

2

3−x = -1

x – 3 = -1(2) x – 3 = -2 x = -2 + 3 x = 1

and 2

2+y = -3

y + 2 = 2(-3) y + 2 = -6 y = -6 -2 y = -8 So the coordinates of T are (1, -6) 3. One endpoint P of segment PS and its midpoint R are given. Use the midpoint formula to find the coordinates of the second segment S.

a. P(3, -4), R(0, 0) b. P(2, 5), R(5, -1) c. P(-6, -3), R(0, 1)

17

Solutions: a. Let S have the coordinates (x, y). Using the midpoint formula, the midpoint of PS

is given as

−+

24,

23 yx

The coordinates of midpoint R are (0, 0)

02

3=

+x

x + 3 = 0 x = -3

02

4=

−y

y -4 = 0 y = 4 The coordinates of R are (-3, 4) b. Coordinates of S are (x, y). Applying the midpoint formula, the coordinates of R

is

++

25,

22 yx

The coordinates of midpoint R are ((5, -1)

52

2=

+x

x + 2 = 10 x = 8

12

5−=

+y

y + 5 = -2 y = -7 The coordinates of R are ( 8, -7) c. Coordinates of S are (x, y). Using the midpoint formula, the coordinates of

midpoint R is given as

−+−+

2)3(,

2)6( yx

The coordinates of midpoint R are (0, 1). Solving for x and y,

02

6=

−x

x – 6 = 0 x = 6

18

12

3=

−y

y – 3 = 2 y = 2 + 3 y = 5 Therefore the coordinates of S are (6, 5) 4. The vertices of ∆XYZ are X(1, 4), Y(6, 2) and Z(-2, -1). Find the length of the median to ZY . Solution: The median of a triangle is a segment joining the vertex and the midpoint of the opposite side . Let the midpoint of XY be point P. Step 1. Get the coordinates of P

P

−−

212,

226

P

21,

24

P

21,2

Step 2. Find the length of XP . X(1, 4), P

21,2

XP = ( )2212 4)12( −+−

= ( )22721 −+

= 4

491+

0

Y(6,2)

X(1,4)

Z(-2,-1)

Y

X 0 P

19

= 453

XP = 253

Hence, the length of median XP is 253 .

5. Find the perimeter of a the triangle formed by joining the midpoints of the sides of a triangle whose vertices are P(-4, 0), Q(2, 3) and R(5, -2). Solution: Step 1. Let A, B and C be the midpoints of PQ , QR and PR . Get the coordinates of each midpoint. For the coordinates of A

P(-4, 0), Q(2, 3)

A

++−

203,

224

= A

−

23,

22

= A

−

23,2

For the coordinates of B Q(2, 3), R(5, -2)

B

−+

223,

252

= B

21,

27

For the coordinates of C P(-4, 0), R(5, -2)

C

−+−

220,

254

= C

−1,

21

0

R(5,-2)

P(-4,0)

Q(2,3)

A

B

C

X

Y

0

20

Step 2. Find the lengths of AB , BC and AC .

AB = 22

21

23

271

−+

−−

= 22

22

29

+

−

= 21481

+

= 4

85

AB = 285

BC = 22

211

27

21

−−+

−

= 22

23

26

−+

−

= 49

436

+

= 445

BC = 2

53

AC = 22

)1(23

211

−−+

−−

= 22

123

23

++

−

= 2

25

49

+

= 425

49+

= 4

34

AC = 234

21

Step 3. Get the sum of the lengths AB + BC + AC

Perimeter of ∆ABC = 285 +

253 +

234

= 2

345385 ++

Try this out A. What are the coordinates of the midpoint of the segment joining each pair of points.

1. (0, 0), (6, 0) 2. (0, 0), (-7, 0 3. (0, 0), (0, -8) 4. (1, 3), (5, 7) 5. (5, -1), (-1, -7) 6. (-8, -2), (0, 0) 7. (-3, -4), (3, -3) 8. (6, -1), (-4, 7) 9. (-1, -1), (-8, -9) 10. (a, b), (c, d)

B. If M is the midpoint of AB , determine the coordinates of B.

11. A(3, 7), M(3, 0) 12. A(5, 2), M(-1, -1) 13. A(-4, -1), M(5, 2) 14. A(3, -4), M(5, 2) 15. A(-5, 6), M(7, 2) 16. A(0, -8), M(4, -4) 17. A(-1, 4), M(1, 1) 18. A(7, 0), M(0, 9) 19. A(-3, -5), M(3, -7) 20. A(a, b), M(c, d)

C. Solve the following problems: 21. Find the coordinates of the midpoint of each side of a triangle with vertices at (3, 5), (6,-4) and (-1, 1).

22. Find the coordinates of the midpoint of each side of a quadrilateral with vertices at (-2, -4), (7, -8), (4, -3) and (_5, 3).

23. Find the length of the median to side RP of ∆RPQ whose vertices are R(-3, 2), P(3, -3) and Q(-1, 6).

22

24. Find the length of each median of a triangle with vertices at (-1, 6), (-3, -2) and (7, -4).

25. A rectangle has vertices R(-3, 4), S(-3, -4), T(2, -4) and U(2, 4). Show that its diagonals have the same midpoint.

26. Use the distance formula to show that X(1, -1) is the midpoint of the segment with endpoints A(4, 1) and B(-2, -3).

27. Given R(5, 2), S(a, -2) and T(-3, b). Find a and b so that S is the midpoint of RT .

28. Show that the points (-1, -2), (2, 1) and (-3, 6) are the vertices of a right triangle. Use the distance formula. 29. Given A(7, 1), R(2, x) and B(-x, 5), find x so that R is the midpoint of AB .

30. Find the perimeter of the triangle in no. 28.

Let’s summarize

1. The distance between two points on the plane is the length of the segment joining the two points. For horizontal distance between points A and B, the formula to be used is

AB = │x2 – x1│, where the y-coordinate is the same.

For vertical distance between points A and B, the distance is denoted by

AB = │y2 – y1│, where the x-coordinate is the same.

2. The distance between two points A(x1, y1) and B(x2, y2) is given by the formula

AB = 212

212 )()( yyxx −+−

3. The midpoint of a segment is a point that divides the segment into two congruent segments.

4. The midpoint of a horizontal segment can be determined by the formula

M = 2

21 xx +

5. The midpoint of a vertical segment is determined by the formula

M = 2

21 yy +

6. The midpoint M of a segment whose endpoints are A(x1, y1) and B(x2, y2) is given

by the formula M

++

2,

22121 yyxx

23

What have you learned Answer the following questions as indicated.

1. What is the length of PQ? P Q

-4 -3 -2 -1 0 1 2 3 4 5 Find the length of each side of a triangle whose vertices are J(0, 3), K(-4, 0) and L(1, -1). 2. JK 3. KL 4. JL What are the coordinates of the midpoint of a segment whose endpoints are: 5. A(7, 3), and X(1, -11)

6. R(-6, 1) , and S(1, -10) 7. Find the perimeter of ∆XYZ in the figure if X, Y and Z are the midpoints of FD , DE and EF respectively. 8. M is the midpoint of PR . If the coordinates of M and P are given, find the coordinates of R. P(-3, 7) and M(1, 1)

9. If the length of RS is 29 , and the coordinates of R are (-5, 1), find the x-coordinate of S if its y-coordinate is 3.

10. Find the coordinate of the intersection of the diagonal of a rectangle whose vertices are M(2, 7), N(6, 3), O(-1, -4) and P(-5, 0)

Y(2,3)

Z(1,-1)

X(-1,1)

D

E

F

X

Y

24


1. 5 2. 5 3. 10 4. 7 5. (5, 3) 6. (-4, 3)

7.

3,

23

8. 5 + 2 2 + 61

9.

2,

21

10. From the figure, you can easily conclude that the quadrilateral is a parallelogram. To determine what kind of parallelogram it is, compute for the slope of any two adjacent sides and compare. Their slopes are the negative reciprocal of each other.

Try this out Lesson 1 A. 1. 2 2. 5

3. 11 4. 13 5. 4 10 6. 130 7. 53 8. 17 9. 85 10. 5 5

B. 11. 16

12. 24 13. 265223 ++ 14. 3410518 ++

(2,3)

(-1,-1)

(3,-4)

(6,0) X

Y

25

15. 10254 + 16. 5424 + 17. 13102981322 ++++

C. 18. x = -16 ; x = 8 19. y = 33± 20. y = -6 ; y = 0 21. x = 5 ; x = 1

D. 22. the triangle is isosceles

23. Let A(3, 0) , B(0, 4) and C(6, -4) be the given points. Find the lengths, AB, BC and AC. Solution: AB = 22 )04()30( −+−

= 22 43 + = 169 + = 25 AB = 5 BC = 22 )44()06( −−+−

= 22 )8(6 −+

= 6436 + = 100 BC = 10 AC = 22 )04()36( −−+−

= 22 )4(3 −+

= 169 + = 25 AC = 5 AB + AC = 5 + 5 = 10 = BC. By definition of betweenness, A is between B and C. Therefore, A, B and C are collinear.

(2,7) (10,7)

(6,3)

Y

X 0

26

24. x = 8; x = 2 25. S(2, 2) 26. y = 6; y = -2 27. x = 6; x = 0

Lesson 2 A. 1. (3, 0)

2.

− 0,

27

3. (0, -4) 4. (3, 5) 5. (2, -4) 6. (-4, -1)

7.

−

27,0

8. (1, 3)

9.

−− 5,

29

10.

++

2,

2dbca

11. (3, -7) 12. (-7, -4) 13. (14, 5) 14. (7, 8) 15. (19, -2) 16. (8, 0) 17. (3, -2) 18. (-7, 18) 19. (9, -9) 20. (2c – a, 2d – b)

21.

21,

29 , ( )3,1,

23,

25

−

22.

−−

−

−

−

21,

27,0,

21,

211,

211,6,

25

23. Length of the median is 2173

27

24. Length of the medians are 53 , 103 and 117

25. Midpoint of RT =

− 0,

21

Midpoint of SU =

− 0,

21

26. AX = 13 , BX = 13

Since AX = BX, therefore, X is the midpoint of AB.

27. a = 1, b = -6 28. s1 = [ ] [ ]22 )2(1)1(2 −−+−−

= 22 33 + = 99 + = 18 s1 = 3 2 s2 = ( ) ( )22 1623 −+−−

= ( ) 22 )5(5 +−

= 2525+ = 50 s2 = 25 s3 = [ ] 22 )62()3(1 −−+−−−

= 22 )8(2 −+

= 644 + = 68 s1

2 + s22 = ( )223 + ( )225

= 18 + 50 = 68 = s32

The square of s3 is equal to the sum of the squares of s1 and s2. 29. x = 3 30. P = s1 + s2 + s3

= 1722523 ++ = 17228 +

28

What have you learned.

1. │PQ│= 5 2. 5 3. 26 4. 17 5. (4, -4)

6.

−−

29,

25

7. 171322 ++ 8. R(5, -5) 9. x = 0; x = -10

10.

23,

21



This module will show you a different kind of proving. The properties of triangles and quadrilaterals will be verified in this module using coordinate plane and the application of the lessons previously discussed in the other modules.

Furthermore, you will also have the chance to do analytical proof and compare it with

the geometric proof that you have been doing since the start of the year. Included in this module are lessons that will discuss in detail properties/relationships of circles with other figures in the coordinate plane.


This module is written for you to

1. Define and illustrate coordinate proof. 2. Verify some properties of triangles and quadrilaterals by using coordinate proof. 3. Determine the difference between geometric proofs and coordinate proof. 4. Illustrate the general form of the equation of the circle in a coordinate plane. 5. Derive the standard form of the equation of the circle from the given general form. 6. Find the coordinate of the center of the circle and its radius given the equation . 7. Determine the equation of a circle given its

a. center and radius b. radius and the point of tangency with the given line

8. Analyze and solve problems involving circles. How much do you know Answer the following questions

Tell which of the axes placements will simplify a coordinate proof for a theorem involving the figure shown.

2

1. a b c 2. a b c 3. a b c Give the center and radius of the following circles:

4. x2 + y2 = 16 5. x2 + y2 – 25 = 0 6. (x – 3)2 + (y + 1)2 = 36 7. x2 + y2 – 4x + 10y + 16 = 0 8. Give the standard form of the circle whose center is at (2, -1) and a radius of 7 units.

Write the equation of the circle satisfying the following conditions. 9. The line segment joining (4, -2) and (-8, -6) is a diameter. 10. The center is at (0, 5) and the circle passes through (6, 1).

3

What you will do

Lesson 1

Coordinate Proof

The coordinates of a point on the coordinate plane are real numbers, thus it is possible to prove theorems on geometric figures by analytic or algebraic method. We call this proof the coordinate proof. In writing coordinate proof, some suggestions have to be taken into considerations. First, we may choose the position of the figure in relation to the coordinate axes. If the figure is a polygon, it is always simpler to put one of the sides on either the x-axis or the y-axis. Second, we may place one of the vertices on the origin. Third, the essential properties of the given figure should be expressed by the coordinates of key points. The proof is accomplished by setting up and simplifying algebraically equations and relations involving these coordinates. Fourth, the figure should never be made special in any way so that the proof will be general and can be applied to all cases. Except for zero (0), numerical coordinate should not be used in the proof. Example 1. Prove analytically that the line segment joining the midpoints of two sides of a

triangle is parallel to the third side and measures one-half of it. Solution:

y A(a, t) X Y X O(0.0) B(b,0)

4

a. Analysis: The figure shows the triangle appropriately placed on rectangular coordinate plane. We have to show two properties here.

1) XY ║OB and (2) XY = 2

1 OB.

b. Since X and Y are midpoints, then the midpoint formula can be used to get their

coordinates. The coordinates of X are

2,

2ta and that of Y are

+

2,

2tba .

The slope m1 of OB = 0000=

−−b

. The slope m2 of XY is

m2 =

22

22baa

tt

+−

− = 0

2

0=

− b

Since the slope of OB = 0 and that of XY = 0, then the two slopes are equal.

Therefore, XY ║OB . Therefore, the segment joining the midpoints of the two sides of a triangle is parallel to the third side.

c. For the second conclusion, get the length of XY and OB.

XY = 22

2222

−+

−

+ ttaba

= 2

2

b

= 2b

OB = 22 )00()0( −+−b

= 2b = b

The computation showed that XY = 2b and OB = b. Therefore, XY =

21 OB.

Hence, the segment joining the midpoints of the two sides of a triangle is one-half the third side.

5

Example 2. Prove that the diagonals of a rectangle are equal.

Given: ABCD is a rectangle Prove: AC = BD Proof:

In this problem we are given not just any quadrilateral but a rectangle. Place one vertex on the origin. In the figure, it was vertex A which was in the origin. Then vertex B on the x-axis, and vertex D on the y-axis.

The following are the coordinates of the vertex : A(0, 0), B(b, 0), C(b, c) and D(0, a).

AC and BD are the two diagonals. We have to determine if the length of AC is equal to the length of BD. AC = 22 )0()0( −+− cb = 22 cb + BD = 22 )0()0( −+− cb = 22)( cb +−

= 2cb + Therefore, from the computed distances or lengths, AC = BD. Hence, the diagonals of rectangle are equal.

y D(0, c) C(b, c) x A(0,0) B(b, 0)

6

Example 3. Prove that the diagonals of a square are perpendicular to each other. Given: HOPE is a square Prove: HP ⊥ OE Proof:

To show the perpendicularity of HP and OE , then their slopes must be the negative reciprocal of each other. Since the vertices of the square were placed on the coordinate plane in such a way that makes proving simpler, then we are now ready to find the slopes of both HP and OE . Computing for the two slopes ,

m HP = 100

==−−

aa

aa

m OE = 10

0−=

−=

−−

aa

aa

From the results you can easily see that the reciprocals of HP and OE are 1 and – 1 which are negative reciprocals.

Therefore, HP and OE are perpendicular to each other.

y E(0, a) P(a, a) H(0,0) O(a, 0)

7

Try this out A. Determine the coordinates of A, B or C. B. Given the following statements, do the following: a. Use the given figure on the rectangular coordinate plane. b. Write the hypothesis. c. Write what is to be proven. d. Prove analytically.

1. y MAIN is a rectangle 2. y BEAM is a parallelogram M(c,b) B

M(0,b) A

A(0,0) E(a,0) N(0,0) I(a, 0) y y 3. LEAD is an isosceles trapezoid 4. ABC is an isosceles triangle C

L(b,c) E(a-b, c) C F x x B(0,0) A(a, 0)

D(0,0) A(a, 0)

8

1. The medians to the legs of an isosceles triangle are equal. y E (a, b)

D F

D F

x A (0, 0) C (2a, 0)

2. The diagonals of an isosceles trapezoid are equal. y D (b, c) C (a-b, c)

x A (0, 0) B (a, 0) 3. The diagonals of a parallelogram bisect each other. y E (b, c) P (a+b, c)

x H (0, 0) O (a, 0)

9

4. The midpoint of the hypotenuse of a right triangle is equidistant from the vertices. y C (0, 2b)

E x A (0, 0) B (2a, 0) 5. The segments joining the midpoints of the opposite sides of a quadrilateral bisect each other. y K (4d, 4e) B

E (4b, 4c)

F

C

x M(0, 0) D A (4a, 0)

Lesson 2

Circles on Coordinate Plane

You know from the previous chapter that the set of all points equidistant from a fixed point is called a circle. The fixed point is called the center. If the circle is on the Cartesian coordinate plane, the distance formula will lead us to write an algebraic condition for a circle.

10

r

y In the figure, if P has coordinates (x, y), using the distance formula will give us 22 )0()0( −+− yx = r

22 yx + = r Squaring both sides, we get x x2 + y2 = r2 Thus the equation of a circle with center at (0, 0) and radius r in standard form is x2 + y2 = r2

Example 1. Find the equation of the circle whose center is the origin and the radius is a. r = 3 b. r = 1 c. r = 5 d. r = 2 2 Solution: a. C(0, 0) r = 3 The equation of the circle is x2 + y2 = 32 x2 + y2 = 9 b. C(0, 0) r = 1 The equation of the circle is x2 + y2 = 12 x2 + y2 = 1 c. C(0, 0) r = 5 The equation of the circle is x2 + y2 = 52 x2 + y2 = 25 d. C(0, 0) r = 2 2 The equation of the circle is x2 + y2 = (2 2 )2 x2 + y2 = 8

(x,y)

c (0,0)

11

Example 2. Determine the center and the radius of the circle whose equations are given. a. x2 + y2 = 49 b. x2 + y2 = 36 c. x2 + y2 = 18 Solutions: The equations are of the form x2 + y2 = r2, hence the center is the origin. The radius in each circle is

a. r = 9 b. r = 6 c. r = 3 2

Not all circles on the rectangular coordinate plane has its center at the origin. In the

given circle below, its center is not the origin. We will represent the center of the circle which is not the origin as (h, k).

y x

In the coordinate plane given above, center C has coordinates (h, k) and radius r. By Pythagorean Theorem, the distance from the center of the circle to a point A(x, y) on the circle can be solved. This will also give us the standard form of the equation of a circle. AC = ( ) ( )22 kyhx −+− = r (x - h)2 + (y – k)2 = r2

The standard form of the equation of the circle with center at (h, k) and radius r is (x – h)2 + (y – k)2 = r2

r

(x,y)

c (h,k)

12

r

Example 1. Find the equation of a circle with center at (1, 5) and a radius of 3 units. Solution: Substitute the following values in the standard form h = 1, k = 5, r = 3 The equation is (x – 1)2 + (y – 5)2 = 32 (x – 1)2 + (y – 5)2 = 9 Example 2. Find the equation of a circle with center at (2, -3) and radius of 5 units. Sketch the figure. h = 2, k = -3, r = 5 The equation is (x – 2)2 + (y + 3)2 = 52 (x – 2)2 + (y + 3)2 = 25 The figure is y x

The standard form of the equation of a circle with center at C(h, k) and radius r can be presented in another form. This is done by squaring the binomials and simplifying the results. (x – h)2 + (y – k)2 = r2 x2 – 2hx + h2 + y2 – 2yk + k2 = r2 x2 + y2 – 2hx – 2yk + h2 + k2 – r2 = 0 By assigning capital letters D, E and F to represent the constants, the equation will now assume this general form. x2 + y2 + Dx + Ey + F = 0

C (2,-3)

13

Example 1. Find the radius and the center of the circle given its equation. x2 + y2 – 4x – 6y -12 = 0 Solution: First isolate the constant term at the right side of the equal sign by applying the addition property of equality x2 + y2 – 4x – 6y – 12 = 0 x2 + y2 – 4x – 6y = 12 Then group the terms with x together and those with y together. (x2 - 4x)+ (y2 - 6y) = 12 Complete each group like in completing the square by adding the third term of the trinomial. Note that what you added to each group should be added to the right side of the equation also. (Application of addition property of equality) (x2 - 4x + 4 ) + (y2 – 6y + 9 ) = 12 + 4 + 9 (x2 - 4x + 4 ) + (y2 – 6y + 9 ) = 25 Rewrite each perfect trinomial square into binomial factors. (x – 2)2 + (y – 3)2 = 25 (x – 2)2 + (y – 3)2 = 52

Since the equation is in center-radius form, then we can determine the coordinates of the center and the radius of the circle. The center is at (2, 3) and the radius is 5 units. Example 2. Find the radius and the coordinates of the center of the circle given the Equation x2 + y2 + 6x – 2y + 6 = 0 Solution: Isolate first the constant term x2 + y2 + 6x – 2y = -6 Then group the x and y together (x2 + 6x) + (y2 – 2y) = - 6 Add constants to each group by completing the square. Add to the right side of the equation what you will add to the left side. (x2 + 6y + 9) + (y2 – 2y + 1) = -6 + 9 + 1 (x2 + 6y + 9) + (y2 – 2y + 1) = 4

14

Write each trinomial as factors or square of binomial.

(x + 3)2 + (y – 1)2 = 4

x + 3)2 + (y – 1)2 = 22

The equation is in center-radius form. So the center of the circle is at (-3, 1) and its radius is 2 units. Example 3. Tell whether the equation x2 + y2 -4x + 8y + 24 determines a circle. Solution: To solve this problem, you should be aware of these fact. The existence of a circle depends on the value of r2. In the standard form of equation of the circle, (x – h)2 + (y – k)2 = r2, one of the following statements is always true if r2 > 0, the graph is a circle. r2 = 0, the graph is a point (We call this the point circle) r2 < 0, the graph or the circle does not exist In the given example, solve for the value of r2. In doing this, you simply follow the procedure in examples 1and 2. x2 + y2 - 4x + 8y + 24 = 0

x2 + y2 – 4x + 8y = - 24

(x2 – 4x) + (y2 + 8y) = - 24

(x2 – 4x + 4 ) + (y2 + 8y + 16 ) = -24 +4 + 16

(x – 2)2 + (y + 4)2 = -4 Since r2 = -4, then the circle does not exist. The following examples discuss of problems that involve circles in the coordinate plane. Each problem is treated differently according to what is given and what is being asked .

15

Example 4. Find the equation of a circle whose center is at (4, 2) and is tangent to y-axis. Sketch the figure. y y x Solution: Since the circle is tangent to y-axis, the radius of the circle is perpendicular to y-axis. It also means that the length of the radius is also the length of the perpendicular segment from the center of the circle to y –axis. From the figure, you can determine that the point of tangency is at (0, 2). To find the length of the radius, use the distance formula. r = ( ) ( )22 2204 −+−

= 042 +

= 16

r = 4 To solve for the equation , use the coordinates of the center (4, 2) and the computed length of radius r = 4. (x - h)2 + (y – k)2 = r2

(x – 4 )2 + (y – 2)2 = 42

x2 – 8x + 16 + y2 – 4y + 4 = 16

x2 + y2 – 8x – 4y + 4 = 0

r

C(4,2)

16

Example 5. Write the equation of the circle with the given condition. (10, 8) and (4, -2) are the endpoints of the diameter. Sketch the figure. y x

Solution:

In a circle, the radius is one-half of the diameter. Since the given are the endpoints of the diameter, then the center of the circle is the midpoint of the diameter.

M

−++

2)2(8,

2410

M

26,

214

M ( )3,7 The next step is to get the length of the radius. Since radius is one-half of the circle, so get the distance from the center to one of the endpoint of the diameter. Any endpoint will do. r = ( ) ( )22 38710 −+−

r = 22 54 +

r = 2516 +

r = 41

●C

(10,8)

(4,-2)

17

To find the equation of the line, use C(7, 3) and r = 41 (x – 7)2 + (y – 3)2 = ( 41 )2 x2 – 14x + 49 + y2 – 6y + 9 = 41 x2 + y2 – 14x – 6y + 17 = 0 Example 6. Write the equation of the circle with center at (-8, 5) and passing through

A(- 6, 4). Solution: Since the circle is passing through A, then the distance from the center to A is the length of the radius of the circle. Compute first the radius of the circle. r = ( )[ ] ( )22 4568 −+−−− . r = 22 1)2( +− r = 14 + r = 5 The equation of the circle is (x + 8)2 + (y – 5)2 = ( 5 )2 x x2 + 16x + 64 + y2 – 10y + 25 = 5 x2 + y2 + 16x – 10y + 84 = 0 y Try this out. A. Determine if the following are equations of a circle, a point or a circle that does not exist. 1. x2 + y2 = 3 2. x2 + y2 - 12 = 0 3. x2 + y2 + 121 = 0 4. (x – 5)2 + y2 = 1 5. x2 + y2 – 10x – 8y + 41 = 0 B. Give the center and the radius of each circle. 1. x2 + y2 = 25 2. x2 + y2 - 12 = 0 3. x2 + (y – 3)2 = 121

C (-8,5)

(-6,4)

18

4. (x – 7)2 + (y – 5)2 = 18 5. (x + 1)2 + (y – 4)2 = 3 6. (x – 8)2 + y2 = 49

7. 2

21

+x + (y – 7)2 = 25

8. x2 + y2 + 6x + 16y – 11 = 0 9. x2 + y2 + 2x – 6y – 8 = 0 10. x2 + y2 – 4x – 12y + 30 = 0 C. Write the equation of a circle in standard form with center C and radius r given. 1. C(0, 0), r = 4 2. C(0, 0), r = 2 3 3. C(1, 1), r = 3 4. C(-2, -5), r = 10 5. C(-3, 4), r = 2 2 6. C(2, -5), r = 5 7. C(0, 6), r = 6 8. C(-4, 0), r = 3.5 9. C(0, -5), r = 5 10. C(3, 0), r = 3 3 D. Solve the following problems. Sketch the figure. Show the complete solution. 1. Write the equation of the circle with center at (3, -1) and tangent to the x-axis. 2. Write the equation of a circle with center at (2, 5) and passing through (-2, 1). 3. The line segment joining ( -2, 5) and (-2, -3) is a diameter. 4. A circle is tangent to both axes and the radius at the first quadrant is 3 units. 5. A circle is tangent to the line 3x – 4y = 24 and the center is at (1, 0).

