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NYS COMMON CORE MATHEMATICS CURRICULUM 7β’3 Lesson 21
Lesson 21: Surface Area
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π π π π π π π π π
Lesson 21: Surface Area
Student Outcomes
Students find the surface area of three-dimensional objects whose surface area is composed of triangles and
quadrilaterals. They use polyhedron nets to understand that surface area is simply the sum of the area of the
lateral faces and the area of the base(s).
Classwork
Opening Exercise (8 minutes): Surface Area of a Right Rectangular Prism
Students use prior knowledge to find the surface area of the given right rectangular prism by decomposing the prism
into the plane figures that represent its individual faces. Students then discuss their methods aloud.
Opening Exercise: Surface Area of a Right Rectangular Prism
On the provided grid, draw a net representing the surfaces of the right rectangular prism
(assume each grid line represents π inch). Then, find the surface area of the prism by finding
the area of the net.
There are six rectangular faces that make up the net.
The four rectangles in the center form one long rectangle that is ππ π’π§. by π π’π§.
ππ«ππ = ππ
ππ«ππ = π π’π§ β ππ π’π§
ππ«ππ = ππ π’π§π
Two rectangles form the wings, both π π’π§ by π π’π§.
ππ«ππ = ππ
ππ«ππ = π π’π§ β π π’π§
ππ«ππ = ππ π’π§π
The area of both wings is π(ππ π’π§π) = ππ π’π§π.
The total area of the net is
π¨ = ππ π’π§π + ππ π’π§π = πππ π’π§π
The net represents all the surfaces of the
rectangular prism, so its area is equal to the
surface area of the prism. The surface area of
the right rectangular prism is πππ π’π§π.
Note: Students may draw any of the variations of nets for the given prism.
Scaffolding:
Students may need to review the meaning of the term net from Grade 6. Prepare a solid right rectangular prism such as a wooden block and a paper net covering the prism to model where a net comes from.
NYS COMMON CORE MATHEMATICS CURRICULUM 7β’3 Lesson 21
Lesson 21: Surface Area
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3 in
6 in
4 in
5 in
Discussion (3 minutes)
What other ways could we have found the surface area of the rectangular prism?
Surface area formula: ππ΄ = 2ππ€ + 2πβ + 2π€β
ππ΄ = 2(3 in β 4 in) + 2(3 in β 6 in) + 2(4 in β 6 in)
ππ΄ = 24 in2 + 36 in2 + 48 in2
ππ΄ = 108 in2
Find the areas of each individual rectangular face:
Area = length Γ width
π΄ = 6 in Γ 3 in π΄ = 4 in Γ 3 in π΄ = 6 in Γ 4 in
π΄ = 18 in2 π΄ = 12 in2 π΄ = 24 in2
There are two of each face, so ππ΄ = 2(18 in2 + 12 in2 + 24 in2)
ππ΄ = 2(54 in2)
ππ΄ = 108 in2.
Discussion (6 minutes): Terminology
A right prism can be described as a solid with two βendβ faces (called its bases) that are exact copies of each other and
rectangular faces that join corresponding edges of the bases (called lateral faces).
Are the bottom and top faces of a right rectangular prism the bases of the
prism?
Not always. Any of its opposite faces can be considered bases because
they are all rectangles.
If we slice the right rectangular prism in half along a diagonal of a base (see
picture), the two halves are called right triangular prisms. Why do you think
they are called triangular prisms?
The bases of each prism are triangles, and prisms are named by their
bases.
Why must the triangular faces be the bases of these prisms?
Because the lateral faces (faces that are not bases) of a right prism have to be rectangles.
6 in 6 in
3 in
3 in
4 in
4 in
Front rectangle
Top rectangle
Side rectangle
NYS COMMON CORE MATHEMATICS CURRICULUM 7β’3 Lesson 21
Lesson 21: Surface Area
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Can the surface area formula for a right rectangular prism (ππ΄ = 2ππ€ + 2πβ + 2π€β) be applied to find the
surface area of a right triangular prism? Why or why not?