Lets summarize

1. Proving theorems in geometry analytically is also known as coordinate proof. 2. In coordinate proof, the location of one of the vertices and one of the sides of the

figure can help in proving easily the theorem. 3. The set of all points equidistant from a fixed point is called a circle. The fixed

point is called the center. 4. The standard form of equation of a circle with center at the origin and radius r is

x2 + y2 = r2

5. The standard form of equation of a circle with center at (h, k) and radius r is (x – h)2 + (y – k)2 = r2

19

6. The general form of equation of a circle is x2 + y2 + Dx + Ey + F = 0

7. The existence of a circle depends upon the value of r2.

If r2 > 0, the circle exist. If r2 = 0, point circle exist. If r2 < 0, the circle does not exist.


1. y D(0, b) C A(0, 0) B(2a, 0) x ABCD is a rectangle, What is the coordinate of C ?

2. In figure in # 1, find the length of AC . 3. In doing coordinate proof, it is always simpler to put one of the vertices of the polygon

or figure on the ___________. 4. What is the standard form of the equation of a circle if the center is the origin and the

radius is r units? What is the center and the radius of the following circles given their equations?

5. x2 + y2 = 64 6. x2 + (y + 1)2 = 25 7. (x – 6)2 + y2 = 1 8. (x – 3)2 + (y + 7)2 = 12 9. (x – 1)2 + (y +1)2 = 49 10. What is the equation of the circle whose center is at (-4, -1) and passing through

the origin? Sketch the figure. Express the answer in general form.

20

Answer Key How much do you know.

1. c 2. a 3. b 4. C(0,0) or origin, r = 4 5. C(0, 0) or origin, r = 5 6. C(3, -1), r = 6 7. C(2, -5), r = 13 8. (x – 2)2 + (y + 1)2 = 49 9. x2 + y2 + 4x + 8y – 20 = 0 10. x2 + y2 – 10y – 27 = 0

Lesson 1 A. 1. A(a, b) 2. B(a + c, b)

3. C

2,

2cb

4. C

ca ,

2

B. 1. y E (a, b)

D F

A(0, 0) C (2a, 0) x Given: ACE is an isosceles triangle. AF and DC are medians Prove: AF = DC Solution: 1. First determine the coordinates of D and F.

21

Since AF and DC are medians, then D and F are midpoints.

D

++

20,

20 ba

D

2,

2ba

F

+

2,

22 baa

F

2,

23 ba

2. After finding the coordinates of D and F, determine the length of AF and DC .

AF = 22

02

02

3

−+

−

ba

= 22

223

+

ba

= 44

9 22 ba+

= 4

9 22 ba +

AF = 2

9 22 ba +

DC = 22

20

22

−+

−

baa

= 22

223

−+

ba

= 44

9 22 ba+

= 4

9 22 ba +

DC = 2

9 22 ba +

22

Based on computations, AF = DC. Therefore, the two medians are equal and we can conclude that that the medians to the legs of an isosceles triangle are equal. 2. y D(b, c) C (a-b, c)

A (0, 0) B (a, 0) x Given: ABCD is an isosceles trapezoid. AC and BD are its diagonals Prove: AC ≅ BD Solution: To prove that AC ≅ BD we have to show that AC = BD. AC = ( ) 22 )0(0 −+−− cba = ( ) 22 cba +−

AC = 222 2 cbaba ++− BD = ( ) ( )22 0 cba −+−

= 222 2 cbaba ++− The computations showed that AC = BD. Thus we can conclude that the diagonals of an isosceles trapezoid are congruent. 3. y E (b, c) P (a+b, c) x H (0,0) O (a, 0)

23

Given: HOPE is a parallelogram HP and EO are the diagonals Prove: HP and EO bisect each other Solution: The simplest way of proving this is to show that the midpoints of the two diagonals are one and the same. 1. Find the midpoints of HP and EO and compare.

M (HP)

++

2,

20 cba

M (HP)

+

2,

2cba

M (OE)

+

2,

2cba

The results showed that HP and EO have the same midpoint. Hence they bisect each other. We can conclude that the diagonals of a parallelogram bisect each other. y 4. C (0, 2b) E x A (0, 0) B (2a, 0) Given: ∆ABC is a right triangle. E is the midpoint of the hypotenuse BC.

Prove: CE = BE = AE Solutions: Since E is the midpoint of BC, then its coordinates are

24

E

22,

22 ba

E(a, b) After finding the coordinates of E, find CE, BE and AE. Then compare the lengths. CE = ( ) ( )22 20 bba −+−

= ( ) ( )22 ba −+−

CE = 22 ba + BE = ( ) 222 baa +−

BE = 22 ba + AE = ( ) ( )22 00 −+− ba

AE = 22 ba + Since the three segments CE, BE and AE have the same lengths, we can therefore conclude that E is equidistant from the vertices of the right triangle. y 5. K (4d, 4e)

E (4b, 4c) B F

C

x M (0, 0) D A (4a, 0)

Given: MAKE is a quadrilateral B, C, D and F are midpoints of the sides Prove: FC and BD bisect each other Solutions:

1. Get the coordinates of F, B, C and D.

F

++

204,

204 cb

25

F ( )cb 2,2

B

++

244,

244 ecdb

B ( )ecdb 22,22 ++

C

+

24,

244 ead

C ( )ead 2,22 +

D

20,

24a

D ( )0,2a 2. Get the midpoints of FC and BD . Midpoint of FC

M1

+++

222,

2222 ecadb

M1 ( )ecadb +++ ,

Midpoint of BD

M2

++++

2022,

2222 ecadb

M2 ( )ecadb +++ , You can see that the midpoints of FC and BD are both (a+b+d, c+e). Therefore, we can conclude that the segments joining the midpoints of opposite sides of a quadrilateral bisect each other. Lesson 2 A. 1. a circle 2. a circle 3. the circle does not exist 4. a circle 5. a point circle

26

●

B. 1. C(0, 0), r = 5 2. C(0, 0), r = 2 3 3. C(0, 3), r = 11 4. C(7, 5), r = 3 2 5. C(-1, 4), r = 3 6. C(8, 0), r = 7

7. C

− 7,

21 , r = 5

8. C(-3, -4), r = 6 9. C(-1, 3), r = 3 2 10. C(2, 6), r = 10 C. 1. x2 + y2 = 16 2. x2 + y2 = 12 3. (x – 1)2 + (y – 1)2 = 9 4. (x + 2)2 + (y + 5)2 = 10 5. (x + 3)2 + (y – 4)2 = 8 6. (x – 2)2 + (y + 5)2 = 25 7. x2 + (y – 6)2 = 36 8. (x + 4)2 + y2 = 12.25 9. x2 + (y + 5) = 5 10. (x – 3)2 + y2 = 27 . D. y 1. C(3, -1) , tangent to x-axis Solution:

Since the circle is tangent to x-axis, then r is ⊥ to x-axis. The distance from the center of the circle to the x-axis or the x length of radius r is I unit. The equation of the circle with center at (3, -1) and r = 1 is (x – 3)2 + (y + 1) = 1 x2 – 6x + 9 + y2 + 2y + 1 = 1 x2 + y2 – 6x + 2y + 10 – 1 = 0 x2 + y2 – 6x + 2y + 9 = 0

27

2. Center at (2, 5) and passing through ( -2, 1) y Solution: Since the center is known what we need is the length of the radius. To find the length of the radius, use the other point as the other end of the radius. r = ( )[ ] ( )22 1522 −+−−

r = ( ) 22 422 ++

r = 22 44 + r = 1616 + x r = 32 r = 4 2 The equation of the circle therefore is (x – 2)2 + (y – 5)2 = (4 2 )2 x2 – 4x + 4 + y2 – 10y + 25 = 32 x2 + y2 – 4x – 10y – 3 = 0 3. Line segment joining (-2, 5) and (-2, -3) is a diameter. Solution: The midpoint of the segment is the center of the circle. Find the coordinates of the midpoint.

M

−−+−

235,

2)2(2

M

−

22,

24

M (-2, 1) Then find the length of the radius using the midpoint and one of the endpoints of the diameter . r = [ ] ( )22 15)2(2 −+−−−

= 22 4)22( ++−

= 240 +

= 24

C(2,5)

(-2,1)

(-2,5)

(-2,1)

28

r

C(3,3)

x r

C(1,0)

3x-4y=24

r = 4 The equation of the circle with center at ( -2, 1) and r = 4 is (x + 2)2 + (y – 1)2 = 42 x2 + 4x + 4 + y2 – 2y + 1 = 16 x2 + y2 + 4x – 2y - 11 = 0 y 4. Solution: Since the circle is tangent to both axes, therefore the center of the circle is at equal distance from both x and y axes. That distance is 3 units since the radius is given at 3 units. The center is also at the first quadrant. The circle passes through (0, 3) and (3, 0). The center is at (3, 3) x The equation of the circle is (x – 3)2 + (y – 3) = 32 x2 – 6x + 9 + y2 – 6y + 9 = 9 x2 + y2 – 6x – 6y + 9 = 0 5. Tangent to the line 3x – 4y = 24 with center at (1, 0).

y Solution: The radius of the circle is equal to the distance of the center (1, 0) from the line 3x – 4y = 24. To find the distance from a point to a line, we use the formula

d = 22

11

BACByAx

+−

++

where A and B are the coefficients of x and y, and C is the constant in the equation of line. x1 and y1 are the coordinates of the point. Therefore in the given, A = 3, B = -4 and C = -24. x1 = 1, y2 = 0

r = 22 43

24)0)(4()1(3+−

−−+

29

C(4,-1)

= 1692403

+−−+

= 521−−

r = 521

The equation of the circle with center at (1, 0) and radius r = 521 is

(x – 1)2 + y2 = 2

521

=

25441


1. C(2a, b) 2. AC = 222 ba + 3. origin 4. x2 + y2 = r2 5. C(0, 0), r = 8 6. C(0, -1), r = 5 7. C(6, 0), r = 1 8. C(3, -7), r = 2 3 9. C(1, -1), r = 7

10. Solution: Find the length of r using the origin and the coordinates of the center (-4, -1). r = 22 )01()04( −−++− y

= 22 )1()4( −+−

= 116 + = 17 The equation is (x + 4)2 + (y + 1)2 = ( 17 )2 x x2 + 8x + 16 + y2 + 2y + 1 = 17 x2 + y2 + 8x + 2y + 17 – 17 = 0 x2 + y2 + 8x + 2y = 0

Module 1 Triangle Congruence

What this module is about This module is about triangle congruence. It will teach you how to prove that two triangles are congruent by SSS, SAS, ASA and SSS congruent postulates.


1. apply the properties of congruence by:

• Reflexive property

• Symmetric property

• Transitive property

2. use inductive skills to prove congruence between triangles 3. show congruence between triangles by:

• SSS congruence

• SAS congruence

• ASA congruence

• SAA congruence

How much do you know For each pair of triangles, identical marks indicate two pairs of corresponding congruent parts. Name the congruence postulate which will prove the triangles congruent.

What you will do

Lesson 1

Properties of Congruence

Lesson 2

Congruent Triangles

Lesson 3

SSS Congruence SSS Congruence

Lesson 4

SAS Congruence

Lesson 5

ASA Congruence

Lesson 6

SAA Congruence

Let’s summarize


Answer key

93o50’

32o

c 4

b C

Module 2 Triangle Trigonometry

What this module is about This module is about law of sines. As you go over this material, you will develop the skills in deriving the law of sines. Moreover, you are also expected to learn how to solve different problems involving the law of sines.

What you are expected to learn This module is designed for you to demonstrate ability to apply the law of sines to solve problems involving triangles.

How much do you know 1. Find the measure of ∠B in ABC if m∠A = 8o and m∠C = 97o.

2. Find the measure of ∠A in ABC if m∠B = 18o14’ and m∠C = 81o41’.

3. Two of the measures of the angles of ABC are A = 8o3’ and B = 59o6’.

Which is the longest side?

4. What equation involving sin 32o will be used to find side b? 5. Given a = 62.5 cm, m∠A = 62o20’ and m∠C = 42o10’. Solve for side b.

6. Solve for the perimeter of ABC if c = 25 cm, m∠A = 35o14’ and m∠B =

68o.

7. Solve ABC if a = 38.12 cm, m∠A = 46o32’ and m∠C = 79o17’.

A

B

2

8. Solve ABC if b = 67.25 mm, c = 56.92 mm and m∠B = 65o16’.

9. From the top of a building 300 m high, the angles of depression of two street signs are 17.5o and 33.2o. If the street signs are due south of the observation point, find the distance between them.

10. The angles of elevation of the top of a tower are 35o from point A and 51o from another point B which is 35 m from the base of the tower. If the base of the tower and the points of observation are on the same level, how far are they from each other? What you will do

Lesson 1

Oblique Triangles

An oblique triangle is a triangle which does not contain a right angle. It contains either three acute angles (acute triangle) or two acute angles and one obtuse angle (obtuse triangle).

Acute triangle Obtuse triangle

If two of the angles of an oblique triangle are known, the third angle can be computed. Recall that the sum of the interior angles in a triangle is 180o.

Examples:

Given two of the angles of ABC, solve for the measure of the third angle. 1. A = 30o, B = 45o.

A + B + C = 180o.

30o + 45o + C = 180o.

C = 180o – (30o + 45o)

C = 180o – 75o

C = 105o

3

2. B = 69.25o, C = 114.5o.

A + B + C = 180o.

A + 69.25o + 114.5o = 180o.

A = 180o – (69.25o + 114.5o)

A = 180o – 183.75o

A = 3.75o

3. A = 56o38’, C = 64o49’.

A + B + C = 180o.

56o38’ + B + 64o49’o = 180o.

B = 180o – (56o38’ + 64o49’)

B = 180o – 121o27’

B = 58o33’

Try this out

Given two of the angles of ABC, solve for the measure of the third angle.

Set A

1. A = 60o, B = 45o 2. B = 76o, C = 64o 3. A = 69.25o, C = 14.5o 4. B = 19.15o, A = 107.05o 5. C = 85o38’, A = 74o23’

Set B

1. A = 57o, B = 118o 2. B = 46o, C = 69o 3. A = 51.63o, C = 70.47o 4. B = 37.3o, A = 77.24o 5. C = 58o6’, A = 47o24’

Set C

1. A = 58o, B = 77o 2. B = 64o, C = 64o 3. A = 18.82o, C = 109.3o 4. B = 56.85o, A = 37.18o

4

5. C = 27o34’, A = 72o43’

Lesson 2

Deriving the Law of Sines; Solving Oblique Triangles Involving Two Angles and a Side Opposite One of Them

Consider ABC with altitude h.

The area of this triangle is

1Area ch2

= (1)

From the right triangle

hsin Ab

= (2)

Solving h in (2) gives H = bsin A (3) Substituting (3) in (1) gives

1Area c(bsin A)2

= (4)

Similarly, two other formulas for the area of ABC can be derived, namely:

1Area a(csin B)2

= (5)

b h a

B A

C

c

5

93o50’

32o

c 4

b C

B

A

and 1Area b(a sin C)2

= (6)

Since (4), (5), and (6) represent the area of one and the same triangle,

then

1 1 1c(bsin A) a(csin B) b(a sin C)2 2 2

= =

or equivalently

sin A sin B sin Ca b c

= = .

This is the Law of Sines which states that “In any triangle ABC, with a, b,

and c as its sides, and A, B and C as its angles, sin A sin B sin Ca b c

= = .”

Note that this is also equivalent to

a b csin A sin B sin C

= = .

The Law of Sines is used to solve different cases of oblique triangles.

One case of oblique triangle that can be solved using the law of sines involves two angles and a side opposite one of them.

Examples:

Solve each triangle ABC described below. 1. a = 4, A = 32o, B = 93o50’

The given parts of the triangle involve two angles A and B, and side a

which is opposite A. The given parts and the unknown parts are shown in the figure below.

To solve for C:

A + B + C = 180o

6

32o + 93o50’ + C = 180

C = 180o – (32o + 93o50’)

C = 180o – 125o50’

C = 54o10’ To solve for b:

a bsin A sin B

=

a sin Bbsin A

=

4(sin 93 50 ')bsin 32

°=

°

4(0.99776)b0.52992

=

b = 7.53 To solve for c:

a csin A sin C

=

a sin Ccsin A

=

4(sin 54 10 ')csin 32

°=

°

4(0.81072)b0.52992

=

b = 6.12

2. b = 2.3, C = 42.5o, B = 84o

The given parts of the triangle involve two angles B and C, and side b which is opposite B. The given parts and the unknown parts are shown in the figure below.

84o

42.5o

a

2.3

c

B

C A

7

To solve for A: A + B + C = 180o

A + 84o + 42.5o = 180o

A = 180o – (84o + 42.5o)

A = 180o – 126.5o

A = 53.5o or 53o30’ To solve for c:

c bsin C sin B

=

bsin Ccsin B

=

2.3(sin 42.5 )csin84

°=

°

2.3(0.67559)c0.99452

=

c = 1.6 To solve for a:

a bsin A sin B

=

bsin Aasin B

=

2.3(sin 53 30 ')asin84

°=

°

2.3(0.80386)a0.99452

=

a =1.86

Try this out

Solve triangle ABC described by the given parts.

Set A

1. a = 12, A = 60o, B = 45o 2. b = 20, B = 32o, C = 24o

8

3. c = 42.5, A= 35.2o, C = 29o 4. a = 9, A = 42.4o, C = 37.8o 5. b = 75, B = 30o45’, C = 32o30’

Set B

1. a = 10, A = 61o, B = 49o 2. c = 80, C = 56o, B = 68o 3. b = 38, B = 47o, A = 30o 4. c = 8, C = 77.2o, A = 88o 5. b = 8, B = 67o15’, A = 33o5’

Set C

1. A = 65o50’, B = 14o, b = 16 2. A = 69o15’, C = 28o40’, a = 80 3. B = 47.5o, A = 80o , a = 15.1 4. B = 63o36’, C = 89o , b = 8 5. C = 57.25o, A = 33.5o , c = 23.5

Lesson 3

Solving Oblique Triangles Involving Two Angles and the Included Side

Another case of oblique triangle that can be solved using the law of sines involves two angles and the included side.

Examples:

1. c = 440, A = 48o, B = 75o

The given parts of the triangle involve two angles and the included side.

This case can also be solved using the law of sines.

To solve for C: A + B + C = 180o

75o

48o

440

b C

a

A

B

9

48o + 75o + C = 180

C = 180o – (48o + 75o)

C = 180o – 123o

C = 57o To solve for a:

a csin A sin C

=

csin Aasin C

=

440(sin 48 )asin 57

°=

°

440(0.74315)a0.83867

=

a = 389.87 To solve for b:

b csin B sin C

=

csin Bbsin C

=

440(sin 48 )bsin 57

°=

°

440(0.74314)b0.96593

=

b = 338.51

2. a = 8.2, B = 38o26’, C = 75o19’ The given parts of the triangle involve two angles and the included side.

This case can also be solved using the law of sines.

38o26’ 75o19’ 8.2

b c

A

B C

10

To solve for A:

A + B + C = 180o

A + 38o26’ + 75o19’ = 180

A = 180o – (38o26’ + 75o19’)

C = 180o – 113o45’

C = 66o15’ To solve for b:

a bsin A sin B

=

a sin Bbsin A

=

8.2(sin 38 26 ')bsin 75 19 '

°=

°

8.2(0.62160)b0.96734

=

b = 5.27 To solve for c:

a csin A sin C

=

a sin Ccsin A

=

8.2(sin 66 15')csin 75 19 '

°=

°

8.2(0.91531)c0.96734

=

c = 7.76

Try this out

Solve each triangle ABC described below.

Set A

1. a = 12, A = 60o, B = 45o 2. b = 20, A = 32o, C = 24o

11

3. c = 42.5, A = 35.2o, B = 79o 4. a = 9, A = 42.4o, B = 37.8o 5. b = 0.751, A = 100o30’, C = 32o45’

Set B

1. a = 248, A = 51o, B = 49o 2. c = 8.5, A = 56o, B = 68o 3. b = 38, A = 47.2o, C = 50.5o 4. a = 8, B = 77o26’, C = 28o40’ 5. c = 1.2, A = 67.3o, B = 33.9o

Set C

1. A = 65o50’, B = 14o, c = 16 2. B = 69.2o, C = 28.8o a = 0.8 3. A = 15o28’, C = 77o, b = 100 4. a = 1.8, B = 36o36’, C = 98o 5. c = 23.5, B = 57.25o, A = 33o33’

Lesson 4

Solving Oblique Triangles Involving Two Sides and an Angle Opposite One of Them

Another case of oblique triangle that can be solved using the law of sines

involves two sides and an angle opposite one of them.

Examples: The given parts and the unknown parts are shown in the figures: 1. b = 15, a = 20, A = 80o25’

To solve for B:

a bsin A sin B

=

C

80o25’

15 20

c B A

12

bsin Asin Ba

=

15(sin80 25')sin B20

°=

15(0.98604)sin B20

=

sin B 0.73953=

B = Arcsin (0.73953)

B = 47o41’29”

To solve for C: A + B + C = 180o

80o25’ + 47o41’ + C = 180

C = 180o – (80o25’ + 47o41’)

C = 180o – 128o6’

C = 51o54’ To solve for c:

a csin A sin C

=

a sin Ccsin A

=

20(sin 51 54 ')csin80 25'

°=

°

20(0.78694)b0.98604

=

b = 15.96 2. a = 2.46, c = 5, C = 70.2o

70.2o

b 2.46

A 5 B

C

13

To solve for A:

a csin A sin C

=

a sin Csin Ac

=

2.46(sin 70.2 )sin A5

°=

2.46(0.94088)sin A5

=

sin A = 0.46291

A = Arcsin (0.46291)

A = 27.58o To solve for B:

A + B + C = 180o

27.58o + B + 70.2o = 180

B = 180o – (27.58o + 70.2o)

B = 180o – 97.78o

B = 82.22o To solve for b:

b csin B sin C

=

csin Bbsin C

=

5(sin82.22 )bsin 70.2

°=

°

5(0.99080)b0.94088

=

b = 5.27

Try this out

Solve each triangle ABC described below.

Set A

1. a = 12, b = 18, B = 45o 2. b = 20, c = 32, B = 24o

14

3. a = 2.5, c = 5.2, C = 79o 4. a = 9, b = 14.4, B = 37.8o 5. b = 0.751, c = 1, C = 32o45’

Set B

1. a = 140, b = 150, B = 49o 2. a = 8.5, c = 5.6, A = 68o 3. a = 47.2, b = 38, A = 50.5o 4. a = 0.8, c = 0.72, C = 28o40’ 5. c = 1.2, b = 6.3, B = 131.5o

Set C

1. A = 65o50’, a = 15, c = 16 2. B = 95.2o, b = 2.8, c = 0.8 3. C = 15o28’, c = 77, b = 100 4. B = 36o36’, a = 1.8, b = 2.98 5. A = 33o33’, a = 33.5, b = 57.25

Lesson 5

Solving Problems Involving Oblique Triangles

The following examples show how the law of sines is used to solve word problems involving oblique triangles.

Examples:

Solve each problem. Show a complete solution. 1. The longer diagonal of a parallelogram is 27 cm. Find the perimeter of the

parallelogram if the angles between the sides and the diagonal are 30o and 40o10’.

The perimeter of any parallelogram is twice the sum of the lengths of the two consecutive sides.

40o10’30o

27 cm y

xθ

15

Let x and y be the consecutive sides as shown in the figure. The angle opposite x is 40o10’. Hence, the angle θ between x and y is

30o + 40o10’ + θ = 180o

θ = 180o – (30o + 40o10’)

θ = 180o – 70o10’

θ = 109o50’ To solve for x:

x 27sin 40 10 ' sin109 50 '

=° °

27sin 40 10 'xsin109 50 '

°=°

27(0.64501)x0.94068

=

x = 18.51 cm To solve for y:

y 27sin 30 sin109 50 '

=° °

27sin 30ysin109 50 '

°=°

27(0.5)y0.94068

=

x = 14.35 cm The perimeter P of the parallelogram is P = 2(x + y)

P = 2(18.51 + 14.35)

P = 2(32.86)

P = 65.72 cm

2. From the top of a 300 m building, the angle of depression of two traffic aides on the street are 17o30’ and 33o12’, respectively. If they are due south of the observation point, find the distance between them.

16

Let B1B2 be the distance between the two traffic aides.

First, find θ1:

θ1 = 33o12’ – 17o30’

θ1 = 15o42’ Then, find OB1:

1

300cos33 12 'OB

° =

1300OB

cos33 12 '=

°

1300OB

0.83676=

OB1 = 358.53

To solve for B1B2:

1 2 1B B OBsin15 42 ' sin17 30 '

=° °

11 2

OB (sin15 42 ')B B

sin17 30 '°

=°

1 2358.53(0.27060)B B

0.30071=

B1B2 = 322.63 m

17o30’

300 m

P B1 B2

33o12’θ1

O

17o30’33o12’

17

Try this out

Solve each problem. Show a complete solution.

Set A

1. The shorter diagonal of a parallelogram is 5.2 m. Find the perimeter of the parallelogram if the angles between the sides and the diagonal are 40o and 30o10’.

2. From the top of a 150 m lighthouse, the angles of depression of two

boats on the shore are 20o and 50o, respectively. If they are due north of the observation point, find the distance between them.

3. Two policemen 122 meters apart are looking at a woman on top of a

tower. One cop is on the east side and the other on the west side. If the angles of elevation of the woman from the cops are 42.5o and 64.8o, how far is she from the two cops?

4. The angle between Rizal St. and Bonifacio St. is 27o and intersect at P.

Jose and Andres leaves P at the same time. Jose jogs at 10 kph on Rizal St. If he is 3 km from Andres after 30 minutes, how fast is Andres running along Bonifacio St?

5. The angle of elevation of the top of a tower is 40o30’ from point X and

55o from another point Y. Point Y is 30 meters from the base of the tower. If the base of the tower and points X and Y are on the same level, find the approximate distance from X and Y.

Set B

1. The vertex of an isosceles triangle is 40o. If the base of the triangle is 18 cm. find its perimeter.

2. An 80-meter building is on top of a hill. From the top of the building, a

nipa hut is sighted with an angle of depression of 54o. From the foot of the building the same nipa hut was sighted with an angle of depression of 45o. Find the height of the hill and the horizontal distance from the building to the nipa hut.

3. The lengths of the diagonals of a parallelogram are 32.5 cm and 45.2

cm. The angle between the longer side and the longer diagonal is 35o. Find the length of the longer side of the parallelogram.

18

4. A diagonal of a parallelogram is 35 cm long and forms angles of 34o and 43o with the sides. Find the perimeter and area of the parallelogram.

5. A wooden post is inclined 85o with the ground. It was broken by a

strong typhoon. If a broken part makes an angle of 25o with the ground, and the topmost part of the post is 40 ft from its base, how long was the post?

Set C

1. The measures of two of the angles of a triangle are 50o and 55o. If its longest side measures 17 cm, find the perimeter of the triangle.

2. Bryan’s (B) and Carl’s (C) houses are along the riverbank. On the

opposite side is Angel’s (A) house which is 275 m away from Carl’s. The angles CAB and ACB are measured and are found to be 125o and 49o, respectively. Find the distance between Angel’s and Bryan’s houses.

3. A park is in the shape of an obtuse triangle. One angle is 45o and the

opposite side is 280-m long and the other angle is 40o. Find the perimeter of the park.

4. Three circles are arranged as shown in the figure. Their centers are

joined to form a triangle. Find angle θ.

5. Two forces are acting on each other at point A at an angle 50o. One force has a magnitude of 60 lbs. If the resultant force is 75 lbs find the magnitude of the second force.

r = 1.5

r = 3

r = 2.5 θ

30o

19

Let’s summarize

1. An oblique triangle is a triangle which does not contain a right angle. It contains either three acute angles (acute triangle) or two acute angles and one obtuse angle (obtuse triangle).

2. If two of the angles of an oblique triangle are known, the third angle can

be computed. The sum of the interior angles in a triangle is 180o.

3. In any triangle ABC, with a, b, and c as its sides, and A, B and C as its

angles, sin A sin B sin Ca b c

= = .

This is also equivalent to

a b csin A sin B sin C

= = .

4. The Law of Sines is used to solve different cases of oblique triangles. This

law is used to solve oblique triangles that involve

a. two angles and a side opposite one of them. b. two angles and the included side c. two sides and the angle opposite one of them


1. Find the measure of ∠B in ABC if m∠A = 8o and m∠C = 97o. 2. Find the measure of ∠A in ABC if m∠B = 18o14’ and m∠C = 81o41’.

3. The measures of the side of ABC are c = 83 cm and b = 59 cm and a =

45 cm. Which is the largest angle?

4. What equation that involves sin 70.2o will be used to find angle A?

70.2o

b 2.46

A 5 B

C

20

5. Given a = 62.5 cm, m∠A = 112o20’ and m∠C = 42o10’. Solve for side b. 6. Solve for the perimeter of ABC if c = 25 cm, m∠A = 35o14’ and m∠B

= 68o.

7. Solve ABC if b = 38.12 cm, m∠B = 46o32’ and m∠A = 79o17’.

8. Solve ABC if b = 67.25 mm, c = 56.92 mm and m∠B = 65o16’.

9. From the top of a building 300 m high, the angles of depression of two street signs are 17.5o and 33.2o. If the street signs are due south of the observation point, find the distance between them.

10. A park is in the shape of an acute triangle. One angle is 55o and the

opposite side is 300-m long and the other angle is 60o. Find the perimeter of the park.

21

Answer Key


1. 75o

2. 80o5’

3. c

4. b 4sin 93 50 ' sin 32

=° °

5. 68.32 cm

6. 14.82 cm

7. B = 54o11’, b = 42.59 cm, c = 51.61 cm

8. C = 50o14’30”, A = 64o29’30”, a = 66.83 mm

9. 155.45 m

10. 26.73 m

Try this out

Lesson 1

Set A

1. 75o

2. 40o

3. 96.25o

4. 53.8o

5. 19o59’

Set B

1. 5o

2. 65o

3. 57.9o

4. 65.46o

5. 74o30’

22

Set C

1. 45 o

2. 52 o

3. 51.88o

4. 85.97o

5. 79o43’

Lesson 2

Set A

1. C = 75o, b = 9.80, c = 13.38

2. A = 124o, a = 31.29, c = 15.35

3. B = 115.8o, a = 50.53, b = 78.92

4. B = 99.8o, b = 13.15, c = 8.18

5. A = 116o45’, a = 130.99, c = 78.81

Set B

1. C = 70o, b = 8.63, c = 10.74

2. A = 56o, a = 80, b = 89.47

3. C = 103o, a = 25.98, c = 50.63

4. B = 14.8o, a = 8.20, b = 2.10

5. C = 79o40’, a = 4.74, c = 8.53

Set C

1. C = 100o10’, a = 60.34, c = 65.10

2. B = 82o5’, b = 84.73, c = 41.04

3. C = 52.5o, b = 11.30, c = 12.16

4. A = 27o24’, a = 4.11, c = 8.93

5. B = 89.25o, a = 15.42, b = 27.94 Lesson 3

Set A

1. C = 75 o, b = 9.80, c = 13.38

2. B = 123 o, a = 12.64, c = 9.70

23

3. C = 65.8 o, a = 26.84, b = 45.74

4. C = 99.8 o, b = 8.18, c = 13.15

5. B = 46o45’, a = 1.014, c = 0.558

Set B

1. C = 80o, b = 240.84, c = 314.27

2. C = 56o, a = 8.50, b = 9.51

3. B = 82.3o, a = 28.14, c = 29.59

4. A = 73o54’, b = 8.13, c = 3.99

5. C = 79o33’, a = 1.13, b = 0.68

Set C

1. C = 100o, a = 14.83, b = 14.83

2. A = 82o, b = 0.76, c = 0.39

3. B = 87o32’, a = 26.69, c = 97.53

4. A = 45o24’, b = 1.51, c = 2.50

5. C = 89o12’, a = 13.00, b = 19.77

Lesson 4

Set A

1. A = 28o7’32”, C = 106o52’28”, c = 10.83

2. A = 115o23’48”, C = 40o36’12”, a = 44.42

3. A = 28o9’36”, B = 72o50’24”, b = 5.06

4. A = 22o31’56”, C = 119o40’4”, c = 20.41

5. A = 123o16’45”, B = 23o58’15”, a = 1.55

Set B

1. A = 44o46’51”, C = 86o13’9”, c = 198.32

2. B = 74o20’56”, C = 37o39’4”, b = 8.83

3. B = 38o24’20”, C = 91o5’40”, c = 61.16

4. A = 32o12’34”, B = 119o7’26”, b = 1.31

5. A = 40.3o, C = 8.2, a = 5.44

24

Set C

1. B = 37o27’57”, C = 76o42’3”, b = 10.00

2. A = 71.21o, C = 16.59o, a = 0.76

3. A = 144o16’12”, B = 20o15’48”, a = 168.61

4. A = 21o6’31”, C = 122o17’29”, c = 4.23

5. B = 70o49’4”, C = 75o37’56”, c = 58.72

Lesson 5

Set A

1. The perimeter is 12.66 m.

2. The distance between them is approximately 286.26 m.

3. The distances of the woman from the two policemen are approximately

86.33 m and 115.62 m.

4. Andres ran approximately at the rate of 3.21 kph.

5. The distance between X and Y is approximately 20.16 m.

Set B

1. The perimeter of the isosceles triangle is approximately 70.62 cm.

2. The horizontal distance of the building to the nipa hut is 212.55 m.

3. The longer side is 28.31 cm.

4. The perimeter of the parallelogram is 89.18 cm and its area is 1,962.21

sq.cm.

5. The wooden post is 60.40 ft long.

Set C

1. The perimeter is 44.90 cm.

2. The distance between Angel’s and Bryan’s houses is 1,985.54 m.

3. The perimeter of the park is 929 m.

4. The angle θ is 115o46’ 12”

5. The second force is approximately 97.83 lbs.

25

How much have you learned

1. 75o

2. 80o5’

3. Angle C

4. sin A sin 70.22.46 5

°=

5. 29.20 cm

6. 63.63 cm

7. C = 54o11’, a = 51.61 cm, c = 42.59 cm

8. C = 50o14’30”, A = 64o29’30”, a = 66.83 mm

9. The distance between the signs is 493.03 m.

10. The perimeter is 949.09 m

Module 3 Triangle Congruence

What this module is about This module is about using triangle congruence to prove congruent segments and angles. You will understand that a correspondence between two triangles is a congruence if the corresponding angle and corresponding sides are congruent.


1. prove congruent segments and angles using the conditions for triangle congruence.

2. solve routine and non-routine problems.

How much do you know 1. Given: HS bisects ∠THE

∠HTS ≅ ∠HES Prove: TS ≅ ES

T

H S

E

2

2. Given: PM // NS; PM ≅ SN

Prove: ML ≅ LN

3. Given: NO ≅ XS; OE ≅ SI; NI ≅ XE

Prove: ∠O ≅ ∠S 4. Given: ∆PNC and ∆TNC are isosceles triangles with common base

NC.

Prove: ∠PNT ≅ ∠PCT

P

N C

T

P M

N S

L

N I

O

E

S

X

3

H

5. Given: AE is the ⊥ bisector of BC

Prove: ∠B ≅ ∠C 6. Given: CH ≅ EH, FH ≅ GH

Prove: ∠C ≅ ∠E

7. Given: ∠A and ∠B are right angles and AP ≅ BP.

Prove: AQ ≅ BQ

A

B C E

D

C E

F G

P

A B

Q

4

8. Given: ∠ 1 ≅ ∠ 2

O is the midpoint of SP Prove: MO ≅ NO

9. Given: LO ≅ LN ∆ JLO and ∆MLN are right ∆ s Prove: ∠ J ≅ ∠K

10. Given: PQ ≅ RS ∠QRP ≅ ∠SRP

Prove: SP ≅ QR

O

J

L

M

N

1 2S

M

O

N

P

S P

R Q

5

What you will do

Lesson

Congruent Segments and Congruent Angles To prove two segments or two angles are congruent, you must show that they are corresponding parts of congruent triangles. For triangle congruence, you have the following:

• SSS congruence • SAS congruence • ASA congruence • SAA congruence

For right triangle congruence, you have the following:

• LL congruence • LA congruence • HyL congruence • HyA congruence

Examples:

Formal Proofs:

1. Given: AB // DC, AD // BC Prove: AB ≅ DC

DA

CB 1 4

3 2

6

Proof:

Statement Reason

1. AB // DC, AD // BC 1. Given 2. ∠ 1 ≅ ∠ 2, ∠ 3 ≅ ∠ 4 2. If two parallel lines are cut by a transversal, then the alternate interior angles are ≅ 3. BD ≅ BD 3. Reflexivity 4. ∆ABD ≅ ∆CDB 4. ASA 5. ∴AB ≅ DC 5. Corresponding parts of ≅ triangles are ≅ or CPCTC

2. Given: LS bisects ∠TLE

∠ LTS ≅ ∠ LES Prove: TS ≅ ES

Proof: Statement Reason

1. LS bisects ∠TLE 1. Given 2. ∠TLS ≅ ∠ELS 2. Definition of angle bisector 3. LS ≅ LS 3. Reflexivity 4. ∠ LTS ≅ ∠ LES 4. Given 5. ∆TSL ≅ ∆ESL 5. SAA congruence 6. TS ≅ ES 6. Corresponding parts of ≅ ∆ s are ≅ or CPCTC

T

S L

E

7

3. Given: MO ≅ YS, OB ≅ SP, MP ≅ YB

Prove: ∠O ≅ ∠S Proof: Statement Reason 1. MO ≅ YS 1. Given OB ≅ SP

MP ≅ YB 2. PB ≅ PB 2. Reflexivity 3. MB ≅ PY 3. By addition 4. ∆MOB ≅ ∆YSP 4. SSS congruence 5. ∠O ≅ ∠S 5. CPCTC

4. Given: AR is the ⊥ bisector of BX.

Prove: ∠B ≅ ∠X

M P

O

B

S

Y

R

A

B X

8

Proof: Statement Reason 1. AR is the ⊥ bisector of BX 1. Given 2. AR ≅ AR 2. Reflexivity 3. BR ≅ XR 3. Definition of ⊥ bisector 4. ∆ARB ≅ ∆ARX 4. LL congruence 5. ∠B ≅ ∠X 5. CPCTC 5. Given: ∠ 1 ≅ ∠ 2

O is the midpoint of SP Prove: DO ≅ SO Proof: Statement Reason 1. ∠ 1 ≅ ∠ 2 1. Given 2. ∠DSO ≅ ∠SPO 2. Supplements of ≅ ∠ s are also ≅ 3. O is the midpoint of SP 3. Given 4. SO ≅ PO 4. Definition of midpoint 5. ∠DOS ≅ ∠SOP 5. Vertical ∠ s are ≅ 6. ∆SDO ≅ ∆PSO 6. ASA congruence 7. DO ≅ SO 7. CPCTC

1 2S

D

O

S

P

9

Try this out

1. Given: ∠SLO and ∠KMO are right angles LO ≅ MO

Prove: ∠S ≅ ∠K

2. Given: XY ≅ RS

∠YXR ∠SRX Prove: SX ≅ YR

3. Given: ∆PNU and ∆TNU are isosceles triangles with common base NU

Prove: ∠PNT ≅ ∠PUT

L

S

O

K

M

YX

S R

P

N u

T

O

10

4. Given: CD ≅ ED, FC ≅ FE

Prove: ∠C ≅ ∠E

5. Given: ∠A ≅ ∠C are right angles AK ≅ MC

Prove: MA ≅ KC

6. Given: MN ⊥ NR, PR ⊥ NR, MR ≅ PN

Prove: MN ≅ NR

D

C

E

F

M C

A K

M P

N R

11

7. Given: AI ≅ BN, BI ≅ AN

Prove: ∠ I ≅ ∠N

8. Given: In the figure, JV and NC bisect each other at O

Prove: ∠ J ≅ ∠V

9. Given: MQ and PN bisect each other at O

Prove: ∠P ≅ ∠N

A B

I N

J C

N V

O

Q

N

O

P

M

12

10. Given: GQ ≅ RS ∠QGR ≅ ∠SRG

Prove: SG ≅ QR

Let’s summarize

To prove two segments or two angles are congruent, you must show that they are corresponding parts of congruent triangles. For triangle congruence, you have the following:

• SSS congruence

• SAS congruence

• ASA congruence

• SAA congruence

For right triangle congruence, you have the following:

• LL congruence

• LA congruence

• HyL congruence

• HyA congruence

QG

S R

13


1. Given: TS bisects ∠MTE

∠TMS ≅ ∠TES Prove: TM ≅ TE

2. Given: PQ // NS; PQ ≅ SN Prove: QL ≅ LN

M

S T

E

P Q

N S

L

14

3. Given: XO ≅ YS; OE ≅ SI; XI ≅ YE Prove: ∠O ≅ ∠S

4. Given: ∆PNQ and ∆TNQ are isosceles triangles with common base NQ. Prove: ∠PNQ ≅ ∠TNQ

5. Given: HE is the ⊥ bisector of BC Prove: ∠B ≅ ∠C

P

N Q

T

H

B C E

X I

O

E

S

Y

15

H

6. Given: CG ≅ EG, FG ≅ HG Prove: ∠C ≅ ∠E

7. Given: ∠R and ∠S are right angles and RP ≅ SP. Prove: RQ ≅ SQ

8. Given: ∠ 1 ≅ ∠ 2 O is the midpoint of SP

Prove: RO ≅ TO

P

R S

Q

12 S

R

O

T

P

D

C E

F H G

16

9. Given: ∠ JLO and ∠KMO are right ∠ s LO and MO

Prove: ∠ J ≅ ∠K

10. Given: PQ ≅ RS ∠QPR ≅ ∠SRP

Prove: SP ≅ QR

Q P

S R

L

J

O

K

M

17

Answer key How much do you know

1. Proof

Statement Reason 1. HS bisects ∠THE 1. Given ∠HTS ≅ ∠HES 2. ∠THS ≅ ∠EHS 2. Definition of ∠ bisector 3. HS ≅ HS 3. Reflexivity 4. ∆THS ≅ ∆EHS 4. SAA congruence 5. TS ≅ ES 5. CPCTC

2. Proof:

Statement Reason 1. PM // NS, PM ≅ SN 1. Given 2. ∠P ≅ ∠S, ∠M ≅ ∠N 2. If 2 // lines cut by a transversal, the alternate interior ∠ s are ≅ 3. ∆PLM ≅ ∆SLN 3. ASA congruence 4. ML ≅ LN 4. CPCTC

3. Proof:

Statement Reason 1. NO ≅ XS, OE ≅ SI 1. Given NI ≅ XE 2. IE ≅ IE 2. Reflexivity 3. NE ≅ IX 3. By addition 4. ∆NOE ≅ ∆XSI 4. SSS congruence 5. ∠O ≅ ∠S 5. CPCTC

4. Proof:

Statement Reason 1. ∆PNC and ∆TNC 1. Given are isosceles ∆ 2. PN ≅ PC, TN ≅ TC 2. Definition of isosceles 3. PT ≅ PT 3. Reflexivity 4. ∆PNT ≅ ∆PCT 4. SSS congruence 5. ∠PNT ≅ ∠PCT 5. CPCTC

18

5. Proof: Statement Reason

1. AE is the ⊥ bisector of BC 1. Given 2. BE ≅ CE 3. Definition of ⊥ bisector 3. AE ≅ AE 2. Reflexivity 4. ∆AEB ≅ ∆AEC 4. LL congruence 5. ∠B ≅ ∠C 5. CPCTC

6. Proof:

Statement Reason 1. CH ≅ EH, FH ≅ GH 1. Given 2. ∠FHC ≅ ∠GHE 2. Vertical ∠ s are ≅ 3. ∆FHC ≅ ∆GHE 3. SAS 4. ∠C ≅ ∠E 4. CPCTC

7. Proof:

Statement Reason 1. ∠A and ∠B are rt. ∠ s 1. Given AP ≅ BP 2. PQ ≅ PQ 2. Reflexivity 3. ∆PAQ ≅ ∆PBQ 3. HyL congruence 4. AQ ≅ BQ 4. CPCTC

8. Proof:

Statement Reason

1. ∠ 1 ≅ ∠ 2 1. Given 2. ∠MSO ≅ ∠NPO 2. Supplements of ≅ ∠ s are also ≅ 3. O is the midpoint of SP 3. Given 4. SO ≅ PO 4. Definition of midpoint 5. ∠MOS ≅ ∠NOP 5. Vertical ∠ s are ≅ 6. ∆MDO ≅ ∆NPO 6. ASA congruence

7. MO ≅ NO 7. CPCTC

19

9. Proof:

Statement Reason 1. LO ≅ LN 1. Given 2. ∠ JLO ≅ ∠MLN 2. Vertical ∠ s are ≅ 3. ∆JLO ≅ ∆MLN 3. LA congruence 4. ∠ J ≅ ∠K 4. CPCTC

10. Proof:

Statement Reason 1. PQ ≅ RS 1. Given ∠QRP ≅ ∠SRP 2. PR ≅ PR 2. Reflexivity 3. ∆PSR ≅ ∆RQP 3. SAS congruence 4. SP ≅ QR 4. CPCTC

Try this out Lesson

1. Proof:

Statement Reason 1. LO ≅ MO 1. Given 2. ∠SOL ≅ ∠KOM 2. Vertical ∠ s are ≅ 3. ∆SLO ≅ ∆KMO 3. LA congruence 4. ∠S ≅ ∠K 4. CPCTC

2. Proof:

Statement Reason 1. XY ≅ RS 1. Given ∠YXR ≅ ∠SRX 2. RX ≅ RX 2. Reflexivity 3. ∆XYR ≅ ∆RSX 3. SAS congruence 4. SX ≅ YR 4. CPCTC

20

3. Proof:

Statement Reason 1. ∆PNU and ∆TNU 1. Given are isosceles ∆ 2. PN ≅ PU, TN ≅ TU 2. Definition of isosceles 3. PT ≅ PT 3. Reflexivity 4. ∆PNT ≅ ∆PUT 4. SSS congruence 5. ∠PNT ≅ ∠PUT 5. CPCTC

4. Proof:

Statement Reason 1. CD ≅ ED, FC ≅ FE 1. Given 2. DF ≅ DF 2. Reflexivity 3. ∆DCF ≅ ∆DEF 3. SSS congruence 4. ∠C ≅ ∠E 4. CPCTC

5. Proof:

Statement Reason 1. ∠A ≅ ∠C 1. Given AK ≅ MC

2. MK ≅ MK 2. Reflexivity 3. ∆MCK ≅ ∆KAM 3. HyL congruence

4. MA ≅ KC 4. CPCTC

6. Proof:

Statement Reason 1. MR ≅ PN 1. Given 2. NR ≅ NR 2. Reflexivity 3. ∆MNR ≅ ∆PRN 3. Hyl congruence 4. MN ≅ PR 4. CPCTC

21

7. Proof:

Statement Reason 1. AI ≅ BN 1. Given BI ≅ AN 2. IN ≅ IN 2. Reflexivity 3. ∆AIN ≅ ∆BNI 3. SSS congruence 4. ∠ I ≅ ∠N 4. CPCTC

8. Proof:

Statement Reason 1. JV and NC bisect 1. Given each other at O 2. JO ≅ VO 2. Definition of Segment Bisector CO ≅ NO 3. ∠ JOC ≅ ∠VON 3. Vertical ∠ s are ≅ 4. ∆JOC ≅ ∆VON 4. SAS 5. ∠ J ≅ ∠V 5. CPCTC

9. Proof:

Statement Reason 1. MQ and PN bisect 1. Given each other at O 2. MO ≅ QO 2. Definition of Segment Bisector PO ≅ NO 3. ∠POM ≅ ∠NOQ 3. Vertical ∠ s are ≅ 4. ∆POM ≅ ∆VOQ 4. SAS 5. ∠P ≅ ∠N 5. CPCTC

10. Proof:

Statement Reason 1. GQ ≅ RS 1. Given ∠QGR ≅ ∠SRG 2. RG ≅ RG 2. Reflexivity 3. ∆QGR ≅ ∆SRG 3. SAS congruence 4. SG ≅ QR 4. CPCTC

22


1. Proof

Statement Reason 1. TS bisects ∠MHE 1. Given ∠TMS ≅ ∠TES 2. ∠MTS ≅ ∠ETS 2. Definition of ∠ bisector 3. TS ≅ TS 3. Reflexivity 4. ∆SMT ≅ ∆SET 4. SAA congruence 5. TM ≅ TE 5. CPCTC

2. Proof:

Statement Reason 1. PQ // SN 1. Given 2. ∠QPL ≅ ∠NSL 2. If 2 // lines cut by a transversal, ∠PQL ≅ ∠SNL the alternate interior ∠ s are ≅ 3. ∆PQL ≅ ∆SNL 3. ASA congruence 4. QL ≅ LN 4. CPCTC

3. Proof:

Statement Reason 1. XO ≅ YS, OE ≅ SI 1. Given XI ≅ YE 2. IE ≅ IE 2. Reflexivity 3. XE ≅ YI 3. By addition 4. ∆XOE ≅ ∆YSI 4. SSS congruence 5. ∠O ≅ ∠S 5. CPCTC

4. Proof:

Statement Reason 1. ∆PNQ and ∆TNQ 1. Given are isosceles ∆ 2. PN ≅ PQ, TN ≅ TQ 2. Definition of isosceles 3. PT ≅ PT 3. Reflexivity 4. ∆PNT ≅ ∆PQT 4. SSS congruence 5. ∠PNQ ≅ ∠TNQ 5. CPCTC

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5. Proof: Statement Reason

1. HE is the ⊥ bisector of BC 1. Given 2. BE ≅ CE 3. Definition of ⊥ bisector 3. HE ≅ HE 2. Reflexivity 4. ∆HEB ≅ ∆HEC 4. LL congruence 5. ∠B ≅ ∠C 5. CPCTC

6. Proof:

Statement Reason 1. CG ≅ EG, FG ≅ HG 1. Given 2. ∠FGC ≅ ∠HGE 2. Vertical ∠ s are ≅ 3. ∆FGC ≅ ∆HGE 3. SAS 4. ∠C ≅ ∠E 4. CPCTC

7. Proof:

Statement Reason 1. RP ≅ SP 1. Given 2. PQ ≅ PQ 2. Reflexivity 3. ∆PRQ ≅ ∆PSQ 3. HyL congruence 4. RQ ≅ SQ 4. CPCTC

8. Proof:

Statement Reason

1. ∠ 1 ≅ ∠ 2 1. Given 2. ∠RSP ≅ ∠TPO 2. Supplements of ≅ ∠ s are also ≅ 3. O is the midpoint of SP 3. Given 4. SO ≅ PO 4. Definition of midpoint 5. ∠SOR ≅ ∠POT 5. Vertical ∠ s are ≅ 6. ∆SOR ≅ ∆POT 6. ASA congruence

7. RO ≅ TO 7. CPCTC

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9. Proof:

Statement Reason 1. LO ≅ MO 1. Given 2. ∠ LOJ ≅ ∠MOK 2. Vertical ∠ s are ≅ 3. ∆LOJ ≅ ∆MOK 3. LA congruence 4. ∠ J ≅ ∠K 4. CPCTC

10. Proof:

Statement Reason 1. PQ ≅ RS 1. Given ∠QPR ≅ ∠SRP 2. PR ≅ PR 2. Reflexivity 3. ∆QPR ≅ ∆SRP 3. SAS congruence 4. SP ≅ QR 4. CPCTC

Module 1 Geometry of Shape and Size


This module is about undefined terms and angles. As you go over the exercises you will learn to name the real-world objects around you that suggest points, lines and planes. You will develop skills in naming a point, a line and its subsets. You will also learn to name the parts of an angle and determine its measure in degrees.