No, because each of the terms in the surface area formula represents the area of a rectangular face. A
right triangular prism has bases that are triangular, not rectangular.
Exercise 1 (8 minutes)
Students find the surface area of the right triangular prism to determine the validity of a given conjecture.
Exercise 1
Marcus thinks that the surface area of the right triangular prism will be half that of the right rectangular prism and wants
to use the modified formula πΊπ¨ =ππ(πππ+ πππ + πππ). Do you agree or disagree with Marcus? Use nets of the prisms
to support your argument.
The surface area of the right rectangular prism is πππ π’π§π, so Marcus believes the surface areas of each right triangular
prism is ππ π’π§π.
Students can make comparisons of the area values depicted in the nets of the prisms and can also compare the physical
areas of the nets either by overlapping the nets on the same grid or using a transparent overlay.
The net of the right triangular prism has one less face than the right
rectangular prism. Two of the rectangular faces on the right triangular prism
(rectangular regions 1 and 2 in the diagram) are the same faces from the right
rectangular prism, so they are the same size. The areas of the triangular bases
(triangular regions 3 and 4 in the diagram) are half the area of their
corresponding rectangular faces of the right rectangular prism. These four
faces of the right triangular prism make up half the surface area of the right
rectangular prism before considering the fifth face; no, Marcus is incorrect.
The areas of rectangular faces 1 and 2, plus the areas of the triangular regions
3 and 4 is ππ π’π§π. The last rectangular region has an area of ππ π’π§π. The total
area of the net is ππ π’π§π + ππ π’π§π or ππ π’π§π, which is far more than half the
surface area of the right rectangular prism.
Use a transparency to show students how the nets overlap where the lateral faces together form a longer rectangular
region, and the bases are represented by βwingsβ on either side of that triangle. Consider using student work for this if
there is a good example. Use this setup in the following discussion.
Discussion (5 minutes)
The surface area formula (ππ΄ = 2ππ€ + 2πβ + 2π€β) for a right rectangular prism cannot be applied to a right
triangular prism. Why?
The formula adds the areas of six rectangular faces. A right triangular prism only has three rectangular
faces and also has two triangular faces (bases).
MP.3
NYS COMMON CORE MATHEMATICS CURRICULUM 7β’3 Lesson 21
Lesson 21: Surface Area
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The area formula for triangles is 1
2 the formula for the area of rectangles or parallelograms. Can the surface
area of a triangular prism be obtained by dividing the surface area formula for a right rectangular prism by 2?
Explain.
No. The right triangular prism in the above example had more than half the surface area of the right
rectangular prism that it was cut from. If this occurs in one case, then it must occur in others as well.
If you compare the nets of the right rectangular prism and the right triangular
prism, what do the nets seem to have in common? (Hint: What do all right
prisms have in common? Answer: Rectangular lateral faces)
Their lateral faces form a larger rectangular region, and the bases are
attached to the side of that rectangle like βwings.β
Will this commonality always exist in right prisms? How do you know?
Yes. Right prisms must have rectangular lateral faces. If we align all the
lateral faces of a right prism in a net, they can always form a larger
rectangular region because they all have the same height as the prism.
How do we determine the total surface area of the prism?
Add the total area of the lateral faces and the areas of the bases.
If we let πΏπ΄ represent the lateral area and let π΅ represent the area of a
base, then the surface area of a right prism can be found using the
formula:
ππ΄ = πΏπ΄ + 2π΅.
Example 1 (6 minutes): Lateral Area of a Right Prism
Students find the lateral areas of right prisms and recognize the pattern of multiplying the height of the right prism (the
distance between its bases) by the perimeter of the prismβs base.