1. describe the ideas of • point • line • plane

2. name the subsets of a line • segment • ray

3. name the parts of an angle 4. determine the measure of an angle using a protractor 5. illustrate different kinds of angles

• acute • right • obtuse

How much do you know Identify the term described.

1. It has no length, width, or thickness. 2. It has length but no width and no thickness. 3. It is a flat surface that extends infinitely in all directions. 4. It is the union of two noncollinear rays with a common endpoint. 5. It is an instrument used to determine the approximate measure of an angle.

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6. An angle with a measure greater than 0 but less than 90 7. It is a subset of a line with two endpoints. 8. An angle with a measure greater than 90 but less than 180. 9. The geometric figure suggested by the ceiling of your room. 10. It is the intersection of two distinct planes.

What you will do

Lesson 1

Undefined Terms The three undefined terms are point, line and plane. These three undefined terms form the foundation of geometry. Although they will not be defined they will however be used in defining other important terms. For example, space is defined as a set of all points. A point is an exact location in space. It has no length, width or thickness. It is represented by a dot. Look at the tip of your pen. It suggests a point. A point is named by using a capital letter. The points below are named points P, Q and R respectively. • P • Q • R A line has infinite length, but no width and no thickness. It is an infinite set of points that extends infinitely in opposite directions. The pen or pencil you are holding right now is a real world object that suggests a line. A line is represented by . The arrow suggests that the line continues without end in both directions. You can name a line in two ways. One way of naming a line is by using two different capital letters. Observe the line below. It is named line AB written as AB . The double-headed arrow placed over AB indicates that the line has no endpoints. A B

• • Example: Give six names for the line below. S T U

• • • Answers: ST, TS, TU, UT, SU, US

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The second method of naming a line is by using a small letter. The line below is named line m. m Like a line, a plane is also a set of infinite points. However, a plane has infinite width and length but no thickness. It is a flat surface that extends infinitely in all directions. The top of your dining table, the wall of your room and even a page of this module are examples of real-world objects that suggest planes. A slanted four sided figure similar to the one below is used to represent a plane. You can name a plane in three ways. You may use a capital letter placed at one of its corners. The plane below is named plane P. P You may use a small letter placed at one of its corners. The plane below is named plane m. m You may named it by using three points not on a straight line. The plane below is named plane PRQ. • P • Q • R

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The three points below are collinear. Points are collinear if they are on the same line. D E F • • • Example: List all sets of three collinear points in the figure.

• D A C

• • • B • E

Answers: : A, B, C and D, B, E Consider the three points below. It is not possible to draw one straight line through the three points A, B and C. These three points are non collinear points. • A •B • C In the figure below, points A, B, and C are in the same plane. Points such as points A, B, and C, which are in the same plane are called coplanar points. In the same figure, points A, B, C and D are not coplanar because they do not lie in the same plane. Points A, B, C lie in plane P, whereas point D lies in plane Q. P Q A • •D B• C•

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The following statements describe some basic relationships among points, lines and planes

1. Two points determine exactly one line. .

a. Through two different points B and C below, you can draw one and only one line.

B C • • In geometry, line means straight line.

b. It is not possible to draw more than one straight line through given two points. In the following illustration, there is only one straight line that passes through points C and D. The other line is a curve line.

C D • •

2. Three collinear points are contained in at least one plane.

. R

• A Q • B P • C

In the figure, points A, B, and C are collinear. They lie in plane P, plane Q and

plane R. In fact they can be contained in an infinite number of planes.

3. Three non collinear points are contained in exactly one plane.

Q •E

P •D

F•

In the figure, points D, E, and F are not collinear. They are contained in exactly one plane P.

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4. The intersection of two distinct lines is a point.

In the figure, line m and line n intersect and their intersection is point A. m A

n

5. The intersection of two distinct planes is a line.

In the figure below, planes P and Q intersect and their intersection is line AB. Q A B P

6. If two points are in a plane, then the line containing the points is in the same plane. If the two points A and B are in plane P, then the line l which contains them lies also in plane P. A B l • • P 7. A line and a point not on the line are contained in exactly one plane. In the figure, point A does not lie on line BC. This point and line BC are contained in one plane P. This is the same as saying they determine exactly one plane P. C • A B P • •

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8. Two intersecting lines are contained in exactly one plane. Example: Lines a and b which intersect at point P are contained in exactly one plane Q. There is no other plane that can contain them. a P b Q 9. If a line not contained in a plane intersects the plane, the intersection is a single point. In the figure, plane P does not contain line m. The intersection of line m and plane P is a single point Q. m Q P Try this out Set A. Determine the undefined term suggested by each of the following.

1. the tip of a pencil 2. the top of a coffee table 3. telephone wires 4. the wall of a room 5. the surface of the page of a book 6. the ruler’s edge 7. the tip of a needle 8. a window pane

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9. the floor of your bedroom 10. the string on a guitar

Set B. Write True or False Use the three-dimensional figure below for exercises 1-10. A F I E J D B H G C

1. Points A, F, B are collinear. 2. Points A, E, B are collinear 3. Points B, G and C are on the same line 4. Points G, C, D are not on the same line. 5. Points A, I, H are coplanar. 6. Points A, F, G are coplanar. 7. Points A, F, G , E are coplanar. 8. Points A, F, B, G are coplanar. 9. Points A. I, C are collinear and coplanar. 10. Points A, F, C are collinear and coplanar.

Set C. Complete the following statements.

1. A ______ is an exact location in space. 2. A ______ has infinite length but no width and no thickness. 3. A ______ has infinite width and length but no thickness. 4. Two points determine exactly one ______ 5. Three _________points are contained in at least one plane. 6. Three _________points are contained in exactly one plane. 7. The intersection of two distinct planes is a ________. 8. The intersection of two distinct lines is a ________. 9. Two intersecting lines determine a _________. 10. If a line not contained in a plane intersects the plane, the intersection is a single

_____.

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Lesson 2

The Subsets of a Line

The subsets of a line are segment and ray. A segment has two endpoints. It is named by its endpoints.

The segment below may be named AC or CA. A vinculum is placed above its name to distinguish it from the name of a line where the same letters are used. A C • • Example: Write the name of each segment. E F M N a. • • b. • • Answers:

a. EF or FE b. MN or NM

The length of a segment is the distance between its endpoints. Example: If the distance between points C and D below is 9 cm. then the length of segment CD is 9 cm. This is written as CD = 9 cm. Notice that there is no vinculum above CD. C D • • A segment may be defined as the union of points A, C together with all the points between them. Illustration: A B C • • • In the above segment, A and C are the endpoints of the segment. There are points between A and C. These points together with the endpoints A and C make a segment. In the above figure, point B is just one of the points between A and C.

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A point such as point B above is between point A and C if and only if (1) A, B, and C are distinct points, (2) they are collinear and (3) AB + BC = AC. These three conditions must be satisfied before it can be said that B is between A and C. The word distinct in the first condition means that the three points are different from one another.

Examples:

1. Draw points C, D, and E on a line. How many different segments are determined? Name them.

C D E • • •

Answers: CD, DE, CE

2. If AB =5 cm, BC = 7 cm, and AC = 12 cm. Is B between A and C? A B C • • • In the figure, A, B, and C are different points on the same line. The sum of the lengths of AB and BC is equal to the length of AC. AB + BC = AC 5 cm + 7 cm = 12 cm Since the three conditions are satisfied, therefore B is between A and C. Example: Is F in the figure below between E and G? F• E • • G

In the figure , points E, F and G are not collinear, hence F is not between point E and point G. Also, EF + FG ≠ EG.

A ray is a subset of a line that has one endpoint and extends forever in one direction.

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Example:

The part of the line from point B that goes on indefinitely to the right is a ray. The part of the line from point B that goes on indefinitely to the left is another ray.

A B C l • • •

The ray which starts from point B that goes on indefinitely to the right is named ray BC denoted by BC. Its endpoint is B. Notice that when you name a ray, you use two capital letters, and its endpoint is written first. The other ray in the above figure is ray BA, denoted by BA.

Example: Write a name for each figure. J K N M a. • • b. • • Answers: a. JK b. MN Another term you should learn in this lesson is the term opposite rays. Two rays are opposite if they are subsets of the same line and have a common endpoint.

A B C • • •

BC and BA are opposite rays. They are parts of the same line l and their common

endpoint is B. F G E D • • • • DE and FG are not opposite rays because they are not subsets of the same line. A B C D l • • • • BA and CD are not opposite rays because they do not have a common endpoint. Example: D Name all the points, segments and rays in the figure. Answers: A B C The points are A, B, C, and D. The segments are AB, BC, AC, and BD. The rays are BA, BC, and BD

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Try this out Set A: Use the figure below for exercises nos. 1-10 A B C D • • • • 1. Name the ray with endpoint at B going in the direction of D. 2. Name the ray with endpoint at C going in the direction of A. 3. Name the segment joining point B with point D. 4. Give two opposite rays with common endpoint C. 5. What is the intersection of ray BD and ray CA? 6. Name the ray opposite BC. 6. Name the ray opposite CA. 7. What point is between points B and D? 8. Give another name for BC 9. Give another name for CB Set B

Write true or false

Use the following figure C• A B D E • • • •

1. AB + BD = AD 2. AB + BE = AE 3. AC + CD = AD 4. B is between A and D 5. C is between B and D 6. A, B, C, D are collinear 7. AB = AD – BD 8. DE and BA are opposite rays. 9. Ray BE can be named BD. 10. Ray DA can be named AD.

Set C

Fill in the blanks 1. A segment has _________endpoints

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2. A __________is a subset of a line with one definite endpoint and extends infinitely in one direction.

3. _________are two collinear rays with a common endpoint.

Use the figure at the right for exercises nos. 4-8

• • • C D E

4. CD + ______ = CE. 5. The ray opposite DE is ________ 6. The ray with endpoint C going in the direction of D is _________ 7. The ray with endpoint E going in the direction C is __________ 8. The point between two other points is ________.

9. If two points P and Q are exactly the same point, then the distance between them is

______

10. The endpoint of each ray in the figure is _______

Lesson 3

Angles

An angle is a union of two noncollinear rays with a common endpoint. The common endpoint is called the vertex of the angle and the two rays are called sides. Example: The figure below is an angle. Its vertex is point B and its two sides are BA and BC. The symbol used for an angle is ∠. The angle in the example can be named ∠ ABC. It can also be called ∠CBA. The letter representing the vertex is written between the other two letters. A • B • C An angle may be written in other ways.

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Example: Angle DFG can also be named ∠EFG, ∠GFD, ∠GFE, ∠F and ∠a. Angle HIJ can be named ∠ 1. D • H • E • a 1 F • I • G J There are times when it is not advisable to use the vertex letter in naming an angle. Using it may result to confusion. Example: Angle ABC below may be named ∠ B

A •

B • C

Angle ABC below should not be named ∠B. In the figure, there are three angles with vertex B. They are ∠ABC, ∠DBC and ∠ABD.

A • C •

B • Example: D Give three different names for the angle shown below.

P • • • Answers: Q R ∠ PQR, ∠ RQP, ∠Q An angle separates a plane into three sets: the points on the angle, the interior of the angle, and the exterior of the angle.

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Example: In the figure, points T and S are on ∠ABC.. Point P is in the interior and points Q and R are in the exterior of the angle. • • Q • Exterior Interior • P B • • • T C • R Exterior The Measure of an angle You can determine the measure of an angle in degrees by means of a protractor. You can do this by placing the center mark of the protractor on the vertex of the angle you want to measure and then placing the 0 degree mark on one side of the angle. Then read the number where the other side crosses the scale. You can also use a protractor in constructing an angle of a given measure. Example: Measure angle ABC below. A

B C The measure of ∠ABC as indicated in the protractor is 90 degrees. This can be

written in two ways. ∠ ABC = 900 (Angle ABC equals 90 degrees.) m∠ABC = 90. (The measure of ∠ABC is 90.)

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In this module the measure of an angle is always greater than 0 degree but less than

180 degrees. This restriction will be followed in this module because of the definition of an angle. Addition of Angles

The measures of two or more angles can be added. Example The measure of ∠A is 500 and the measure of ∠ B is 600. Find the sum of their measures. A 500 B 600 m ∠ A + m ∠ B = 500 + 600. = 1100

Example ∠ABD and ∠CBD are two coplanar angles with a common side BD. If m ∠ ABD = 40 and m ∠ CBD = 30, find the measure of angle ABC. A D 400 B 300 C m ∠ ABD + m ∠ CBD = 400 + 300 = 700 Example: If m∠ABC = 120, m∠ABD = 2x + 10, and m∠CBD =3 x [Use the preceding figure] Find m∠ABD.

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m∠ABD + m∠CBD = m∠ABC 2x + 10 + 3x = 120 2x + 3x = 120 – 10 5x = 110 x = 22 2x + 10 = 2(22) + 10 = 44 + 10 = 54 Kinds of Angles

There are three kinds of angles according to measure. They are the following. 1. Acute angle- is an angle with a measure grater than 0 but less than 90.

∠ABC below is an acute angle.

A 450 C B

2. Right angle- is an angle with a measure of 90. ∠ABC below is a right angle. D E F The symbol in the corner of a right of the figure indicates that the measure of the angle is 90.

3. Obtuse angle – is an angle with a measure greater than 90 but less than 180,

G 1100 H I

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Try this out Set A.

1. Name the angle below in three ways. A

B C

2. Which is the vertex letter in angle STG? 3. Name the three angles in the figure below.

D F E G 4. What are the sides of ∠ BET? 5. What is the common side of ∠ABD and ∠CBD? 6. Into how many sets does an angle separate a plane? 7. Is the vertex of an angle in its interior? 8. How many angles are there in the figure?

E

A B C D

9. Is the figure below an angle? Why? Why not?

A B C 10. Explain why it is not correct to name the angle below ∠ Q? P R Q S

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Set B. Use the figure below for exercises 1-10. The three angles in the figure are coplanar.

A D C B

1. If m∠ABD = 80 and . m∠CBD = 40, find the m∠ ABC. 2. If m∠CBD = 30 and m∠ABD = 85, find the m∠ABC 3. If m∠ABD =45.5 and m∠CBD= 44, find the m∠ABC. 4. If the m∠CBD = 30.5 and m∠ ABD = 65, find the m∠ABC. 5. If m∠ABC =110 and . m∠CBD = 40, find the m∠ ABD. 6. If m∠ABC =115 and . m∠ABD = 40, find the m∠ CBD 7. If m∠ABC =84 and . m∠CBD = 2x, and m∠ABD = 4x, find m∠ABD 8. If m∠ABC =96 and . m∠CBD = x, and m∠ABD = 2x, find m∠CBD

Use a protractor for exercises 9-10

9. Construct an angle with a measure of 45 degrees. 10. Construct an angle with measure of 125 degrees.

Set C. Use the figure below for exercise 1 –3.

A • D • • F • K •G • • • B E C •H

1. Name all the points in the interior of ∠ABC. 2. Name all the points in the exterior of ∠ABC. 3. Name all the points that are neither on the exterior nor interior of ∠ABC. Use the figure below for exercises 4-6 E D A C

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In the figure BC and BA are opposite rays.

4. Name all the angles determined in the figure. 5. Tell whether the angles in the figure are acute, right or obtuse. 6. Name the two angles with the same measure. 7. In the figure below, BA, BD and BC are coplanar rays. If ∠ABC is a right angle, find x

A

D (3x)0 x0 B C

8. Which of the following angles is an acute angle a. b. c.

9. Using your protractor, find the measure of each angle below.

a. b.

10. Draw angles with the following measures. a. 1250 c. 900 b. 350 d. 1400

Let’s summarize

1. The three undefined terms in geometry are point, line and plane. 2. A line is an exact location in space. It has no length, width or thickness. 3. A line has infinite length, but no width and no thickness. 4. A plane has infinite width and length but no thickness. 5. Two points determine exactly one line. 6. Two distinct lines intersect in only one point 7. Collinear points are points on the same line. 8. Coplanar points are points on the same plane. 9. Three collinear points are contained in at least one plane. 10. Three noncollinear point are contained in exactly one plane.

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11. The intersection of two distinct planes is a line 12. If two points are in a plane, then the line containing the points is in the same line. 13. A line and a point not on the line, are contained in exactly one plane. 14. Two intersecting lines are contained in exactly one plane. 15. If a line not contained in a plane intersects the plane, the intersection is a single

point. 16. A segment is a subset of a line that consists of two endpoints and all the points

between them. 17. A ray is a subset of a line with a definite endpoint and extends infinitely in one

direction. 18. An angle is the union of two noncollinear rays with a common endpoint. 19. An angle separates the plane into three sets: the points in the interior of the angle,

the points in the exterior of the angle and the points on the angle itself. 20. A protractor is used to measure an angle in degrees. 21. An angle with a measure greater than 0 but les than 90 is an acute angle. 22. An angle with a measure of 90 is a right angle. 23. An angle with a measure greater than 90 but less than 180 is an obtuse angle.


1. It is flat surface that extends infinitely in all directions. A. Point C. Plane B. Line D. rectangle

2. It is a set of points that extends forever in opposite directions. A. Point C. Plane B. Line D. Space.

3. Which of the following is false? A. Exactly one plane contains two intersecting lines. B. Two points determine a line. C. The intersection of two distinct planes is a line D. Three collinear points are contained in exactly one plane

4. Which of the following real objects suggest a point? A. The edge of the beam of a building B. The corner of the Main street and the 1st Ave. C. The floor of a newly constructed building. D. The wall of your room.

5. It is a subset of a line with a definite endpoint and extends infinitely in one direction. A. Ray C. Opposite Rays

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B. Segment D. Plane

6. It is the union of two noncollinear rays with a common endpoint. A. Plane C. Space B. Angle D. Segment

7. It is an angle with a measure greater than 0 but less than 90. A. Acute angle C. Obtuse angle B. Right Angle D. non of these

8. It is angle with a measure of 90. A. Acute angle C. Obtuse angle

B. Right Angle D. none of these

9. Which of the following angles is obtuse?

I III. II. IV. A. I only C. II only B. I and II D. I and III 10. It is used to measure an angle in degrees.

A. Compass C. protractor B. Ruler D. tape measure

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Answer Key


1. point 2. line 3. plane 4. angle 5. protractor 6. acute angle 7. segment 8. obtuse angle 9. plane 10. line


1. point 2. plane 3. line 4. plane 5. plane 6. line 7. point 8. plane 9. plane 10. line

Set B

1. True 2. False 3. True 4. True 5. True 6. True 7. False 8. True 9. True 10. False

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Set C

1. point 2. line 3. plane 4. line 5. non-collinear 6. collinear 7. line 8. point 9. plane 10. point

Lesson 2 Set A .

1. BC or BD 2. CA or CB 3. BD 4. CD and CA or CD or CB 5. BC 6. BA 7. CD 8. C 9. CB 10. BC, AB

Set B 1. True 2. True 3. False 4. True 5. False 6. False 7. True 8. False 9. True 10. False

Set C

1. two 2. ray

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3. opposite rays 4. DE 5. DC 6. CD or CE 7. EC or ED 8. D 9. 0 10. B

Lesson 3 Set A

1. ∠ABC, ∠CBA, ∠B 2. T 3. ∠DEF, ∠GEF, ∠DEG 4. EB, ET 5. BC 6. Three sets including itself 7. No. 8. 4 9. No 10. Q is the vertex of the three angles. ∠Q may mean ∠PQR, ∠RQS, and ∠PQS

Set B

1. 120 2. 115 3. 89.5 4. 95.5 5. 70 6. 75 7. 56 8. 32 9. Use your protractor 10. Use your protractor

Set C.

1. F, K 2. G, H 3. A, D, B, E, C 4. ∠ABE, ∠ABD, ∠DBE, ∠EBC, ∠DBC 5. ∠ABE and ∠DBE are acute angles ∠ABD and ∠DBC are right angles ∠EBC is obtuse

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6. ∠ABD and ∠DBC. Both are right angles with measure of 900 each. 7. 22.5 8. a 9. Use your protractor 10. Use your protractor

What have you learned 1. C 2. B 3. D 4. B 5. A 6. B 7. A 8. B 9. C 10. C

Module 2 GEOMETRY OF SHAPE and SIZE

What this module is about There are different shapes around us. Everywhere you look, you see varied geometric figures. Many of these figures are called polygons. This module will help you be on familiar terms with them. You will also learn to distinguish types of triangles; knowledge of which will help you communicate ideas about real life situations.

What you are expected to learn This module is designed to

1. illustrate different kinds of polygons and identify the parts of a regular polygon;

2. differentiate convex and non-convex polygons; 3. illustrate a triangle, its basic and secondary parts and 4. classify triangles according to angles and sides.


Select the correct answer for each question: 1. Which of the following is a polygon?

a. b. c. d.

2. A pentagon is a polygon with

a. 4 b. 5 c. 6 d. 7 sides

3. One of the following is NOT a convex polygon. Which one is it? a. b. c. d.

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4. In ∆ABC, AB, AC and BC are called

a. sides b. interior angles c. vertices d. exterior angles

5. Using the figure at the right, BD is a /an B

a. median b. altitude c. ∠ bisector d. ⊥ bisector A D C

6. In ∆DEF, m∠E = 120. ∆DEF is

a. acute b. right c. obtuse d. equilateral triangle

7. It is a triangle with all sides congruent

a. scalene b. isosceles c. obtuse d. equilateral

8. The figure below is a regular hexagon. ∠COD is a / an

a. inscribed angle b. central angle c. obtuse angle d. interior angle C O

9. The longest side in a right triangle is known as D

a. leg b. hypotenuse c. vertex d. base

10. In isosceles triangle ABC, AB = BC. AB and BC are called

a. bases b. vertices c. hypotenuse d. legs

11. A polygon with no given number of sides can be named as

a. dodecagon b. undecagon c. n-gon d. gon

12. The sides of a polygon is made up of

a. segments b. rays c. lines d. planes

13. A line segment from a vertex of a triangle to the midpoint of the opposite side.

a. altitude b. perpendicular bisector c. median d. angle bisector

14. A right triangle with congruent legs is

a. equilateral b. isosceles c. scalene d. isosceles right triangle

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15. An equilateral triangle is also isosceles. The statement is a. always true b. sometimes true c. never true d. can’t be determined

What you will do

Lesson 1

Different Kinds of Polygons and the Parts of a Regular Polygon

The word polygon is from 2 Greek words poly (many) and gon (sides). The following are polygons:

Each of the figures above is closed, made up of segments and the segments or sides intersect only at their endpoints. A polygon is a closed figure made up of segments that intersect at their endpoints and no two consecutive segments are collinear. Each line segment is a side of the polygon and each endpoint is a vertex.

The following are NOT polygons: The figure is made up of segments but it is not closed The figure is closed but not entirely made up of segments Curved portion The figure is closed but it is curved

4

The figure is closed, made up of segments but 2 sides

intersect at another segment. The intersection should be at the endpoints only. Look at the figures below. Can you identify the parts/portion that are polygons?

Polygons have special names depending on their number of sides.

Name of polygon Number of sides Triangle 3 Quadrilateral 4 Pentagon 5 Hexagon 6 Heptagon 7 Octagon 8 Nonagon 9 Decagon 10

Examples:

Triangle Quadrilateral Pentagon

Hexagon Heptagon Octagon

Nonagon Decagon

5

11 – gon 24 - gon

Polygons with more than 10 sides are often referred to as 11-gon, 12-gon, 13-gon and so on. When the number of sides is not given, the polygon is simply called n-gon.

Drawing Tip: To draw a polygon easily,

Step 1: Lightly sketch a circle. Step 2: Place the points you need on the circle. Step 3: Then connect the points to form your polygon. Step 4: Erase the circle.

. . . . . . . . . . . . . . . . Step 1 Step 2 Step 3 Step 4

Regular Polygons: Polygons with equal sides and equal angles are called regular polygons. Study the following illustrations: I II

Equal sides equal sides and = angles not equal angles

600 600 III IV

equal angles equal sides not = sides equal angles

6

Figures I and IV are regular polygons. A regular polygon has the following

parts: A B Vertex angle F o C

Exterior angle central angle

E D G ABCDEF is a regular polygon. All sides are equal and all angles have the same measure.

Vertex Angle A vertex angle is an angle formed by the intersection of two sides of the polygon. ∠B is a vertex angle. ∠C is also a vertex angle. Can you name 4 more vertex angles? They are ∠D, ∠E, ∠F, and ∠G.

Central Angle ∠COD is a central angle because the center of the polygon is also the vertex of the angle. Name some more central angles. There are more than 6. When you name them, be sure that the middle letter is O.

Exterior Angle If you extend one side of the polygon, you form an exterior angle. In the above figure, ∠CDG is an exterior angle. Let’s have some more exterior angles: 1 4 5 8 2 3 6 7

Lesson 2

Differentiate convex and non-convex polygons Polygons are also classified as convex and non-convex. Let’s find out how they differ by studying these examples:

7

Convex Polygons: Non-Convex Polygons: A polygon is convex if a segment joining any two interior points lies completely within the polygon. within outside

convex non-convex Try this out A. Which of the following is a polygon? 1 2 3 4 5 6 7 B. Tell whether the polygon is convex or non-convex:

___________ ___________ ___________ ___________ _________ ___________ ___________ ___________ ___________ _______

8

C. Name the polygon: 1 2 3 4 5 6 7 8 D. If possible, draw a polygon that fits each description

1. A regular quadrilateral. 2. A non-convex pentagon 3. An equilateral octagon.

4. An equilateral nonagon. 5. A convex 15-gon

Let’s summarize

1. A polygon is a closed figure made up of segments that intersect at their endpoints and no two consecutive segments are collinear.

2. Each line segment is a side of the polygon and each endpoint is a vertex.

3. Polygons are named based on the number of their sides 4. Polygons with equal sides and equal angles are called regular polygons.

5. A polygon is convex if a segment joining any two interior points lies

completely within the polygon.

Lesson 3

A Triangle. Its Basic and Secondary Parts You have learned that the triangle is the simplest polygon because it has the least number of sides.

But how do we name triangles? Triangles are named by using the letters at their vertices. Starting from any vertex, go clockwise or counterclockwise.

9

Examples:

A Clockwise naming of triangle C B Triangle ABC is the same as ∆BCA, ∆CAB. A Counterclockwise C B ∆ACB = ∆CBA = ∆BAC. The basic parts of a triangle are sides, vertices and angles. 1. The sides of ∆ABC are AB, BC, and AC. A Side AC side AB C B

Side BC 2. The vertices are the endpoints A, B, and C. A

vertex A

vertex C vertex B

C B 3. The angles. Can you name the 3 angles of the figure below?

10

They are ∠ABC, ∠BCA, and ∠CAB.

A ∠CAB C B ∠ACB ∠CBA Name Triangle I and II in different ways and then complete the table below. P I II Q S R Name Sides Vertices Angles Triangle I ∆QPR,__,__ QP,__,__ Q,__.__ ∠PQR,__,__ Triangle II ∆PSR,__,__ PS,__,__ P,__,__ ∠PSR,__,__

Secondary parts of a triangle a. Median

A median is a segment whose endpoints are a vertex of a triangle and the midpoint of the opposite side. (A midpoint divides a segment into two equal segments).

P P P S T Q V R Q R Q R

PV is a median of ∆PQR. V is the midpoint of QR. QS is a median of ∆PQR. S is the midpoint of PR. RT is a median of ∆ PQR. T is the midpoint of PQ.

11

b. Altitude

An altitude is a segment from a vertex of the triangle perpendicular to the opposite side or to the line containing the opposite side. (Perpendicular means the two segments form a right angle). X U X Y U Y V X V U V Y

In all the figures above, XY is an altitude. c. Angle Bisector M C C M C M MC bisects angle M in all the triangles above. MC is called angle bisector. You will notice that a triangle has 3 medians, 3 altitudes and 3 angle bisectors.

If the triangle is equilateral, the median is the same as the angle bisector and altitude. See the figure below: C J K L

12

Lesson 4

Classification of triangles according to sides and angles Triangles are classified by their sides into 3 categories:

1. Scalene – all sides have different lengths 2. Isosceles – 2 sides have the same length 3. Equilateral – all 3 sides have the same length

10 7

8 scalene ∆ isosceles ∆ equilateral ∆ Triangles are classified by their angles into 3 categories:

1. Acute ∆ - all 3 angles are acute 2. Right ∆ - one angle is right 3. Obtuse ∆ - one angle is obtuse

All angles are acute Acute ∆ 600

600 600

An acute triangle is equiangular if all 3 angles have the same measure.

Right angle Right ∆

13

Obtuse angle

Obtuse ∆

Parts of special triangles: Triangle MON is an isosceles triangle. M Vertex angle leg

base angle O N Base Legs – the congruent sides Base – the third side Vertex angle – the angle opposite the base Base angles – the angles at the base Triangle ACB is a right triangle A Hypotenuse legs

C B Hypotenuse – the longest side; the side opposite the right angle Legs – the sides forming right angle Try this out

A. Using the figure at the right, identify the following. A 1. AB, AC, BC 2. C, B, A 3. AE F D 4. BD 5. CF

B C E

14

B. Match each triangle with all words that describe it: a. scalene b. isosceles c. equilateral d. acute e. right f. obtuse g. equiangular

(1) (2) (3)

600 6 1000 8 600 600 500 300

9 (4) (5)

C. Tell if each statement is true or false. Draw a figure to justify your answer. 1. A triangle can be isosceles and acute. 2. A triangle can have two obtuse angles.

3. A triangle can be obtuse and scalene.

4. A right triangle can be equilateral.

5. A triangle can be right and isosceles.

6. A triangle can have two right angles.

7. A triangle can have at most 3 acute angles.

8. A triangle can have at least one (1) acute angle.

9. An equilateral triangle is also an acute triangle.

10. An equiangular triangle can never be a right triangle.

Tessellations A tessellation is a design in which congruent copies of a figure are arranged to fill the plane in such a way that no figures overlap and there are no gaps.

15

Examples: Try to make a tessellation for each figure below. Which figure could not be used for tessellation?

You can also combine figures to make a tessellation: To make an original pattern for a tessellation, start with a parallelogram. Make congruent changes on one or both pairs of sides:

16

Let’s summarize

1. The basic parts of a triangle are sides, angles and vertices while the

secondary parts are median, altitude and angle bisector.

2. A median is a segment whose endpoints are a vertex of a triangle and the midpoint of the opposite side.

3. An altitude is a segment from a vertex of the triangle perpendicular to the

opposite side or to the line containing the opposite side.

4. Triangles are classified according to sides and angles. a. Scalene – all sides have different lengths b. Isosceles – 2 sides have the same length c. Equilateral – all 3 sides have the same length d. Acute ∆ - all 3 angles are acute e. Right ∆ - one angle is right f. Obtuse ∆ - one angle is obtuse

What have you learned Select the correct answer for each question:

1. Which of the following is a polygon?

a b c d

2. A heptagon is a polygon with

a. 4 b. 5 c. 6 d. 7 sides

3. One of the following is NOT a convex polygon. Which one is it?

a b c d

4. In ∆ABC, A, C and B are called

a. sides b. interior angles c. vertices d. exterior angles

17

5. Using the figure at the right, BD is a /an B

a. median b. altitude c. ∠ bisector d. ⊥ bisector A D C

6. In ∆DEF, m∠E = 90. ∆DEF is

a. acute b. right c. obtuse d. equilateral triangle

7. It is a triangle with all angles congruent

a. scalene b. isosceles c. obtuse d. equiangular

8. ∠CBA is a B A

a. inscribed angle b. central angle C O c. obtuse angle d. vertex angle D

9. The longest side in a right triangle is known as

a. leg b. hypotenuse c. vertex d. base

10. In isosceles triangle ABC, AB = BC. AB and BC are called a. bases b. vertices c. hypotenuse d. legs

11. A polygon with no given number of sides can be named as a. dodecagon b. undecagon c. n-gon d. gon 12. The sides of a polygon is made up of a. segments b. rays c. lines d. planes 13. A line segment from a vertex of a triangle to the midpoint of the opposite

side. a. altitude b. perpendicular bisector c. median d. angle bisector 14. A right triangle with congruent legs is a. equilateral b. isosceles c. scalene d. isosceles right triangle 15. An isosceles triangle is also equilateral. The statement is a. always true b. sometimes true c. never true d. can’t be determined

18

Answer Key


1. C 2. B 3. A 4. A 5. B 6. C 7. D 8. B 9. B 10. D 11. C 12. A 13. C 14. D 15. A

Try this out Lesson 2 A.

1. Polygon 2. not 3. not 4. not 5. Polygon 6. Polygon 7. not

B. 1. non-convex 2. convex 3. non-convex 4. convex 5. non-convex 6. non-convex 7. non-convex 8. convex 9. convex 10. non-convex

19

C. 1. Nonagon 2. Heptagon 3. Quadrilateral 4. Octagon 5. Heptagon 6. Pentagon 7. Hexagon 8. Quadrilateral

D.

1. Draw a square

2. Any 5-sided polygon that is not convex

3.

4. – 5. Refer to the drawing tip

Try this out Lesson 4 A.

1. Sides 2. Vertex / vertices 3. altitude 4. angle bisector 5. median

B. 1. isosceles right triangle 2. scalene right triangle 3. isosceles acute triangle 4. equilateral / equiangular 5. scalene obtuse triangle

C.

1. true 2. false

20

3. true 4. false 5. true 6. false 7. true 8. false 9. true 10. true


1. B 2. D 3. C 4. C 5. B 6. B 7. D 8. D 9. B 10. D 11. C 12. A 13. C 14. D 15. C

Module 3 Quadrilaterals

What this module is about This module is about Quadrilaterals. As you go over the exercises, you will develop skills in identifying Quadrilaterals and their parts and ability to appreciate their application in daily life. Treat the lesson with fun and take time to go back if you think you are at a loss. What you are expected to learn This module is designed for you to learn

1. illustrate a Quadrilateral and its parts 2. illustrate the different kinds of Quadrilaterals

How much do you know? Write the letter of the correct answer Z E Y N O

1. Quadrilateral ZENY is a ____. a. parallelogram b. trapezoid c. trapezium d. rectangle

2. A diagonal of quadrilateral ZENY is ____. a. ZE b. ZN c. ZO d. ZY

3. Y∠ is opposite angle ____. a. Z∠ b. E∠ c. N∠ d. O∠ 4. DH & ET are the ____ of the Quadrilateral DETH a. median b. altitude c. bases d. legs 5. If BH ET≅ then BETH is a/an ____. a. rectangle b. trapezium c. isosceles trapezoid d. square

6. BH ET≅ are the _____ of Quadrilateral BETH. a. legs b. bases c. median d. altitude

7. whish is the median of quadrilateral DETH a. BH b. MP c. EA d. DH

8. What kind of parallelogram is quadrilateral DETH? a. rhombus b. square c. parallelogram d. rectangle

9. Using the figure at the right which is a rhombus? a. ACGH b. CEFG c. BDGH d. AEFH A B C D E

H G F

10. Which is a rhombus and a rectangle? a. AEFH b. CEFG c. ACGH d. BDGH

What you will do

Lesson 1

Identifying and Naming Quadrilaterals.

A quadrilateral is a polygon of four sides Example: M O D E O O L A D E R P M E You ca name a quadrilateral by its vertices. The order of vertives is very important. You read or write the four letters clockwise or counterclockwise. Examples: R O You can name a quadrilateral at the Right as. ROSE or OSER or SERO or EROS Or RESO or ESOR or SORE E S The name of this quadrilateral can be: K L

KLMN or LMNK or MNKL or NKLM Or NMLK or MLKN or LKNM or KNML N M Try this out.

A. Which of the following is a quadrilateral or not. 1) 3) 5) 2) 4) 6)

7) 9)

8) 10)

B. Name the following quadrilaterals. C 1) O 2) H O D L E P 3)B E 4) O P H A T S 5) A L R 6) B M A Y Y 7)M A 8)A B

Y N D C 9) O R 10) S P R Q S E

C. Name & Identify the Quadrilaterals in the figure.

A B C D EEE hhhdhjhsjsjj Name 10 Quadrilaterals.

Lesson 2

Parts of a Quadrilateral

A quadrilateral has the following parts: 4 sides 4 vertices 4 angles 2 diagonals You can take a look on quadrilateral LOVE

• the sides are: LO , OV , EV , LE • the vertices are: L, O, V, E • the angles are L∠ , O∠ , V∠ , E∠ • the diagonals are LV & OE

diagonals are segments joining opposite vertices.

The vertices E and O; L and V are opposite vertices. Vertices L and O, O and V, V and E, E and L are consecutive vertices.

Two sides with a common vertex like LO and OV are consecutive sides.

So, OV and VE , VE and EL , EL and LO are other pairs of consecutive sides. On the otherhand, LO and EV , OV and LE are opposite sides.

Two angles with a common side like L∠ and O∠ are consecutive angles,

the others are O∠ and V∠ , V∠ and E∠ , E∠ and L∠ , on the other hand, E∠ and O∠ ; L∠ and V∠ are opposite angles.

Try this out A. Using quadrilateral COLA, identify

1. two pairs of opposite vertices 2. two pairs of opposite angles 3. two pairs of opposite sides 4. one pair of diagonals 5. four pairs of consecutive vertices 6. four pairs of consecutive angles 7. four pairs of consecutive sides 8. four sides of quadrilateral POEM 9. four angles of quadrilateral POEM

EM

P

L

O

O

E V

C

LAO

10. two diagonals that can be drawn in quadrilateral POEM B. Fill the blanks:

1) ______ the vertex opposite S 2) ______ is the opposite side of PS 3) Q∠ is opposite _________ 4) PQ and QR are _______ sides. 5) The diagonals that can be drawn are QS and ______ 6) BC and CD are ___________ sides. 7) B and C are consecutive vertices, B and _____ are also consecutive

vertices. 8) AC and _____ are the diagonals of quadrilateral ABCD. 9) A∠ and D∠ are ______ angles.

10) A∠ and _____ are opposite angles. C. Choose the letter of the correct answer. Use quadrilateral D E T H

1. How many diagonals has quadrilateral DETH? a. one b. two c. three d. four

2. The sides of quadrilateral DETH are DH , HT , TE and ______ a. HE b. DT c. DE d. DO

3. LH∠ is opposite of what angle? a. E∠ b. D∠ c. T∠ d. O∠

4. The opposite side of ET is _________ a. DO b. DE c. HT d. DH

5. The diagonals of quadrilateral DETH are _________ a. DO & OT b. HE & DT c. DH & ET d. DE & HT

Given: quadrilateral PQRS

6. How many vertices are there in

Quadrilateral PQRS?

E D

A D

C

B

RS

QP

O

H I

Q R

S P

a. one b. two c. three d. four 7. A pair of consecutive sides is PQ and ______

a. QR b. RS c. PR d. QS 8. A pair of opposite vertices is P and ______

a. R b. Q c. S d. M 9. How many pairs of opposite s∠ has quadrilateral PQRS?

a. one b. two c. three d. four 10. How many pairs of consecutive sides has quadrilateral PQRS?

a. one b. two c. three d. four

Quadrilateral

Trapezium Parallelogram Trapezoid

Lesson 3 Parallels & Perpendiculars

Kinds of Quadrilaterals As you can see the diagram of the different kinds of quadrilaterals, you can notice the characteristics of the sides of each quadrilateral. Before you proceed to the definition of each quadrilateral, you must know first the meaning of the following: i. parallel lines two lines are parallel if they are coplanar and they do

not meet. ii. perpendicular lines two lines are perpendicular if they intersect and form

a right angle. Examples:

A B

C D

L1 L2

X

S R

Y

O

N A

1. two lines are parallel if they are coplanar

and they do not meet. 2. l1 is parallel to l2. In symbol l1 // l2 3. If XY intersects RS at O and XOS∠ is

a right angle, then XY is perpendicular to RS , in symbol: XY ⊥ RS

4, If MAN∠ is a right angle, then

ANAM ⊥ Try this out.

A. Which of the following seem to be parallel? Write yes if it is and no if not. 1) 7) 2) 8) 3) 9) 4) 10) 5) 6) B. Write parallel or perpendicular.

K L

1) 4) 2) 5) 3) 6) A pair of lines of your pad paper 7) The corner of the blackboard 8) Railroad tracks 9) The grills 10) A pair of consecutive sides of a picture frame C. Are we parallel or perpendicular? 1) Two lines on a plane which do not meet. 2) Two intersecting lines which form a right angle. 3) 4) BDAB & 5) EHFG & 6) FGKI &

I J GF

E HA

B

C

D

7) TUPT & 8) PSQR & 9) TUPV & 10) BOBY &

P V Q

R S

T U

B O

DY

Lesson 4

Kinds of Quadrilaterals

1. Trapezium If a quadrilateral has no parallel sides, then it is a trapezium. MORE is a

trapezium. 2. Trapezoid

If a quadrilateral has exactly a pair of parallel sides, then it is a trapezoid. If EPHO // , then HOPE is a trapezoid.

3. Parallelogram

If a quadrilateral has two pairs of parallel sides, then it is a parallelogram. If POST // and POSP // then STOP is a parallelogram.

A. Identify: ______ 1. A quadrilateral whose opposite sides are parallel. ______ 2. A quadrilateral with no parallel sides. ______ 3. A quadrilateral with a pair of parallel sides. ______ 4. Quadrilateral WHEN.

D R

E M

H O

PE

O

TS

P

W H

N E

______ 5. Quadrilateral COLD. ______ 6. Quadrilateral SORE ______ 7. Quadrilateral PEAS ______ 8. Quadrilateral TOME ______ 9. Quadrilateral PEAK ______ 10. Quadrilateral LOAF

C O

LD

C O

LD

S O

RE

XX

X

X

X

X

P

S

E

A

T O

ME

X

X

X

X

REX

S A

T OX

P E

AK

ME X

REX

S A

T OX

L

O

A

F

B. Select the correct word from the set (trapezium, trapezoid,

parallelogram)

1. quadrilateral DETH 2. quadrilateral DANS 3. quadrilateral SNLU 4. quadrilateral LOVU 5. quadrilateral JMRK 6. quadrilateral OMPV 7. quadrilateral RWCR 8. quadrilateral WGFC 9. quadrilateral PWGY 10. quadrilateral TERY C. Draw the following figures: 1. trapezium ZENY 2. trapezoid BETH 3. parallelogram LOVE 4. parallelogram with diagonals BWDL & 5. trapezoid with diagonals MCIE &

A

N

DS

H T Y H G F

EU V

R P W C

LO

M R Q

K J

Two tents are fixed above. Give me: 6. 7. three parallelograms 8. 9. 10.

A

ED

F B

C

H

G J H

K LN

M

Two trapezoids



This module is about polygons, specifically angles of the polygon. This module will teach you how to find the sum of the interior angles of the polygon. You will also discover other interesting facts about the interior and exterior angles of the polygon. You can also form your own generalization regarding the sum of the measures of the interior and exterior angles of a convex polygon.


This module will help you:

1. Determine the sum of the measures of the interior and exterior angles of the triangle.

2. Determine the sum of the measures of the interior and exterior angles of a quadrilateral.

3. Make generalizations on the sum of the measures of the interior and exterior angles of a polygon.


1. What is the measure of each angle of an equilateral triangle? 2. If the sum of the two angles of a triangle is 130º, what is the measure of the third

angle? 3. One acute angle of a right triangle is 27º. What is the measure of the other acute

angle? 4. What is the measure of an exterior angle of an equilateral triangle? 5. If the measure of the vertex angle of an isosceles triangle is 50º, what is the

measure of each base angles? 6. What is the sum of the measures of the interior angles of a quadrilateral? 7. How many sides does a regular polygon have if each interior angle is 120º? 8. The measures of the four angles of a pentagon are 140º, 75º, 120º, and 115º.

Find the measure of the fifth angle. 9. If the radius of the circle is 12 cm, what is the diameter of the circle? 10. What is the appropriate name of the given circle?

2

What you will do

Lesson 1

Determining the Sum of the Measures of the Interior Angles of a Triangle.

In any given triangle, say ∆ABC, there

are three interior angles and by observation, A there are 6 exterior angles. In the given figure, 1 2 ∠A, ∠B, and ∠C are the interior angles of ∆ABC. Each interior angle has two angles adjacent to it. For ∠A, ∠ 1 and ∠ 2, for ∠B, ∠ 3 and ∠ 4, and for ∠C, ∠ 5 and ∠ 6. 6 3 Aside from being adjacent, ∠ 1 C 5 4 B and ∠ 2 are both supplementary to ∠A. Those six angles which are adjacent to the interior angles of the triangle are called exterior angles of ∆ABC. Measures A of the exterior angles will be discussed in the next lesson.

To get the sum of the measures of ∠A

∠B and ∠C, you can use the protractor and then add the sum of their measures. There is another way of doing this. Here are the steps: C B

A 1. Prepare a cut out of any triangle of any size. 2. Cut off the three angles as in the figure 3. On the given line l, align the 3 cut out angles with all the vertices coinciding with point O. 4. All three vertices should fit perfectly on one side of the line. 5. The measure in degree of the angles about a point on the same side of a line is 180. B C

From the two experiment that you did, you can conclude that the m∠A + m∠B + m∠C = 180. B l A C Examples: o 1. What is the measure of the third angle of a triangle if the measures of the two angles are

a. 46º, 82º

3

b. 5121 º, 67

21 º

Solutions: Let x be the measure of the third angle

a. x + (46º + 82º) = 180º x + 128º = 180º x = 180º - 128º x = 52º, measure of the third angle

b. x + (51

21 º + 67

21 º) = 180º

x + 119º = 180º x = 180º - 119º x = 61º , measure of the third angle P

2. The measure of one acute angle of right triangle PQR is 33º. What are the measures of the other angles? R Q Solution:

Since ∆PQR and ∠R = 90º , and ∠Q = 33º, then ∠P =

33º + 90º + ∠P = 180º 23º + ∠P = 180º ∠P = 180º - 123º ∠P = 57º Alternate Solution:

Remember, the two acute angles of a right triangle are complementary, hence the sum of their measures is 90º. In other words, the sum of the measures of the two acute angles of a right triangle is 90. You can use this knowledge in solving the given problem.

33º + ∠P = 90º ∠P = 90º - 33º ∠P = 57º A 3. In an isosceles triangle, one base angle measures 49º. Find the measures of the other two angles. Solution: Let ∆MAN be isosceles. ∠M ≅ ∠N

m∠M = 49º m∠N = m∠M = 49º M N

m∠M + m∠N + m∠A = 180º 49º + 49º + m∠A = 180º 98º + m∠A = 180º m∠A = 180º - 98º m∠A = 82º

4

Try this out A. Given the measures of the two angles of a triangle, determine the measure of the third. 1. 36º, 70º 2. 58º, 50º 3. 90º, 25º 5. 64

21 º, 59º

4. 35º, 35º B. Given the measure of one base angle of an isosceles triangle, determine the measures of the other two angles. 1. 45º 4. 37

21 º

2. 50º 5. 6341 º

3. 53º C. Given the measure of the vertex angle of an isosceles triangle, find the measure of a base angle. 1. 75º 4. 88º 2. 100º 5. 97º 3. 39º D. In a right triangle, the measure of one acute angle is given. Find the measure of the other acute angle. 1. 45º 4. 43º 2. 36º 5. 61

21 º

3. 52º E. Solve the following problems.

1. The measure of one acute angle of a right triangle is twice that of the other acute angle. Find the measures of the three angles.

2. Find the measures of the angles of a triangle that are in the ratio 1:2:3.

3. The vertex angle of an isosceles triangle is 15 degrees more than three times the

base angle. Find the measures of all the three angles of the triangle.

4. n a triangle, one angle is ten degrees more than twice the smallest angle and the third angle is four less than three times the smallest angle. Find the measures of the three angles.

5. What are the measures of the angles of an isosceles right triangle?

5

Lesson 2

Sum of the Measures of Exterior Angles of a Triangle. Sum of the Interior Angles of a Quadrilateral

To find the sum of the measures of the exterior angles of a triangle, it is necessary to apply the previous lesson; the sum of the measures of the interior angles of the triangle is 180º. The phrase sum of the exterior angles of a triangle means you will use one exterior angle from each vertex of the triangle.

Thus in the given figure, A to find the sum of the exterior angles 1 of ∆ABC, ∠ 1, ∠ 2 and ∠ 3 are the designated exterior angles. Each of the exterior angle is adjacent to one of the interior angle of the triangle. 3 2

C B Study the following procedure:

m∠A + m∠B + m∠C = 180 Sum of the angles of a triangle is 180 m∠ 1 + m∠A = 180 Definition of exterior angle of a triangle m∠ 2 + m∠B = 180 Definition of exterior angle of a triangle m∠ 3 + m∠C = 180 Definition of exterior angle of a triangle m∠ 1 + m∠ 2 + m∠ 3 + m∠A + m∠B + m∠C = 540 APE

m∠A + m∠B + m∠C = 180 by subtraction m∠ 1 + m∠ 2 + m∠ 3 = 360

Therefore, the sum of the measures of the exterior angles of a triangle one on each vertex is equal to 360.

What about a quadrilateral? How do you get the sum of the measures of the interior angles of a four-sided polygon? Consider quadrilateral ABCD. B What is the sum of the measures of A 1 2 ∠A, ∠B, ∠C and ∠D? To determine this, draw a diagonal, say BD. Two triangles are formed, ∆ABD and ∆BCD. 4 3 D In each triangle, the sum of the C interior angles is 180.

6

To determine the sum of the four angles, study the following procedures.

In ∆ABD, m∠A + m∠ 1 + m∠ 4 = 180 In ∆BCD, m∠C + m∠ 2 + m∠ 3 = 180 m∠A + m∠ 1 + m∠ 2 + m∠C + m∠ 3 + m∠ 4 = 360 by APE

But m∠B = m∠ 1 + m∠ 2 and m∠D = m∠ 3 + m∠ 4 By substitution, m∠A + m∠B + m∠C + m∠D = 360.

Therefore, the conclusion is that, the sum of the measures of the interior angles of a quadrilateral is 360. How about the sum of the exterior angles of a quadrilateral? Is it the same as that of the sum of the exterior angles of the triangle? Exploration: Given the figure at the right. Quadrilateral ABCD has one exterior angle in each vertex. Those are ∠ 1, ∠ 2, ∠ 3 and∠ 4. Let us name the interior angles, ∠A, ∠B, ∠C and ∠D. To determine the sum of the measures of B ∠ 1, ∠ 2, ∠ 3 and ∠ 4, study the following steps. 1 2

A From the previous paragraph, it is known that m∠A + m∠B + m∠C + m∠D = 360.