Example 1: Lateral Area of a Right Prism
A right triangular prism, a right rectangular prism, and a right pentagonal prism are pictured below, and all have equal
heights of π.
a. Write an expression that represents the lateral area of the right triangular prism as the sum of the areas of its
lateral faces.
π β π + π β π + π β π
b. Write an expression that represents the lateral area of the right rectangular prism as the sum of the areas of
its lateral faces.
π β π + π β π + π β π + π β π
Scaffolding:
The teacher may need to assist students in finding the commonality between the nets of right prisms by showing examples of various right prisms and pointing out the fact that they all have rectangular lateral faces. The rectangular faces may be described as βconnectorsβ between the bases of a right prism.
MP.8
NYS COMMON CORE MATHEMATICS CURRICULUM 7β’3 Lesson 21
Lesson 21: Surface Area
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c. Write an expression that represents the lateral area of the right pentagonal prism as the sum of the areas of
its lateral faces.
π β π + π β π + π β π + π β π + π β π
d. What value appears often in each expression and why?
π; Each prism has a height of π; therefore, each lateral face has a height of h.
e. Rewrite each expression in factored form using the distributive property and the height
of each lateral face.
π(π + π + π) π(π + π + π + π) π(π + π + π + π + π)
f. What do the parentheses in each case represent with respect to the right prisms?
π (π + π + π)β π©ππ«π’π¦ππππ«
π (π + π + π + π)β π©ππ«π’π¦ππππ«
π (π + π + π + π + π)β π©ππ«π’π¦ππππ«
The perimeter of the base of the corresponding prism.
g. How can we generalize the lateral area of a right prism into a formula that applies to all right prisms?
If π³π¨ represents the lateral area of a right prism, π· represents the perimeter of the right prismβs base, and π
represents the distance between the right prismβs bases, then:
π³π¨ = π·πππ¬π β π.
Closing (5 minutes)
The vocabulary below contains the precise definitions of the visual and colloquial descriptions used in the lesson. Please
read through the definitions aloud with your students, asking questions that compare the visual and colloquial
descriptions used in the lesson with the precise definitions.
Relevant Vocabulary
RIGHT PRISM: Let π¬ and π¬β² be two parallel planes. Let π© be a triangular or rectangular region or a region that is the union
of such regions in the plane π¬. At each point π· of π©, consider the segment π·π·β² perpendicular to π¬, joining π· to a point π·β²
of the plane π¬β². The union of all these segments is a solid called a right prism.
There is a region π©β² in π¬β² that is an exact copy of the region π©. The regions π© and π©β² are called the base faces (or just
bases) of the prism. The rectangular regions between two corresponding sides of the bases are called lateral faces of the
prism. In all, the boundary of a right rectangular prism has π faces: π base faces and π lateral faces. All adjacent faces
intersect along segments called edges (base edges and lateral edges).
Scaffolding:
Example 1 can be explored further by assigning numbers to represent the lengths of the sides of the bases of each prism. If students represent the lateral area as the sum of the areas of the lateral faces without evaluating, the common factor in each term will be evident and can then be factored out to reveal the same relationship.
MP.8
NYS COMMON CORE MATHEMATICS CURRICULUM 7β’3 Lesson 21
Lesson 21: Surface Area
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CUBE: A cube is a right rectangular prism all of whose edges are of equal length.
SURFACE: The surface of a prism is the union of all of its faces (the base faces and lateral faces).
NET: A net is a two-dimensional diagram of the surface of a prism.
1. Why are the lateral faces of right prisms always rectangular regions?
Because along a base edge, the line segments π·π·β² are always perpendicular to the edge, forming a rectangular
region.
2. What is the name of the right prism whose bases are rectangles?
Right rectangular prism
3. How does this definition of right prism include the interior of the prism?
The union of all the line segments fills out the interior.
Exit Ticket (4 minutes)
Lesson Summary
The surface area of a right prism can be obtained by adding the areas of the lateral faces to the area of the bases.