By definition of exterior angle, 4

m∠A + m∠ 1 = 180 D 3 C m∠B + m∠ 2 = 180 m∠C + m∠ 3 = 180 m∠D + m∠ 4 = 180

By Addition Property of Equality or APE (m∠A + m∠B + m∠C + m∠D) + (m∠ 1 + m∠ 2 + m∠ 3 + m∠ 4) = 720 Subtract: (m∠A + m∠ B+ m∠C + m∠D) = 360 m∠ 1 + m∠ 2 + m∠ 3 + m∠ 4 = 360 Therefore, the sum of exterior angles of a quadrilateral is equal to 360. Examples:

1. The measures of the three angles of a quadrilateral are 75º, 101º and 83º. Find the measure of the fourth angle.

7

Solution: Let x = the measure of the fourth angle.

x + 75º + 101º + 83º = 360º x + 259º = 360º x = 360º - 259º x = 101º S 2. Given quadrilateral PRST. Using the figure, R 3x - 23

find x, m∠P, m∠R, m∠S and m∠T. 2x

Solution: Write the representations for each angle. m∠P = 2x – 10 m∠R = 2x m∠S = 3x – 23 2x - 10 m∠T = x + 33 P x + 33 T Using the representations for each angle, substitute to the equation m∠P + m∠R + m∠S + m∠T = 360 (2x – 10) + (2x) + (3x – 23) + (x + 33) = 360 8x + 33 – 33 = 360 8x + 0 = 360 8x = 360

x = 8

360

x = 45 m∠P = 2x – 10 = 2(45) – 10 = 90 – 10 m∠P = 80 m∠R = 2x = 2(45) m∠R = 90 m∠S = 3x – 23 = 3(45) – 23 = 135 – 23 m∠S = 112 m∠T = x + 33 = 45 + 33 m∠T = 78 To check: m∠P + m∠R + m∠S + m∠T = _____

80 + 90 + 112 + 78 = 360

8

Try this out A. Use the given information to find the measure of each 4 5

numbered angle in the figure. ∠ 1 ≅ ∠ 5. 1. m∠ 2 = 106, m∠ 6 = 85 2. m∠ 1 = 43, m∠ 7 = 112 3. m∠ 2 = 115, m∠ 4 = 31 1 2 3 6 4. m∠ 5 = 51, m∠ 3 = 72 7

B. In the figure, determine the measure of the numbered angles. 460

4. m∠ 1 820 5. m∠ 2 2 6. m∠ 3 1 7. m∠ 4 3 390 270 4

C. In quadrilateral MNOP, ∠M ≅ ∠O, M 2x–7 x + 16 N ∠N ≅ ∠P. Using the given in the figure, find the measures of all the angles. P O D. Use the given information to find the measures of the following angles. 3 2 410 5 1. ∠ 1 2. ∠ 2 3. ∠ 3 580 4. ∠ 4 4 450 750 5. ∠ 5 6. ∠ 6

Lesson 3

Finding the Sum of the Interior Angles and Exterior Angles of Any Polygon. To determine the sum of the measures of interior angles and exterior angles of any polygon (polygon of n sides) like the given pentagon, you have to find out the number of triangles that can be formed by drawing the diagonals without intersecting each other except at their endpoints.

9

In quad ABCD, draw diagonal AC. A B With the introduction of diagonal AC, two triangles are formed. The same thing happens if diagonal BD is drawn. If the sum of the measures of the interior angles C of each triangle is 180, then for quadrilaterals, the sum of the interior angles is 360. D The polygon at the right has five (5) sides. If the diagonal is drawn from one vertex, say D, D E

only two diagonals can be drawn, DF andDG . With these two diagonals, three non-overlapping H triangles are formed, ∆DEF, ∆DFG and ∆DGH. F

If in each triangle, the sum of the measures of interior angles is 180, then in pentagon DEFGH, the sum of the measures of the five angles is equal to 3(180) or 540. G This findings can be summarized using the formula Sa = (n – 2) 180, where Sq is the sum of the measures of interior angles of the polygon, n is the number of sides of the polygon. This formula can be applied to convex polygons only. You also have to remember this: The sum of the measures of interior angles of a regular polygon with n sides is equal to Sa = (n – 2) 180.

Regular polygons are polygons which are both equilateral and equiangular. A polygon is convex if and only if the lines containing the sides of the polygon do not

contain points in its interior. To get the measure of each interior angle of a regular polygon you simply remember

this. The measure of each interior angle of a regular polygon with n sides is equal to Ia =

nn 180)2( − , where, Ia is an interior angle, and n is the number of sides of the polygon.

If you can get the measure of each interior angle of a regular polygon, then you can

also compute for the measure of each exterior angle of a regular polygon. You only have to remember these formulas. From the generalizations that you had made in the earlier part of the lesson, the sum of the measures of the exterior angle of the convex polygon is equal to 360. This includes the regular polygons too.

10

The measure of each exterior angle of a regular polygon with n sides is given by the

formula: Ea = n

360 , where Ea is an exterior angle of a regular polygon, and n is the number

of sides of the polygon. Examples: 1. What is the sum of the measures of the interior angles of a convex polygon with

a. 11 sides b. 15 sides

Solutions: a. Sa = (n – 2) 180

= (11 – 2) 180 = 9(180) = 1620

b. Sa = (n – 2) 180 = (15 – 2) 180 = 13(180) = 2340 2. How many sides does a convex polygon have if the sum of the measures of its interior angles is 1440? Solution: Sa = (n – 2) 180 1440 = (n – 2) 180 1440 = 180n – 360

n = 180

3601440 +

n = 1801800

n = 10 The polygon has 10 sides. 3. Find the sum of the measures of the interior angles of a convex heptagon. Solution: Sa = (n – 2) 180 = (7 – 2) 180 = 5(180) = 900

11

4. Find the measure of each interior angle of a regular 11-sided polygon. Solution:

Measure of each interior angle is Ia = n

n 180)2( −

Ia = 11

)180)(211( −

Ia = 11

)180(9

Ia = 11

1620

5. How many degrees are there in each of the exterior angle of a regular hexagon. Solution:

Ea = n

360

Ea = 6

360

= 60 6. If each exterior angle of a polygon is 36º, how many sides does the polygon have? Solution:

Ea = n

360

36 = n

360

36n = 360

n = 36360

n = 10 7. If each interior angle of a polygon is 150º, how many sides does the polygon have? Solution: m + 150 = 180 m = 180 – 150 m = 30 Using the figure, with m as the exterior angle of the polygon,

Ea = n

360

12

30 = n

360

30n = 360

n = 30360 m 150

n = 12 Try this out A. Complete the table below.

Polygon

No. of Sides

No. of

Vertices

No. of Diagonals from

one vertex

No of

triangles

Sum of the Measures of

interior Angles 1 Quadrilateral 2 Pentagon 3 Hexagon 4 Nonagon 5 Dodecagon 6 N-gon

B. Find the number of sides in each regular polygon with exterior angle of the given measure. 1. 90º 2. 45º 3. 36º 4. 60º C. How many sides does a regular polygon have if each angle measures: 5. 120º 6. 90º 7. 135º 8. 150º D. Find the sum of the measures of the exterior angles one at each vertex, of each of the following convex polygon. 9. decagon 10. hexagon 11. octagon 12. dodecagon E. The sum of the measures of the interior angles of a polygon is given. Find the number of sides of the convex polygon. 13. 1260º 14. 1620º 15. 2520º 16. 3060º

13

F. The sum of the measures of the interior angles of a regular polygon is given. Find the measure of each angle. 17. 1260º 18. 2520º 19. 3060º

20. 720º

G. Solve the following problems. 620 21. Find y in the given figure. 680 y 22. The measures of the two angles of a quadrilateral are 105º and 107º. If the remaining two angles are congruent, find the measure of each angle. 23. In a convex quadrilateral WXYZ, the measure of ∠X is twice the measure of ∠Y. If ∠Y ≅ ∠ Z ≅ ∠W, find the measure of each angle. 24. The number of diagonals from a vertex in a regular polygon is 7. How many sides does the polygon have? What is the measure of each interior angle? 25. Each interior angle of a regular polygon is twice the measure of each exterior angle. How many sides does the polygon have? M

N

26. In the figure, like markings indicate 420 congruent parts. Find the measure of each unknown angle. P 420

O

A

27. ∠ABC is a right angle, ∠CDB is D a right angle. If the m∠C = 32, find the measures of the following angles: ∠A ∠ADB ∠ABD C B ∠CBD S

7

28. In the given figure, RV = UV, RU = SU, 4 ST = TU, m∠RVU = 92, m∠STU = 106, R 6 m∠U = 137, find the measures of all 1 the numbered angles. T 1060 5 3

V 920 2 U

14

Let’s summarize

1. The sum of the measures of the interior angles of a triangle is 180. 2. The sum of the measures of the interior angles of a quadrilateral is 360.

3. The sum of the measures of the interior angles of a convex polygon is given by the

formula (n -2)180, where n is the number of sides of the polygon.

4. The sum of the measures of the exterior angle one on each vertex of a polygon is 360.

5. The measure of each interior angle of a regular polygon of n sides is given by the

formula Ia = n

n 180)2( − , where n is the number of sides of the polygon.

6. The measure of each exterior angle of a regular polygon of n sides is given by the

formula Ea = n

360 , where n is the number of sides of the polygon.

What have you learned Answer each question as indicated.

1. What is the measure of each angle of a regular pentagon? 2. The sum of the measures of the two angles of a triangle is 127. What is the measure

of the third angle?

3. The measure of one base angle of an isosceles triangle is 67. Find the measure of the vertex angle.

A B 4. Given quadrilateral ABCD. If m∠C = 60,

and if ∠A ≅ ∠B, what is the m∠D if its measure is 27 less than the measure of ∠A. D C

5. What is the sum of the measures of the interior angles of an octagon? 6. If each interior angle of a polygon is 160º, how many sides does the polygon have?

15

P Q 7. Find the value of x using the given in the 2x + 10 3x

figure x R

2x x T S 8. The angles of a triangle are in the ratio 1:3:5. Find the measures of the three angles. 9. If each interior angle of a regular polygon measures 150º, how many sides does the

polygon have? 10. In the figure, ≅AN XN . If ∠A = 62º, find the measure of ∠ 1.

A

1 X N

16

Answer Key

How much do you know.

1. 60º 2. 50º 3. 63º 4. 120º 5. 65º 6. 360º 7. 6 8. 90 9. 24 cm 10. circle A

Try this out Lesson 1 A. 1. 74º 2. 72º 3. 65º 4. 110º 5. 56

21 º

B. 1. 45º, 90º 2. 50º, 80º 3. 53º, 74º

4. 3721 º, 105º

5. 63 41 º, 53

21 º

C. 1. 52

21 º

2. 40º

3. 7021 º

4. 46º

5. 4121 º

17

D. 1. 45º 2. 54º 3. 38º 4. 47º 5. 28

21 º

E. 1. 30º, 60º, 90º 2. 30º, 60º, 90º 3. 33º, 33º, 114º 4. 29º, 68º, 83º 5. 45º, 45º, 90º Lesson 2 A. 1. m∠ 3 = 74, m∠ 4 = 53, m∠ 5 = 21, m∠ 1 = 21, m∠ 7 = 95 2. m∠ 2 = 11, m∠ 3 = 69, m∠ 4 = 26, m∠ 6 = 68, m∠ 5 = 43 3. m∠ 1 = 34, m∠ 3 = 65, m∠ 5 = 34, m∠ 6 = 81, m∠ 7 = 99 4. m∠ 1 = 51, m∠ 2 = 108, m∠ 4 = 21, m∠ 6 = 57, m∠ 7 = 123 B. 5. m∠ 1 = 114 6. m∠ 2 = 66 7. m∠ 3 = 68 8. m∠ 4 = 32 C. m∠M = m∠O = 2(57) – 7 = 107 m∠P = m∠N = 57 + 16 = 73 D. 1. m∠ 1 = 120 2. m∠ 2 = 19 3. m∠ 3 = 62 4. m∠ 4 = 15 5. m∠ 5 = 45 6. m∠ 6 = 60

18

Lesson 3 A.

Polygon

No. of Sides

No. of

Vertices

No. of Diagonals from one

vertex

No of

triangles

Sum of the Measures Of interior

Angles 1 Quadrilateral 4 4 1 2 360 2 Pentagon 5 5 2 3 540 3 Hexagon 6 6 3 4 720 4 Nonagon 9 9 6 7 1260 5 Dodecagon 12 12 9 10 1800 6 N - gon n N N – 3 N – 2 (n – 2 )180

B. 1. 4 sides 2. 8 sides 3. 10 sides 4. 6 sides C. 5. 6 sides 6. 4 sides 7. 8 sides 8. 12 sides D. 9. 360 10. 360 11. 360 12. 360 E. 13. 9 sides 14. 11 sides 15. 16 sides 16. 19 sides F. 17. 140º 18. 157.5º 19. 161.05º 20. 120º G. 21. y = 130º 22. Each of the remaining angle measures 74º. 23. m∠W = 72, m∠X = 144, m∠Y = 72, m∠Z = 72 24. 10 sides, 144 25. 6 sides 26. m∠N = 69, m∠MON = 69, m∠P = 69, m∠PMO = 69

19

27. m∠A = 58, m∠ADB = 90, m∠ABD = 32, m∠CBD = 58 28. m∠ 1 = 44, m∠ 2 = 44, m∠ 3 = 37, m∠ 4 = 37, m∠ 5 = 56, m∠ 6 = 62 m∠ 7 = 62 What have you learned

1. 108º 2. 53º 3. 46º 4. 82º 5. 1080º 6. 18 sides 7. x = 50º 8. 20º, 60º, 100º 9. 12 sides 10. m∠ 1 = 124


What this module is about This module will discuss the perimeter of the commonly used polygons in geometry such as triangles, quadrilaterals and others (with their corresponding formulas). In addition, this module will also expound on the circumference of the circle. Furthermore, this module will help you apply these concepts in solving problems associated with real life. What you are expected to learn This module is written for you to

1. Recall the different plane figures and their properties which are commonly used in geometry.

2. Name the properties of the sides of different polygons. 3. Define perimeter and determine the different formulas of getting the perimeter of the

different polygons . 4. Define a circle. 5. Identify the lines and segments associated with circles. 6. Define circumference of a circle and determine the formula for getting the

circumference of the circle. 7. Use the formulas for finding perimeter and circumference in solving real life

problems.

How much do you know Answer the following questions as indicated.

Find the perimeter of a regular polygon(indicated) given the length of a side. 1. triangle, s = 13 cm 2. square, s = 10 cm 3. rhombus, s = 11.5 dm 4. pentagon, s = 9.25 cm 5. hexagon, s = 12.73cm

2

6. The circumference of a circle is 66 cm. Find the radius of the circle. Use π = 722 .

7. The length of a rectangle is 13 more than twice its width. If the perimeter is 116 cm, find the dimensions of the rectangle.

8. One of the sides of a rhombus is 2x + 1. What is its perimeter? 9. The diameter of a circle is 14 cm. Find its circumference. Use π = 3.14. 10. If the perimeter of a regular hexagon is 69 cm, what is the length of each side?

What you will do

Lesson 1

Perimeters of Polygon The perimeter of a polygon is the distance around it. It can be computed by getting the sum of the length or measures of all the sides. If the polygon is identified as regular, then the perimeter is computed by simply multiplying the given measure of the side with the number of sides. Let us recall the different formulas for finding the perimeter of different polygons. Triangle: For a general triangle, the perimeter (P) is a b P = a + b + c , where a, b, and c are the measures of the sides. c For isosceles triangle, perimeter is P = m + m + n, where m is the m m length of one of the legs, and n is the length of the base. n For equilateral triangle, the perimeter is P = 3s, where s is the length of s one of the equal sides.

3

For polygon of four sides or quadrilaterals, consider the given figures. x Given quad ABCD, then its perimeter is P = w + x + y + z, where w, x, y and z are the length of the sides w y z For a square, its perimeter is P = s + s + s + s, where s is the length of a side. Therefore, s P = 4s l For a rectangle, the perimeter is P = l + l + w + w, where l is the length and w is the width. Therefore, w w P = 2l + 2w or P = 2 (l + w) l For a parallelogram, the perimeter is a P = a + a + b + b, where a, and b are the lengths of the two consecutive sides

of the parallelogram. Thus b b P = 2a + 2b P = 2(a + b) a For other figures like pentagon and hexagon or those polygons with more than five

sides, the formula is practically the same. Get the sum of all the length of the sides. a For the given pentagon, the perimeter is P = a + b + c + d + e, where a, b, c, b c d, and e are the measures of the sides. e f For a regular pentagon with a as the length of a side, the perimeter is P = 5a s r For a hexagon, the perimeter is t P = r + s + t + u + v + w, where r, s, t, u, v, and w are the length w of the sides. u v

4

For a regular hexagon whose length of side is denoted by r, the perimeter is given by the formula

P = 6r Example 1. Find the perimeter of the following figures. 1. 2. 3. 3cm 5 cm

6 cm

6 cm 5 cm

4 cm 4. 5. 7 cm 6. 7 cm

5 cm 3 cm 5 cm

9 cm 6 cm

Solutions: 1. P = (3 + 5 + 6)cm = 14 cm 2. P = [2(6) + 4] cm = 12 + 4 = 16 cm 3. P = 3(5 cm) = 15 cm 4. P = 4(6 cm) = 24 cm 5. P = [2(7cm) + 2(5cm)] = 14cm + 10cm = 24cm 6. P = (3 + 7 + 5 + 9)cm = 24cm Example 2. Find the perimeter of the following regular polygons given the measure of a side (s). 1. triangle, s = 10 cm

5

2. square, s = 7cm 3. pentagon, s = 13.5cm 4. hexagon, s = 11.25 cm 5. nonagon, s = 9.3 cm Solutions: 1. P = 3s = 3 (10cm) = 30 cm 2. P = 4s = 4(7cm) = 28cm 3. P = 5s = 5(13.5cm) = 67.5 cm 4. P = 6s = 6(11.25cm) = 67.50cm 5. P = 9s = 9(9.3cm) = 83.7cm Example 3 The width of a rectangular garden is 5m. If the length is 2m more than its width, how many meters of fencing materials are needed to enclose the whole garden? How much will the cost of fencing material be if the owner pays P59 per meter? Solution: Width (w) = 5m Length (l) = 2m more than the width Length (l) = 5m + 2m = 7m Get the perimeter: P = 2(l + w) = 2(7m + 5m) = 2(12m) = 24m Cost of materials = 24m (P59 /m) = P1416.00

6

Example 4. The perimeter of a regular pentagon is 120 m. What is the length of each side? Solution: Pentagon has 5 sides, and since it is a regular pentagon, the sides are equal. P = 5s 5s = 120 m

s = 5

120m

s = 24m. Example 5. A saleslady is preparing to wrap a box whose dimensions are 30cm by 20cm by 7cm. If she is going to tie a ribbon around the box as in the figure, how long should the ribbon be if she will allow 25cm for the design at the top? How much will the ribbon cost if a meter is P8.25. Solution: To find the length of the ribbon, find the two perimeters. P1 = 2(30cm) + 2(7cm) = 60cm + 14cm = 74 cm P2 = 2(20cm) + 2(7cm) = 40cm + 14cm = 54 cm Total length = P1 + P2 + 25cm = 74cm + 54cm + 25cm = 128cm = 1.28 m Total cost = 1.28 m(P8.25) = P12.6225 ≈ P12.65 Try this out A. Find the perimeter of the following using the given information in the figure. 1. 2. 3. 9 5 7 15.3

10 10

7

4. 21 5. 12

15 10 16

B. The sides of an isosceles triangle are given. Find the perimeter.

1. 8, 8, 13 2. 5, 4, 4 3. 7, 7, 5 4. 6, 7, 6 5. 1.2, 1.2, 1.3 6. 8.13, 8.14, 8.13 7. 12.75, 12.75, 10 8. 1, 2, 2

C. Given are the length and width of a rectangle. Find its perimeter.

1. 4.5, 8.3 2. 12.01, 19.22 3. 18.3, 21.5 4. 2.03, 5.43 5. 9.75, 12.25 6. 3 , 35 7. 54 , 58 8. a, 3a 9. x + 1, 5x + 4 10. x2 + 1, x2 + 25

D. Find the perimeter of the following

1. A square with side of 25cm. 2. A parallelogram whose consecutive sides are 21cm and 27.5 cm respectively. 3. A regular hexagon whose side is ( )̀732 + cm. 4. A rectangle 17m by 11m. 5. A rhombus with side of 18.23dm.

E. Solve the following problems.

1. The length of a rectangle is 4 less than three time its width. If the perimeter of the rectangle is 272, what are its dimensions?

8

●A

2. The perimeter of a regular pentagon is 215 cm. Find the length of each side.

3. In an isosceles trapezoid, the length of one leg is 53dm and the median is 63dm. Find its perimeter.

4. James always jogs around his rectangular pool, 10m by 6m. If he jogs around it 5

times, what is the distance covered by James? 5. Find the perimeter of the given quadrilateral.

4x - 9

2x + 1

3x - 4

6. When the side of a square is increased by 2cm, its perimeter becomes 40 cm. What is the length of the original square?

7. When the side of a square is reduced by 7cm, the perimeter becomes 84. What is

the perimeter of the original square?

8. The length of a rectangle is 64 cm. Its width is 13cm more than half its length. What is the perimeter of the rectangle?

9. The base of an isosceles triangle is 15 more than one-third of the length of the leg. If

a leg measures 57 dm, find the perimeter of the triangle.

10. If the side of a square is increased by 25%, by how many percent will the perimeter increase?

Lesson 2

Circle and its Circumference A circle is defined as the set of points equidistant from a fixed point called the center.

Though the center is not a part of the circle, it is essential that every circle has a center A circle on a plane can be represented geometrically or algebraically. In this part of the lesson, geometric circles will be the focus of our discussion. In the given circle, A is the center of the circle, thus we can name the circle, circle A. There are other lines and segments associated with a circle like radius, chord, diameter, secant and tangent.

9

In circle O, OA is a segment from the center to the circle. OA therefore is a radius of the circle. Aside from OA , B other radii are OB and OC . BC is a segment whose endpoints are points on the circle and it passes through 0 the center of the circle. Therefore, BC is called the diameter of the circle. If the endpoints of a segment are points on the circle, it is called chord of the circle. A C By this definition, the diameter is also a chord of a circle. By inspection, it is very easy to recognize that the length of the diameter is twice that of the radius. Therefore, in the figure, OCOBBC += . Other lines and segments associated with a circle D E are shown in the figure. They are DEBC, and line t. BC has its endpoints on the circle. Thus it is an example of chord ●A C of the circle. It is not a diameter since it did not pass through B the center of the circle. DE intersects the circle at two points D and E. It is called a secant of the circle. Line t intersects the T X circle at only one point X. Line t is called tangent of the circle. Example 1. Given circle A and points B, H, C, D and E on it. Name: B H

1. 4 radii 2. 2 chords A C 3. a diameter 4. a secant 5. a tangent E

D F Solutions:

1. AEAHACAB ,,, 2. ECEH , 3. EH 4. EC 5. DF

Every polygon has its own perimeter. Likewise a circle has its own perimeter too. To distinguish it from those of polygons, we call the distance around the circle circumference. There is also a formula for finding the circumference of a circle. The letter “C” will be used to represent the circumference of the circle. The formula that we will use is

10

C = 2π r, where r is the radius of the circle and π is an irrational number whose

value is approximated at 3.1416 or 722 .

An alternative formula for circumference can be used utilizing the diameter instead of the radius of the circle. C = π D, where D is the length of the diameter of the circle. Example 1. Find the circumference of a circle of radius 9cm. Solution: C = 2π r C = 2π (9cm) C = 18π cm

If there is assigned value for π , say 722 , then

C = 2(722 )(9cm)

C = 7

396 cm

C ≈ 56.57cm If π ≈ 3.1416, then C = 2(3.1416)(9cm) C ≈ 56.55cm Example 2. The diameter of a circle is 12.6 dm. What is its circumference? Express the answer in terms of π . Solution: C = π D C = π (12.6dm) C = 12.6π dm Example 3. The circumference of a circle is 39.27cm. Find the radius of the circle. Use π =3.1416. Solution: C = 2π r

r = π2C

r = )1416.3(2

27.39

r = 6.25cm

11

Try this out A. Find the length of the diameter of a circle given its radius.

1. 24cm 2. 19 m 3. 3.96dm 4. .08 cm

5. 74 dm

6. 18 m 7. (x+2) cm

8. 2x km

9. nm + m

10. 4

2 ba − dm

B. Use the figure and answer each of the following: B C

1. What is the center of the circle? D 2. Name the circle. E 3. Name 5 radii 4. Name 2 diameters A F 5. Name all the chords 6. Name a tangent of the circle H G 7. Name a secant

C. Find the circumference of the circle given the radius of the circle. Express answers in terms of π .

1. 19cm 2. 25dm 3. 11cm 4. 8cm 5. 13.75cm 6. 2.03cm

7. m75

8. cm5 9. a dm 10. (x + 3)cm

12

D. Find the circumference of the circle if the diameter is given. Use π = 3.1416.

1. 24cm 2. 50dm 3. 35m

4. km43

5. 12.56m 6. 19.75cm

7. km31

8. 18.7cm 9. 25.76cm 10. 10.705cm

11. m52

12. 105.031cm 13. (a+b)dm 14. ( )12 + km 15. 56 dm 16. 3x2 cm 17. (5x – 1) cm 18. 47 +x dm 19. 3.1416m

20. 35

3+

m

E. Given the circumference of the circle, find the radius and the diameter of the circle. Use the appropriate value of π .

1. 36π 2. 58π 3. 126π 4. 39π 5. 101π 6. 37.24π 7. 65.78π 8. 137.5π 9. 89.93π 10. 452.76π 11. 14.25 12. 549.78 13. 298.452 14. 314.16

13

15. 75.5 16. 78.54 17. π3 18. 18456.9 19. 72 π 20. (10x – 4y) π

F. Solve the following problems.

1. An artificial lake has a diameter of 34 m. If Teena jogs around it 6 times, how many meters will that be? (Use π =3.14)

2. At one point in a race, Joseph was 15 m behind Allan and 18 m ahead of Brad. Brad

was trailing Nick by 30m. Allan was ahead of Nick by how many meters? 3. What is the perimeter of a square inscribed in a circle of radius 10 cm?

Let’s summarize

1. Perimeter of a polygon is the distance around the polygon. 2. The general formula for finding the perimeter of a polygon is

Perimeter = s1 + s2 + s3 +. . . + sn-1 + sn , where sn is the measure of the sides and n is the number of sides of the polygon.

3. For regular polygons, the perimeter is equal to the measure of a side multiplied by the number of sides.

4. A circle is the set of points equidistant from a fixed point called the center. The

center of the circle defines the name of the circle.

5. Every circle must have a center.

6. Lines and segments associated with circle are the following:

Radius – the segment joining the center and any point on the circle. Chord – segment joining any two points on the circle. Diameter – chord which passes through the center. Secant – a line intersecting a circle at two points. Tangent – a line intersecting a circle at only one point.

7. Circumference of a circle is the distance around a circle.

14

8. The formula for finding the circumference of a circle is C = 2π r , where r is the radius

of the circle and π is an irrational number approximately equal to 3.1416 or 722 .


1. What is the perimeter of a regular heptagon with side of 23 cm? 2. The length of a leg of an isosceles trapezoid is 15 dm. The length of the median is

35dm. What is the perimeter of the trapezoid? 3x - 7

3. Find the perimeter of the given rectangle. 2x + 9 4. The circumference of a circle is 141π . Find the radius of the circle. 5. If the side of a square is increased by 4 cm, the perimeter becomes 136 cm. Find the

length of the side of the original square. 6. The perimeter of a regular pentagon is 17.5 m. What is the length of each side? 7. What is the longest chord in a circle? 8. If the diameter of a circle is 25 cm , what is the length of the radius of the circle?

9. The base of an isosceles triangle is 27 dm. If the length of a leg is 11 more than

one-third of the base, find the perimeter of the triangle. D 3x - 7 C

10. ABCD is a parallelogram. Using the given in the figure, find x if the perimeter is 130 cm. 2x + 1

A B

15


1. 52 cm 2. 40 cm 3. 46 dm 4. 46.25 cm 5. 76.38 cm 6. 10.5 cm 7. length = 43, width = 15 8. 8x + 4 9. 43.96 cm 10. 11.5 cm

Lesson 1 A.

1. 24 2. 24 3. 61.2 4. 72 5. 48

B.

1. 29 2. 13 3. 19 4. 19 5. 3.7 6. 24.4 7. 35.5 8. 5

C.

1. 25.6 2. 62.46 3. 79.6 4. 14.92 5. 44 6. 12 3 7. 24 5 8. 8a 9. 12x + 10 10. 4x2 + 52

16

D. 1. 100 cm 2. 97 cm 3. ( )cm42312 + 4. 56 m 5. 72.92 dm

E.

1. length = 101, width = 35 2. 43 cm 3. 232 dm 4. 160 m 5. 11x - 11 6. 8 cm 7. 112 cm 8. 218 cm 9. 148 dm 10. 25%

Lesson 2 A.

1. 48 cm 2. 38 m 3. 7.92 dm 4. 0.16 cm

5. dm78

6. 26 m 7. (2x + 4)cm 8. x cm 9. nm +2 m

10. 2

2 ba − dm

B.

1. E 2. circle E 3. EAEGEFEDEC ,,,, 4. CGAD, 5. CGADAB ,,

6. HG 7. AB

17

C.

1. 38π cm 2. 50π dm 3. 22π cm 4. 16π cm 5. 27.5π cm 6. 4.06π cm

7. π7

10 m

8. 52 cm 9. 2a dm 10. (2x + 6) cm

D.

1. 75.3984 cm 2. 157.08dm 3. 109.956 m 4. 2.3562 km 5. 39.458 m 6. 62.0466 cm 7. 1.0472 km 8. 58.74797 cm 9. 80.93 cm 10. 33.63 cm 11. 1.2566 m 12. 329.96 cm 13. 3.1416(a+b) dm 14. 7.58 km 15. 42.15 dm 16. 9.4248x2cm 17. (15.708x – 3.1416) cm 18. 3.1416( 47 +x ) dm 19. 9.8696 m 20. 1.8 m

E.

1. r = 18, d = 36 2. r = 29, d = 58 3. r = 63, d = 126 4. r = 19.5 d = 39 5. r = 50.5, d = 101 6. r = 18.62, d = 37.24 7. r = 32.89, d = 65.78 8. r = 68.75, d = 137 9. r = 44.965, d = 89.93

18

10. r = 226.38, d = 452.76 11. r = 2.268, d = 4.536 12. r = 87.34, d = 174.68 13. r = 47.5, d =95 14. r = 50, d =100 15. r = 12.02, d = 24.04 16. r = 12.5, d = 25

17. r = 23 , d = 3

18. 2937.5, d = 5875 19. 23 , d = 26 20. 5x – 2y, d = 10x – 4y

F.

1. 640.56 m 2. 3 meters 3. 40 2 cm

What have you learned 1. 161 cm 2. 100 dm 3. 10x + 4 4. 70.5 5. 30 cm 6. 3.5 m 7. diameter 8. 12.5 cm 9. 67 dm 10. 14.2 cm


What this module is about This module is about areas of plane figures. In this module you will study the areas of squares, rectangles, parallelograms, triangles, trapezoids and circles, and learn to compute them.


1. Apply the formulas for the measurements of the following plane figures

a. square, b. rectangle, c. parallelogram, d. triangle, e. trapezoid, and f. circle.

2. Solve problems involving areas of plane figures


1. Find the area of a square whose side is 9 cm. 2. Find the area of the triangle below.

8 cm 6 cm

2

3. Find the area of a rectangle whose length and width are 12 cm and 5 cm respectively.

4. Find the area of the trapezoid ABCD below

B 6 cm C 4 cm A D 10 cm

5. Find the area of a circle whose radius is 7 mm. 6. Find the base of a triangle if the altitude is 4 cm and the area is 16 cm2.

7. Find the area of a parallelogram with base 12 cm and height 8 cm.

8. The area of the parallelogram ABCD below is 96 cm2. Find x.

B C

X A D 12 cm

9. Find the area of the figure

9 cm 5 cm 3 cm 6 cm 1 cm

10. Beth’s garden is 4 meters wide and 6 meters long. Find the area of the garden.

3

What you will do

Lesson 1

Areas of Rectangles and Squares The first figure below is a rectangle and the second is a rectangular region.

Rectangle Rectangular region

A rectangular region is a union of a rectangle and its interior. When you are asked to find the area of a rectangle, you are actually asked to determine the area of a rectangular region. The area of a region is the number of square units contained in the region. A square unit is a square with a side 1 unit in length.

I unit l unit

Example 1 In the rectangle below, each small square is one unit in length. Find the area of the rectangle.

The area can be determined by counting the number of small squares. Since there are 24 small squares, therefore, the area is 24 square units.

The standard units of area are square units, such as square centimeters, square

decimeters and square meters.

4

Example 2

Find the area of the rectangle below.

12 cm 9 cm

Solution:

The length ( l ) of the rectangle is 12 cm and the width (w) is 9 cm. Substitute these data in the formula.

A = lw = (12 )( 9) = 108 cm2

The area is 108 square centimeters. Example 3 Find the area of a rectangle whose length and width are 14 cm and 12 cm respectively. Solution: Step 1. Draw and label the figure. 12 cm 14 cm

Step 2. Substitute the data in the formula. The figure shows that l = 14 cm and w = 12 cm.

A = lw = (14)(12) = 168 cm2

The area is 168 square centimeters. .

5

Example 4 Find the width of the rectangle with an area of 80 cm2 and length equal to 10 cm. Solution: Step 1. Draw and label the figure. 10 cm

A = 80 cm2 Step 2. A = 80 cm 2 , l = 10 cm. Substitute these data in the formula. A = lw 80 = 10 w 10w = 80 w = 8 cm The width is 8 cm Example 5 The area of rectangle EFGH below is 48 cm2. Find its width. 8 cm (x + 2) cm Solution: In the figure, l= 8 cm and w = (x + 2) cm. Substitute 8 in place of l and (x + 2) in place of w in the formula A = lw. Replace A by 48 since the area of the rectangle is given as 48 cm2. Then solve the resulting equation for x. A = lw. 48 = 8 ( x + 2 )

48 = 8x + 16 48 –16 = 8x 32 = 8x or 8x = 32 x = 4 Since the width is represented by x + 2, then the width is 4 + 2 or 6 cm

6

The Area of a Square When the length and width of a rectangle are equal, the figure is a square. The formula for finding the area of a square is A = s2 where s = length of a side. D s C s A B

Square AB = BC = CD = DA Example 1 Find the area of the square ABCD.

Solution: D C A = s2 = 62 s = 6 cm = (6)(6) = 36 cm2 A B The area is 36 square centimeters Example 2 Find the side of a square whose area is 25 cm2. Solution: Step 1. Draw and label the figure. Step 2. Substitute the data in the formula. A = 25 cm2 s = ? A = s2 25 = s2 s2 = 25 2s = 25 s = 5 cm The side is 5 centimeters

7

Example 3 Find the area of the shaded region. 20 cm 15 cm 8 cm 8 cm

Solution:

Step 1. Find the area of the rectangle A = lw = (20)(15) = 300 cm2

Step 2. Find the area of the square

A = s2 = 82 = (8)(8) = 64 cm2

Step 3. Find the area of the shaded region by subtracting the area of the square from

the area of the rectangle. A = lw - s2 A = 300 – 64 = 236 cm2

The area of the shaded region is 236 square centimeters.

Another important thing you will learn from this lesson is that areas of plane figures can be added as long as the figures do not overlap.

8

The area of the entire figure in the following illustration is the sum of the areas of the rectangle and the square. Example: Find the area of the figure below. B C 2 E F 6 4 A D 4 G Notice that polygon ABCD is a rectangle and polygon DEFG is a square. The area of the entire figure is the sum of the areas of the rectangle and the square.

Solution:

Step 1. Find the area of the rectangle ABCD A1 = lw = (10)(6) = 60

Step 2. Find the area of the square DEFG. A2 = s2 = 42 = (4)(4) = 16

Step 3. Find the area of the entire figure by adding the area of the rectangle to the area of the square.

A3 = A1 + A2 = 60 + 16 = 76 cm2

The area of the entire figure is 76 square centimeters

9

Try this out Set A In exercises 1-4, each plane figure is divided into small squares each with a side 1 unit in length. 1. Find the area. A = _______ square units.

2. Find the area. A = ________ square units

3. Find the area.

A = ________ square units

10

4. Find the area. A = _________ square units Find the area of each rectangle or square described below. 5. 5 cm 7. 19 cm 5 cm 16 cm 8 cm 5.5 m 6. 8. 16 cm 5.5 m 9. A rectangle with a length of 14 meters and a width of 10 meters. 10. A square with a side of 7 mm. Set B

1. Find the area of a square whose side is 15 cm. 2. Find the area of a rectangle whose length and width are 12 m and 7 m respectively. 3. Find the area of a square whose side is 8.5 m 4. Find the area of a rectangle with length of 8 cm and width of 5 cm. 5. A side of a square is 13 cm. Find its area. 6. The length and width of a rectangle are 18 cm and 10 cm respectively. Find its area.

11

6 6

Find the area of the following figure. 7. 2 cm 8. 3 cm

3 cm 3 cm 2 cm 2 cm 5 cm 4 cm Find the area of the shaded region in the following figures 12 9. 10 8 10 8 Set C Find the area of each square 1. side = 4 m 2. side = 4.5 cm Find the area of each rectangle 3. Length = 11 cm and width = 8 cm 4. length = 9 cm and width = 8 cm Find the length of a side of each square. 5. Area = 36 cm2 6. Area = 81 c m2 Find the length of each rectangle 7. Area = 112 c m2 and width = 8 cm 8. Area = 135 c m2 and width = 9 cm

6 4

12

Find the width of each rectangle 9. Area = 176 c m2 and length = 16 cm 10 Area = 216 c m2 and length = 18 cm

Lesson 2

Areas of Parallelograms and Triangles

The area of a parallelogram is equal to the product of the base times the height. The formula is A = bh Example 1 Find the area of a parallelogram with a base of 5 cm and a height of 3 cm.

Solution: Step 1. Draw and label the figure. The base AB has length of 5 cm and the height or altitude is 3 cm. D C 3 cm A 5 cm B Step 2. Substitute the data in the formula A = bh = 5(3) = 15 cm2

The area is 15 square centimeters Example 2 Find the area of the parallelogram below h = 5.5 15.2

13

The figure shows that b = 15.s cm and h = 5.5 cm. Substitute these data in the formula.

Solution: A = bh = (15.2)(5.5) = 83.6 cm2

The area is 83.6 square centimeters Example 3 The area of a parallelogram is 84 m2. The lengths of a base is 6 m. Find the length of the corresponding altitude.

Solution: Step 1. Draw and label the figure. A = 84 m2 h = ? 6 m Step 2. Substitute the data in the formula A = bh 84 = 6h 6h = 84 h = 84/6 h = 14 m

The height is 14 m Example 4 Find the height of a parallelogram that has a base of 14 cm and an area of 126 cm2

Solution: Step 1. Draw and label the figure. A = 126 cm2 h = ? 14 cm

14

Step 2. Substitute the data in the formula A = bh 126 = 14 h 14 h = 126 h = 126/14 h = 9 cm The height is 9 centimeters The Area of a Triangle The diagonal separates the parallelogram into two congruent triangles. D C D C D A B B

A B The diagonal BD divides the parallelogram ABCD into two congruent triangles. Two congruent triangles have equal areas. The area of ∆ABD is equal to the area of ∆CDB. Since the formula for finding the area of a parallelogram is A = bh, therefore, the formula for finding the area of a triangle is A =

21 bh.

Example 1 Find the area of the triangle shown below.

h = 4 m 9 m

Solution:

A = 21 bh

= 21 ( 9 ) ( 4 )

= 18 m2

The area is 18 square meters

15

Example 2 Find the area of the triangle below. 12 dm 20 dm

Solution

A = 21 bh

= 21 (20) (12)

= 120 dm2

The area is 120 square decimeters

Example 3 Find the area of the triangle shown below. 5.5 m 7.2 m

Solution:

A = 21 bh

= 21 ( 7.2) ( 5.5)

= 19.8 m2 The area is 19.8 square meters Example 4 Find the area the triangle with base equal to 12 cm and altitude equal to 10 cm.

16

Solution: Step 1. Draw and label the triangle

h = 10 cm 12 cm Step 2. Substitute the data in the formula

A = 21 bh

= 21 (12) (10)

= 60 cm The area is 60 square centimeters Example 5 Find the base of the triangle with an altitude of 12 cm and an area of 66 cm2. Solution Step 1. Draw and label the figure h = 12 cm b = ? Step 2. Substitute the data in the formula.

A = 21 bh

66 = 21 (b) (12)

66 = 6 b 6 b = 66 b = 11

The base is 11 centimeters.

17

Try this out Set A Find the area of each parallelogram. 1. 3. 4 mm 4 m 13 mm 6 m 2. 4.

6 cm 8 m 5.2 cm 6 m Find the area of each triangle 5. 8. 12 mm 2 15 mm 4 6. 9. 5 20 m 3 25 m 7. 10. 5 cm 6 m 6 cm 11 m

18

Set B. Complete the table below

Base Altitude Area of the Parallelogram

1 16 5

2 10 cm 180 cm2

3 14 km 168 km2

4 8 m 92 m2

5 4.5 cm 36 cm2

. Complete the table below

base Altitude Area of the Triangle 6 14 cm 84 cm2 7 16 m 80 m2 8 16 cm 64 cm2 9 7.6 cm 4.5 cm __cm2

10 5.6 m 23.8 m2 Set C Find the area of the parallelogram in which

1. b = 30, h=6

2. b = 321 , h = 6.5

3. b = 24. h = 12 Find the area of each triangle described below.

4. b = 40, h = 24 5. b = 26, h = 14 6. b = 28, h = 16

Solve the following

7. Find the area of a parallelogram with base 17 m and height 14 m. 8. Find the area of a triangle with base of 16 m and height of 12 m. 9. A piece of cardboard is in the form of a parallelogram. Find its area if the base is 17

cm and the altitude is 12 cm. 10. A piece of paper is in the form of a triangle. What is its area if the base is 18 cm and

its altitude is 8.6 cm?

19

Lesson 3

Area of a Trapezoid Recall that a trapezoid is a quadrilateral with two bases which are parallel. The area of a trapezoid is one half the product of the length of an altitude and the sum of the lengths of the two bases. D b2 C A b1 B

A = 21 h ( b1 + b2)

Where h = altitude b1 = lower base b2 = upper base Example 1 Find the area of trapezoid ABCD. D 6 C h =4 A B 8

A = 21 h ( b1 + b2)

= 21 ( 4) (8 + 6 )

= 21 ( 4 ) ( 14 )

= 28 The area is 28 square units.

20

Example 2 In the following trapezoid, HG = 6 cm and EF = 12 cm. If its area is 36 cm2, find its altitude h. H 6 cm G h = ? E F 12 cm

Solution:

A = 21 h ( b1 + b2)

36 = 21 h ( 6 + 12)

36 = 21 h ( 18)

36(2) = 18 h 18h = 72 h = 4 cm

The altitude h is 4 cm. Example 3 Find the longer base of a trapezoid with shorter base 5, height 4 and area 24.. Step 1 Draw and label the figure. Represent the longer leg by x.

5 A = 24 4 x Step 2. Substitute the data in the formula.

A = 21 h( b1 + b2)

24 = 21 (4)( x + 5)

24 = 2( x + 5) 2(x + 5 ) = 24 2x + 10 = 24 2x = 24 – 10 2x = 14 x = 7

The longer base is 7 .

21

Try this out Set A ABCD is a trapezoid. D C A B

1. If AB = 16, DC = 8, DE = 6, find the area 2. If AB = 17, DC = 9 , DE = 8, find the area 3. If AB = 20, DC = 18, DE = 10, find the area 4. If AB = 30, DC = 20, DE = 10, find the area 5. If AB = 40, DC = 30, DE = 20, find the area EFGH is a trapezoid. H G

E F

6. If EF = 8, HG = 10, HE = 6, find the area. 7. If EF = 6, HG = 8, HE = 4, find the area 8. If EF = 12, HG =16, HE = 6, find the area 9. If EF = 10, HG =12, HE = 9, find the area 10. If EF = 12, HG =14, HE = 7, find the area

Set B Supply the missing information for each trapezoid

H b1 b2 Area 1 4 8 11 2 8 10 13 3 12 10 14 4 12 16 140 5 10 18 112

22

Find the area of each trapezoid 8 6. 7. 7 cm 5 5 cm 11 13 cm

8. Given a trapezoid with bases 16 and 20 and height 9, Find the area. 9. The bases of a trapezoid are 8 and 20 and the area is 84. Find the height.

8

h = ? A = 84 20

10. The height of a trapezoid is 12 and the area is 162. If one of the bases is 16, find the other.

Set C

1. The height of a trapezoid is 4. The bases are 5 and 7. Find the area. 2. The height of a trapezoid is 6. The bases are 7 and 9. What is the area? 3. The bases of a trapezoid are 4 and 8 and the area is 36. Find the height. 4. The bases of a trapezoid are 6 and 8 and the area is 56. Find the altitude. 5. The altitude of a trapezoid is 8 and the area is 64. If one base is 5, find the other

base. 6. The height of a trapezoid is 2 and the area is 16. If the upper base is 9, find the lower

base. 7. A trapezoid has bases 9 and 10 and the area 38. Find its altitude. 8. A trapezoid has bases 8 and 12 and area 80. Find its height. 9. Find the longer base of a trapezoid with shorter base 5, height 6, and area 48. 10. Find the shorter base of a trapezoid with longer base 12 , altitude 6, and area 54.

Lesson 4

Area of a Circle Recall that a circle is a set of points in a plane that have the same distance from a given point in the plane. The given point is called the center, and the distance from the center to any given point on the circle is called the radius.

23

d = 6 cm

The area of a circle is the measure of the space bounded by it. The ratio of the circumference to the diameter of any circle is equal to the same number, represented by the Greek letter π (pi) The approximate value of π is 3.14 or 22/7. The formula for the area of a circle with radius r units is A = πr2 Where A = area r = radius

Example 1

The radius of a circle is 2 cm. Find its area.

Solution: A = πr2 2 cm

≈ 3.14 (2)2 ● ≈ 12.56 cm2

The area is 12.56 square cm.

Example 2 The diameter of a circle is 6 cm. Find its area.

Solution: Step 1. Find the radius Radius( r ) = diameter (d) divided by 2 r = 6 ÷ 2 r = 3 cm The radius is 3 cm. ●

Step 2. Find the area. A = πr2 ≈ 3.14 (3)2 ≈ 3.14 (9)

≈ 28.26 cm2

The area is 28.26 square centimeters

24

Example 3 Find the area of the circle in terms of π if its radius is 2 m.

Solution

A = πr2 2 m ≈ π(2)2 ● ≈ 4π

Example 4 Find the area of the shaded region. 22 cm 8 cm

Solution: Step 1. Find the area of the rectangle. A = lw = (22) ( 8) = 176 cm2

Step 2. Find the area of the circle

The diameter of the circle is the width of the rectangle.

r = 2d

r = 28

r = 4 cm2 Substitute 4 in the area formula for circles A = πr2 ≈ 3.14 ( 4)2 ≈ 3.14 ( 16) ≈ 50.24 Step 3. Subtract the area of the circle from the area of the rectangle. Area of the shaded region = area of the rectangle – area of the circle = 176 –50.24 = 125.76 cm2 The area of the shaded region is 125.76 square centimeters.

25

Try this out Set A.

Find the area of each circle with the given diameter or radius Use 3.14 for π. 1. radius = 5 cm 2. radius = 1.5 mm 3. diameter = 4 cm 4. diameter = 12 dm 5. radius = 4.6 m 6. radius = 2.2 cm 7. diameter = 4.8 dm 8. diameter = 6.4 cm 9. radius = 4.8 m 10. radius = 3.4 dm

Set B Find the area of each circle described below. Give the answers in terms of π.

1. 3.

d = 10 ● ● r = 8 2. 4. ● ● d = 11 r = 9

5. The radius is 7 cm 6. The radius is 12 cm 7. The diameter is 14 mm 8. The diameter is 26 mm 9. The radius is 3.4 km 10. The radius is 2.6 km

26

Set C.

Find the area of each circle in terms of π 1. r = 8 dm 2. r = 18 dm 3. d = 5 cm 4. d = 14 cm 5. r = 4.6 mm 6. r = 4.4 mm

7. The students performed their dance number in a circular platform 20 m in diameter.

Find the area of the platform.

8. The radius of a circular garden is 40 m. Half of the garden will be planted with roses. How many square meters will be planted with roses.

Find the area of the shaded part of the figure. Give the answers in terms of π 9

r of the bigger circle is 8 cm ● r of the smaller circle is 6 cm. 10. ● 14 cm 12 cm

27

Let’s Summarize

1. The area of a region is the number of square units contained in the region. 2. A square unit is a square with a side I unit in length. 3. The area (A) of a rectangle is the product of its length (l) and its width (w).

A = lw 4. The area (A) of a square is the square of the length of a side (s). A = s2 5. The area (A) of a parallelogram is equal to the product of the base (b) and the height

(h). A = bh 6. The diagonal separates the parallelogram into two congruent triangles. 7. The area (A) of a triangle equals half the product of the base (b) and the height (h).

A = 21 bh. Sometimes altitude is used instead of height.

8. The area (A) of a trapezoid is one half the product of the length of its altitude and the

sum of the lengths of the two bases. A = 21 h (b1 + b1).

9. A circle is a set of points in a plane that have the same distance from a given point in the plane.

10. In all circles the ratio of the circumference to the diameter is always equal to the same number, represented by the Greek letter π.

11. The formula for finding the area of a circle with a radius of r units is: A = πr2


1. Find the area of a square whose side is 15 dm. 2. Find the area of the triangle below.

9 cm

10 cm

3. Find the area of a rectangle whose length and width are 14 dm and 8 dm respectively.

28

4. Find the area of the trapezoid ABCD below

B 8 cm C

6 cm A 12 cm D

5. Find the area of a circle whose radius is 6 cm. 6. Find the base of a triangle if the altitude is 6 cm and the area is 36 cm2. 7. Find the area of a parallelogram with base 14 cm and height 6 cm. 8. The area of the parallelogram ABCD below is 66 cm2. Find h.

B C h = ? A 11 cm D

9. Find the area of the figure

7 5 3 2 3

10. Gen’s garden is 5 meters wide and 7 meters long. Find the area of the garden.

29


1. 81 cm2 2. 24 cm2 3. 60 cm2 4. 32 cm 2

5. 153.86 mm2

6. 8 cm 7. 96 cm2 8. 8 cm 9. 44 cm2 10. 24 m

Try this out Lesson 1 Set A.

1. 10 2. 18 3. 14 4. 16

5. 25 cm2

6. 128 cm2 7. 304 cm2 8. 30.25 m2 9. 140 10. 49 mm2

Set B.

1. 225 cm2 2. 84 m2 3. 72.25 m2 4. 40 cm2

5. 169 cm2

6. 180 cm2 7. 10 cm2 8. 13 cm2 9. 28 sq. units 10. 96 sq. units

Set C

1. 16 m2 2. 20.25 cm2 3. 88 cm2 4. 72 cm2 5. 6 cm

6. 9 cm 7. 14 cm 8. 15 cm 9. 11 cm 10. 12 cm

Lesson 2 Set A

1. 24 m2 2. 31.2 cm2 3. 52 mm2 4. 48 m2 5. 4 sq. units

6. 7.5 sq. units 7. 15 cm2 8. 90 mm2 9. 250 m2 10. 33 m2

30

Set B

1. 80 sq. units 2. 18 cm 3. 12 km 4. 11.5 m 5. 8 cm

6. 12 cm 7. 10 m 8. 8 cm 9. 17.1 cm 10. 8.5 m

Set C

1. 180 sq. units 2. 22.75 sq. units 3. 288 sq. units 4. 480 sq. units 5. 182 sq. units

6. 224 sq. units 7. 238 m2 8. 96 m2 9. 204 cm2 10. 77.4 cm2

Lesson 3 Set A

1. 72 sq. units 2. 104 sq. units 3. 190 sq. units 4. 250 sq. units 5. 700 sq. units

6. 54 sq. units 7. 28 sq. units 8. 84 sq. units 9. 99 sq. units 10. 91 sq. Units

Set B

1. 38 sq. units 2. 92 sq. units 3. 144 sq. units 4. 10 5. 8

6. 47.5 sq. units 7. 50 cm2 8. 162 sq. units 9. 6 10. 11

Set C.

1. 24 sq. units 2. 48 sq. units 3. 6 4. 8 5. 11

6. 7 7. 4 8. 8 9. 11 10. 6

Lesson 4 Set A

1. 78.5 cm2 2. 7.065 mm2 3. 12.56 cm2 4. 113.04 dm2 5. 66.4424 m2

6. 15.1976cm2 7. 18.0864 dm2 8. 32.1536 cm2 9. 72.3456 m2

10. 36.2984 dm2

31

Set B

1. 64 π 2. 81 π 3. 25 π 4. 30.25 π

5. 49 π cm2

6. 144 π cm2 7. 49 π cm2 8. 169 π mm2 9. 11.56 π km2 10. 6.76 π km2

Set C

1. 64 π dm2 2. 324 π dm2 3. 6.25 π cm2 4. 49 π cm2 5. 21.16 π mm2

6. 19. 36 π mm2 7. 100 π m2 8. 800 π m2 9. 64 π cm2– 36 π cm2 = 28 π cm2 10. 196 cm2 – 49 π cm2

What have you learned?

1. 225 dm2 2. 45 cm2 3. 112 dm2 4. 60 cm2 5. 36 π cm2

6. 12 cm 7. 84 cm2 8. 6 cm 9. 29 sq. units 10. 35 m2


What this module is about This module is about surface area of solids. As you go over the exercises, you will develop skills in solving surface area of different solid figures. Treat the lesson with fun and take time to go back if you think you are at a loss. What you are expected to learn This module is designed for you to

1. define surface area of solids. 2. find the surface area of solids such as

• cube • prism ( rectangular, triangular ) • pyramid ( square, rectangular, triangular ) • cylinder • cone • sphere

3. solve problems involving surface area of solids.

How much do you know Find the surface area of each solid. 1. A cube with side (s) = 2.2 cm.

2. A cylinder with h = 15 cm, r = 3.2 cm. 3. A rectangular prism with l = 12 cm, w = 7 cm, h = 6 cm. 4. A square pyramid with s = 4.2 cm, h = 7 cm (slant height).

2

5. A cone with r = 5 cm , s = 12 cm (slant height). 6. A triangular prism with height 14 cm, base (a right triangle with sides 6, 8

and 10 cm and the right angle between shorter sides). 7. A ball with radius of 6 cm (Use π = 3.14). 8. A triangular pyramid with b = 4 cm, h = 8.2 cm (altitude of the base), s = 7

cm (slant height). 9. A rectangular pyramid with l = 5 cm, w = 3.3 cm, h = 9 cm (slant height). 10. A cylindrical tank is 2.6 meters high. If the radius of its base is 0.92

meters, what is its surface area? 11. find the surface area of a rectangular prism which is 45 cm long, 36 cm

wide and 24 cm high. 12. Find the surface area of a pyramid with a square base if the length of the

sides of the base is 1.4 m and the height of the triangular face is 1.9 m. 13. Cube with edge 4 2/3 cm

14. Cylinder with radius of base 6.7 cm and height 14 cm

15. Rectangular prism with base 9m by 10m by 12m

What you will do

Lesson 1

Finding the Surface Area of a Cube, Prism and Pyramid

The surface area of a three-dimensional figure is the total area of its exterior surface. For three-dimensional figures having bases, the surface area is the lateral surface area plus the area of the bases.

3

Examples :

1. Find the surface area of a cube with side of 5 cm.

Figure : Cube If you open this up, This solid is composed of 6 squares or 6 faces. Since the area of a square

is the square of its side or s2, then the surface area of a cube is 6 s2. SA of a Cube = 6s2 Substituting s by 5 cm: SA = 6(52)

= 6(25)

SA = 150 cm2

edgelateral face

face

face

face

facefaceface

4

2. Find the surface area of a rectangular prism whose length is 7 cm, width is 5 cm and thickness is 4 cm. Figure : To find the surface area of a Rectangular prism, add the areas of its flat surfaces.

Area of top and bottom rectangles (bases) plus area of left and right rectangles and area of back and front rectangles (lateral areas).

SA of rectangular prism = 2B + LA

Solution:

SA = 2(7cm x 5 cm) + 2(4 cm x 5 cm) + 2(7cm x 4 cm) = 70 cm2 + 40 cm2 + 56 cm2 SA = 166 cm2 3. Find the surface area of a triangular prism.

Figure :

Triangular Prism

7 cm

5 cm

4 cm base

base

Rectangular Prism

5 cm 5 cm

4.5 cm

2.8 cm

3.9 cm

5

3 cm

3 cm

5 cm

SA of a triangular prism = 2B + LA

Solution:

SA = 2 (21 bh) + LA

SA = bh + LA

= (4.5cm x 3.9cm) + 2( 5cm x 2.8 cm) + (4.5 cm x 2.8cm)

= 17.55 cm2 + 28 cm2 + 12.6 cm2

SA = 58.15 cm2 4. Find the surface area of a square pyramid with a side of the base

as 3 cm and the height of a triangle as 5 cm. Figure:

To find the surface area of a square pyramid, add the area of the square base and the areas of the four face triangles.

Solution:

SA of Square Pyramid = B + 4 (21 bh)

SA = 32 + 4(21 x 3 x 5)

= 9 + 30 SA = 39 cm2

6

5 cm

8 cm

6 cm

5. Find the surface area of the rectangular pyramid with the given dimensions. Figure : Rectangular Pyramid SA of Rectangular Pyramid = B + 2 A1 + 2 A2 Solution:

SA = bh + 2 (21 b1h1) + 2 (

21 b2h2)

= bh + (b1h1) + ( b2h2)

= (8 cm x 5 cm)+ (8 cm x 6 cm) + (5 cm x 6 cm)

= 40 cm2 + 48 cm2 + 30 cm2

SA = 118 cm2 6. Find the surface area of a triangular pyramid with the given dimensions. Figure : Triangular Pyramid

8 cm (slant height)

20 cm

5 cm 20 cm 20 cm

7

1.6 cm

1.6 cm

1.6 cm

52 mm

180 mm

204 mm

6.6 cm

6.6 cm

9.4 cm

SA of Triangular Pyramid = Area of the base + Area of the 3 triangular faces

Solution:

SA = (21 x 20 cm x 5 cm) + (10 cm x 8 cm) + (

21 x 20 cm x 8 cm)

= 50 cm2 + 80 cm2 + 80 cm2

SA = 210 cm2 Try this out Find the surface area of each solid. 1. 2. 3.

8

6 cm

9 cm

7 cm

4. 5. 6.

4 cm 3 cm

5 cm

10 cm

10 cm

5 cm 5 cm

3 cm (altitude of the base)

7 cm (slant height)

9

7. What is the total surface area of a cardboard box that is 1.2 m long, 0.6 m

wide, and 0.3 m high? Figure: 8. Find the surface area of a cube whose side measures 10 cm. Figure: 9. Find the surface area of a triangular chocolate box . Figure:

0.3 cm

1.2 cm

0.6 cm

10 cm

10 cm

10 cm

8 cm

6 cm 6 cm

6 2 cm

10

10. Find the surface area of a camping tent in a square pyramid shape with a

side of the base as 5 cm and the height of a triangle as 7 cm. Figure : 11. A pyramid has a rectangular base whose length and width are 5.5 cm and 3.2

cm respectively. Find its surface area. Figure: 12. Find the surface area of a tetra pack juice drink in triangular pyramid shape

with the given dimensions. Figure:

5.5 cm

3.2 cm

10 cm (slant height)

5 cm

5 cm

7 cm (slant height)

10 cm (slant height)

6 cm

4.5 cm (altitude of the base)

6 cm 6 cm

11

Lesson 2

Finding the Surface Area of Cylinder, Cone and Sphere

Surface Area of a Cylinder Figure: To find the surface are of a cylinder, add the areas of the circular bases and the area of the rectangular region which is the body of the cylinder. SA = Area of 2 Circular Bases + Area of a rectangle

SA = 2πr2 + 2πrh Example:

Find the surface area of a cylinder which has a radius of 5 cm and a body length of 20 cm. ( Use 3.14 for π)

Solution: SA = 2πr2 + 2πrh

= 2(3.14)(5) 2 + 2(3.14)(5)(20)

= (6.28)(25) + (6.28)(100)

= 157 + 628

SA = 785 cm2

base radius (r)

altitude (h) h

base base

base circumference of the base = 2 rπ r

Area of the 2 bases = 2 rπ r2

12

h = s

b = rπ r 2 rπ r

Surface Area of a Cone Figure:

SA = Area of the circular base + Area of the region which resembles a parallelogram SA = πr2 + πrs Example: Find the surface area of a cone if the radius of its base is 3.5 cm and its slant height is 7.25 cm. (Use π = 3.14) Solution :

SA = πr2 + πrs

= (3.14)(3.5)2 + (3.14)(3.5)(7.25)

= (3.14)(12.25) + (10.99)(7.25)

= 38.465 + 79.6775

SA = 118.1425 cm2

Consider an ice-cream cone with its curved surface opened out to resemble a fan.

The fan-shaped surface can be cut into smaller pieces and rearranged to resemble a parallelogram with base b = πr and height h equal to side s.

radius

Slant height

s

r = 3.5 cm

s = 7.25 cm

13

Surface Area of a Sphere Figure:

SA = area of 4 circles

SA = 4πr2 Example: What is the surface area of a ball with radius equal to 7 cm? (Use π = 3.14)

Solution :

SA = 4πr2

= 4(3.14)(7)2

= (12.56)(49)

SA = 615.44 cm2 Try this out Find the surface area of each solid. 1.

A sphere is a solid where every point is equally distant from its center. This distance is the length of the radius of

radius

r = 7

1.8 m

height = 2.4 m

14

2. 3. 4.

2.6 m

s = 12.3 m

r =7.8 cm

8.2 cm

7.6 cm

15

5. 6. 7. A cylindrical water tank is 2.2 meters high. If the radius of its base is 0.8

meter, what is its surface area? Figure:

r = 8 mm

s = 20 mm

r = 6 cm

r = 0.8 m

h = 2.2 m

16

8. The radius of a ball is 19 cm. What is its surface area? Figure: 9. Find the surface area of a conic solid whose radius is 2.5cm and its height is

3.3 cm. Figure: 10. Find the surface area of a spherical tank whose radius is 0.7meter. Figure:

r = 19 cm

r = 2.5 cm

s = 3.3 cm

r = 0.7 m

17

11. A cone with a diameter of 10 cm and height of 8 cm. Find its surface area. Figure :

12. A can whose height is 20 cm and 12 cm is the diameter. Find its surface

area. Figure:

r = 12 cm

h = 20 cm

r = 10 cm

s = 8 cm

18

Let’s summarize

The surface area of a three-dimensional figure is the total area of its exterior surface. For three-dimensional figures having bases, the surface area is the lateral surface area plus the area of the bases.

• Surface Area of a Cube = 6s2 • Surface Area of Rectangular Prism = 2B + LA

• Surface Area of Rectangular Prism = 2B + LA

• Surface Area of Square Pyramid = B + 4 (21 bh)

• Surface Area of Rectangular Pyramid = B + 2 A1 + 2 A2

• Surface Area of Triangular Pyramid = Area of the base + Area of

the 3 triangular faces

• Surface Area of a Cylinder = Area of 2 Circular Bases + Area of a rectangle

SA = 2πr2 + 2πrh

• Surface Area of a Cone = Area of the circular base + Area of the region which resembles a parallelogram

SA = πr2 + πrs

• Surface Area of a Sphere = area of 4 circles

SA = 4πr2

19


Find the surface area of each solid.

1. A cube with side (s) = 4.3 cm 2. A cylinder with h = 13 cm, r = 4.1 cm 3. A rectangular prism with l = 15 cm, w = 8 cm, h = 7 cm 4. A square pyramid with s = 5.4 cm, h = 9 cm (slant height). 5. A cone with r = 4cm , s = 9 cm (slant height).

6. A triangular prism with height 16 cm, base (a right triangle with sides 7 cm,

8 cm and 10 cm and the right angle between shorter sides). 7. A ball with radius of 12 cm. (Use π = 3.14) 8. A triangular pyramid with b = 5 cm, h = 7.2 cm (altitude of the base), s = 8

cm (slant height) 9. A rectangular pyramid with l = 6 cm, w = 6.3 cm, h = 8 cm (slant height) 10. A cylindrical tank is 3.6 meters high. If the radius of its base is 1.9 meters,

what is its surface area? 11. Find the surface area of a rectangular prism which is 40 cm long, 35 cm

wide and 23 cm high. 12. Find the surface area of a pyramid with a square base if the length of the

sides of the base is 1.6 m and the height of the triangular face is 2.8 m.

13. Cube with edge 5 21 cm.

14. Cylinder with radius of base 8.7 cm and height 13 cm.

15. Rectangular prism with base 10m by 12m by 14m.

20


1. 29.04 cm2 2. 365.74742 cm2 3. 396 cm2 4. 135.24 cm2 5. 266.9 cm2 6. 384 cm2 7. 452.16 cm2 8. 100.4 cm2 9. 91.2 cm2 10. 20.337152 cm2 11. 7, 128 cm2 12. 12.6 m2

13. 130 32 cm2

14. 870.9732 cm2 15. 636 m2


1. 15.36 cm2 2. 113, 376 mm2 3. 167.64 cm2

21

4. 159 cm2 5. 132 cm2 6. 85 cm2 7. 2.52 m2 8. 600 cm2 9. 181.88 cm2 10. 95 cm2 11. 104.6 cm2 12. 78.3 cm2

Lesson 2

1. 18.6516 m2 2. 21.2264 m2 3. 492.2892 cm2 4. 301.3576 cm2 5. 703.36 mm2 6. 452.16 cm2 7. 15.072 m2 8. 4,534.16 cm2 9. 45.53 cm2 10. 6.1544 m2 11. 204.1 cm2 12. 979.68 cm2

22


1. 110.94 cm2 2. 131.3148 cm2 3. 562 cm2 4. 101.16 cm2 5. 163.24 cm2 6. 228 cm2 7. 1,808.64 cm2 8. 138 cm2 9. 136.2 cm2 10. 65.626 m2 11. 6,250 cm2 12. 20.48 m2 13. 181.5 cm2 14. 1,185.6012 cm2 15. 856 m2


What this module is about This module is about volume of solids. The volume of a solid is the number of cubic units contained in the solid. If measures are given in centimeter, the volume is stated in cubic cm, written as cm3.


1. define volume of solids. 2. find the volume of solids such as:

• cube

• prism (rectangular, triangular)

• pyramid (square, rectangular, triangular)

• cylinder

• cone

• sphere

3. solve problems involving volume of solids.

How much do you know Find the volume of each solid: 1. a cube with side (s) = 2.4 m

2. a cylinder with h = 20 cm, r = 22 cm.

3. a rectangular prism with l = 25 cm, w = 17 cm, h = 30 cm

2

4. a square pyramid with s = 5 m, h = 6 m.

5. a cone with r = 2 cm, h = 6 cm.

6. a triangular prism with height 10 cm, base (a right triangle with sides 3, 4 and 5 cm and the right angle between shorter sides).

7. a ball with radius of 17 cm (use = 3.14)

8. a triangular pyramid with b = 4 cm, h = 8.2 cm (altitude of the base), h = 7 cm (height of the pyramid).

9. a rectangular pyramid with l = 6 cm, w = 4.3 cm, h = 8 cm (height of the pyramid)

10. a cylindrical tank is 5.3 meters high. If the radius of its base is 2.8 meters, what is its volume?

11. Find the volume of a rectangular prism which is 46 cm long, 37 cm wide and 25 cm high.

12. Find the volume of a pyramid with a square base if the length of the sides of the base is 2.4 m and the height of the triangular face is 3.5 m.

13. cube with edge of 6 32 cm.

14. cylinder with radius of base 8.7 cm and height 12 cm.

15. rectangular prism with base 8 m by 10 m by 15 m.

What you will do

Lesson 1

Finding the Volume of a Cube, Prism and Pyramid

One problem with rooms that have high ceilings is that they are hard to heat and cool. The amount of air in a room determines the heating or cooling power needed. To find the amount of air in a room, you need to find the volume of the room. In finding volume of solids, you have to consider the area of a face and height of the solid. If the base is triangular, you have to make use of the area of a triangle, if rectangular, make use of the area of a rectangle and so on.

3

The next examples will help you to understand more about volume or the amount of space in three – dimensional figures. Volume of a cube

Example:

Find the volume of a cube with edge ( e ) of 3 cm.

Figure: Solution:

V = e3 Substituting e by 3 cm:

V = 33 V = 27 cm3

The volume V of a cube with edge e is the

cube of e. That is, V = e3.

3 cm

3 cm

3 cm

Cube

4

Volume of Prism Example:

Find the volume of a rectangular prism whose length is 7.5 cm, width is 4.3 cm and thickness is 5.1 cm.

Figure: Solution:

V = lwh V = (7.5 cm)(4.3 cm)(5.1 cm) V = 164.475 cm3

The volume of a prism can also be expressed in terms of area of the base, B.

The volume V of a rectangular prism is the product

of its altitude h, the length l and the width w of the base. That is,

V = lwh.

The volume V of a prism is the product of its

altitude h and area B of the base. That is,

V = Bh.

7.5 cm

5.1 cm

4.3 cm

5

Example:

Find the volume of a triangular prism whose dimensions is given in the figure below.

Solution:

Let B = area of the triangular base

B = 21 bh

= 21 (4.5 cm)(3.9 cm)

B = 8.775 cm2 Finding the volume of the prism: V = Bh = 8.775 cm2 (2.8 cm) = 24.57 cm3

3.9 cm

4.5 cm

2.8 cm

6

Volume of Pyramids Consider a pyramid and a prism having equal altitudes and bases with equal areas. If the pyramid is filled with water or sand and its contents poured into a prism, only a third of the prism will be filled. Thus the volume of a pyramid

is 31 the volume of the prism.

Example:

1. Find the volume of the rectangular pyramid with the given dimensions. Figure: Solution:

Let B = the area of the rectangular base

B = lw

= (9 cm)(4 cm)

= 36 cm2 Finding the volume V:

V = 31 Bh

= 31 (36 cm2) (6 cm)

= 72 cm3

The volume V of a pyramid is one third the product

of its altitude h and the area B of its base. That is,

V = 31 Bh.

w = 4 cm

l = 9 cm

h = 6 cm

base

7

2. Find the volume of a square pyramid with a side of the base as 4 cm and

the height of a pyramid as 6 cm.

Figure: Solution: Let B = area of the square base

= s2

= (4 cm)2

B = 16 cm2 Finding the volume of the pyramid:

V = 31 Bh

= 31 (16 cm2 )(6 cm)

= 32 cm3

s = 4 cm

h = 6 cm

8

3. Find the volume of a triangular pyramid with the given dimensions.

Figure: Solution:

Let B = area of the triangular base

B = 21 bh

= 21 (20 cm)(5 cm)

B = 50 cm2

Finding the volume of the pyramid: V = Bh

= (50 cm2)(8cm)

V = 400 cm3

h = 8 cm (height of the pyramid)

b = 20 cm h = 5 cm (height of the base)

9

Try this out Find the volume of each solid: 1. 2. 3.

6 cm

6 cm

6 cm

12 mm

6 mm

9 mm

1.5 cm

15 cm

h = 3 cm

10

Find the volume of each pyramid: 4. 5. 6.

h = 15 cm

w = 10 cm

l = 25 cm

h = 7 m

3.5 m

3.5 m

h = 10.4 cm (pyramid)

h = 4.3 cm (base)

b = 16.2 cm

11

7. What is the volume of a cardboard box that is 9 m long, 6 m wide, and 3 m

high? 8. Find the volume of a cube with side of 8 cm. 9. Find the volume of a triangular chocolate box.

9 m

3 m

6 m

8 cm

8 cm

8 cm

4 cm3 cm

h = 7 cm

12

10. Find the volume of a camping tent in a square pyramid shape with a side of the base as 5 cm and the height of a triangle as 7 cm.

11. A pyramid has a rectangular base whose length and width are 15.5 cm

and 3.3 cm respectively. The height of the pyramid is 4 cm. Find its volume.

12. Find the volume of a tetra pack juice drink in triangular pyramid shape

with the given dimensions.

h = 8.5 cm

6.3 cm

6.3 cm

h = 4 cm

w = 3.3 cm

l = 15.5 cm

h = 10 cm (pyramid)

h = 5 cm (base)

13

Lesson 2

Finding the Volume of a Cylinder, Cone and Sphere

A cylinder has 2 congruent circular bases. The volume of a cylinder is just like finding the volume of a prism. Figure: Volume of a Cylinder

Example: Find the volume of a cylinder which has a radius of 8 cm and a height of 15 cm. (Use 3.14 for π )

The volume V of a circular cylinder is the product

of the altitude h and the area B of the base. That is,

V = Bh or V = π r2h.

r

h

r = 8 cm

h = 15 cm

14

Solution: Let B = area of the circular base

= π r2

= (3.14)(8 cm)

B = 200.96 cm2 Finding the volume of the cylinder: V = Bh

= (200.96 cm2)(15 cm)

= 3014.4 cm3 Volume of a Cone If a cone is filled with water or sand, and then its content is poured into the cylinder (the cone and cylinder have equal areas) only a third of the cylinder will

be filled. This shows that the volume of a cone is 31 that of the cylinder.

The volume V of a circular cone is one third the

product of the altitude h and the area B of the base. That is,

V = 31 Bh or V =

31 π r2h

r

h

15

Example: Find the volume of a cone if the radius of its base is 4.5 cm and its height is 8.75 cm (Use π = 3.14) Solution:

V = 31 π r2h

= 31 (3.14)(4.5 cm)2(8.75 cm)

V = 185.46 cm3 Volume of a Sphere

Fill a cylinder with water. Push the sphere into the cylinder and determine

the amount of water displaced. About 32 of the water will be displaced, so the

volume of the sphere is 32 that of the cylinder.

Figure: In the figure, the height of the cylinder is equal to the diameter of the sphere, the volume of the cylinder will now be equal to 2π r3. Since the volume

of the sphere is 32 that of the cylinder and the height of the cylinder = 2r, then

V = 32 (2π r3) =

34 π r3.

The volume V of a sphere = 34 π r3

r = 4.5 cm

h = 8.75 cm

radius

radius

16

Example: What is the volume of a ball with radius equal to7.8 cm? Solution:

V = 34 π r3

= 34 (3.14)(7.8 cm)3

= 1,986.79 cm3 Try this out Find the volume of each solid. 1. 2.

r = 7.8 cm

h = 8.2 cm

r = 4.3 cm

r = 5.4 cm

17

3. 4. 5.

r = 4.4 cm

h = 7.6 cm

r = 1.7 cm

r = 2.7 cm

h = 6.3 cm

18

6. 7. A cylindrical water tank is 6.2 meters high. If the radius of its base is 1.8 m, what is its volume. 8. The radius of a ball is 5.2 cm. What is its volume?

r = 1.3 cm

h = 3 cm

h = 8.5 cm

h = 2.6 cm

r = 8.3 cm

19

9. Find the volume of a conic solid whose radius is 6.3 cm and its height is 13.5

cm. 10. Find the volume of a spherical tank whose radius is 1.7 meters.

r = 6.3 cm

h = 13.5 cm

r = 1.7 m

20

11. A cone with a diameter of 12 cm and height of 6 cm. Find its volume. 12. A can of milk has a diameter of 12 cm and a height of 17.3 cm. Find its

volume.

diameter = 12 cm

h = 6 cm

h = 17.3 cm

diameter = 12 cm

21

Let’s summarize The volume of a three dimensional figure is the amount of space it encloses.

The volume V of a cube with edge e is the cube of e. That is,

V = e3. The volume V of a rectangular prism is the product of its altitude h, the

length l and the width w of the base. That is,

V = lwh. The volume of a prism can be expressed in terms of area of the base, B. The volume V of a prism is the product of its altitude h and area B of the

base. That is,

V = Bh. The volume V of a pyramid is one third the product of its altitude h and the

area B of its base. That is,

V = 31 Bh.

The volume V of a circular cylinder is the product of the altitude h and the

area B of the base. That is,

V = Bh or V = π r2h. The volume V of a circular cone is one third the product of the altitude h

and the area B of the base. That is,

V = 31 Bh or V =

31 π r2h

The volume V of a sphere = 34 π r3

22

What have you learned Find the volume of each solid: 1. A cube with edge ( e) = 6.3 cm. 2. A cylinder with h = 15 cm, r = 7.1 cm.

3. A rectangular prism with l = 18 cm, w = 7 cm, h = 5 cm.

4. A square pyramid with s = 8.5 cm, h = 6 cm.

5. A cone with r = 3.8 cm, h = 7.2 cm.

6. A triangular prism with height 16 cm, base ( a right triangle with sides

3, 4 and 5 cm and the right angle between shorter sides).

7. A ball with radius of 13 cm. 8. A triangular pyramid with b = 5 cm, h = 7.2 cm (altitude of the base), h = 8 cm (height of the pyramid). 9. A rectangular pyramid with l = 9 cm, w = 6.3 cm, h = 8 cm (height of

the pyramid). 10. A cylindrical tank is 5.4 m high. If the radius of its base is 4.9 m, what

is its volume? 11. Find the volume of a rectangular prism which is 42 cm long, 38 cm

wide and 22 cm high. 12. Find the volume of a pyramid with a square base if the length of the

sides of the base is 3.6 m and a height of 1.8 m. 13. Cube with edge 10.5 cm. 14. Cylinder with radius of base 9.7 cm and height of 12 cm.

15. Rectangular prism with base 12 m by 14.6 m and height of 9.1 m.

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Answer key How much do you know

1. 13.82 m3

2. 30,395.20 cm3

3. 12,750 cm3

4. 50 m3

5. 25.12 cm3

6. 60 cm3

7. 20,569 cm3

8. 38.27 cm3

9. 68.8 cm3

10. 130.47 m3

11. 42,550 cm3

12. 6.72 m3

13. 295.408 cm3

14. 2,851.99 cm3

15. 1,200 m3


1. 216 cm3

2. 648 mm3

3. 33.75 cm3

4. 1,250 cm3

5. 28.58 m3

6. 120.74 cm3

7. 162 m3

8. 512 cm3

9. 42 cm3

10. 112.46 cm3

11. 68.2 cm3

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12. 125 cm3 Lesson 2

1. 158.69 cm3

2. 659.25 cm3

3. 462.01 cm3

4. 20.57 cm3

5. 48.07 cm3

6. 15.92 m3

7. 180.42 cm3

8. 2,393.88 cm3

9. 560.82 cm3

10. 20.57 m3

11. 226.08 cm3

12. 1,955.59 cm3 What have you learned

1. 250.047 cm3

2. 2.374.311 cm3

3. 630 cm3

4. 144.5 cm3

5. 108.82 cm3

6. 96 cm3

7. 2,547.168 cm3

8. 48 cm3

9. 151.2 cm3

10. 407.11356 m3

11. 35,112 cm3

12. 7.776 m3

13. 1,157.625 cm3

14. 3,545.3112 cm3

15. 1,594.32 m3

Grade 9 (Alternate) Mathematics III - Learning Modules for EASE Program of DepEd

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Transcript of Grade 9 (Alternate) Mathematics III - Learning Modules for EASE Program of DepEd