The formula for the surface area of a right prism is πΊπ¨ = π³π¨+ ππ©, where πΊπ¨ represents the surface area of the
prism, π³π¨ represents the area of the lateral faces, and π© represents the area of one base. The lateral area π³π¨ can be
obtained by multiplying the perimeter of the base of the prism times the height of the prism.
NYS COMMON CORE MATHEMATICS CURRICULUM 7β’3 Lesson 21
Lesson 21: Surface Area
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Name Date
Lesson 21: Surface Area
Exit Ticket
Find the surface area of the right trapezoidal prism. Show all necessary work.
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Lesson 21: Surface Area
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Exit Ticket Sample Solutions
Find the surface area of the right trapezoidal prism. Show all necessary work.
πΊπ¨ = π³π¨+ ππ©
π³π¨ = π· β π
π³π¨ = (π ππ¦ + π ππ¦ + π ππ¦ + ππ ππ¦ ) β π ππ¦
π³π¨ = ππ ππ¦ β π ππ¦
π³π¨ = πππ ππ¦π
Each base consists of a π ππ¦ by π ππ¦ rectangle and right triangle with a
base of π ππ¦ and a height of π ππ¦. Therefore, the area of each base:
π© = π¨π + π¨π
π© = ππ+ππππ
π© = (π ππ¦ β π ππ¦) + (ππβ π ππ¦ β π ππ¦)
π© = ππ ππ¦π + π ππ¦π
π© = ππ ππ¦π
πΊπ¨ = π³π¨+ ππ©
πΊπ¨ = πππ ππ¦π + π(ππ ππ¦π)
πΊπ¨ = πππ ππ¦π + ππ ππ¦π
πΊπ¨ = πππ ππ¦π
The surface of the right trapezoidal prism is πππ ππ¦π.
Problem Set Sample Solutions
1. For each of the following nets, highlight the perimeter of the lateral area, draw the solid represented by the net,
indicate the type of solid, and then find the solidβs surface area.
a. Right rectangular prism
πΊπ¨ = π³π¨+ ππ©
π³π¨ = π· β π π© = ππ
π³π¨ = (πππ ππ¦ + π
ππ ππ¦ + π
ππ ππ¦ + π
ππ ππ¦) β π ππ¦ π© = π
ππ ππ¦ β π
ππ ππ¦
π³π¨ = ππ ππ¦ β π ππ¦ π© =ππππ¦ β
πππππ¦
π³π¨ = πππ ππ¦π π© =πππ ππ¦π
πΊπ¨ = πππ ππ¦π + π(πππ
ππ¦π)
πΊπ¨ = πππ ππ¦π + ππ. π ππ¦π
πΊπ¨ = πππ. π ππ¦π
The surface area of the right rectangular prism is πππ. π ππ¦π
(3-Dimensional Form)
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b. Right triangular prism
πΊπ¨ = π³π¨+ ππ©
π³π¨ = π· β π π© =ππππ
π³π¨ = (ππ π’π§.+ π π’π§.+ ππ π’π§. ) β ππ π’π§. π© =ππ(π π’π§. ) (π
ππ π’π§. )
π³π¨ = ππ π’π§.β ππ π’π§. π© = π π’π§. (πππ π’π§. )
π³π¨ = πππ π’π§π π© = (ππ +ππ) π’π§π
π© = ππππ π’π§π
πΊπ¨ = πππ πππ + π(ππππ πππ)
πΊπ¨ = πππ πππ + (ππ +ππ) πππ
πΊπ¨ = πππ πππ + ππππππ
πΊπ¨ = πππππ πππ
The surface area of the right triangular prism is πππππ πππ.
2. Given a cube with edges that are ππ
inch long:
a. Find the surface area of the cube.
πΊπ¨ = πππ
πΊπ¨ = π(π
π π’π§. )
π
πΊπ¨ = π(π
π π’π§. ) β (
π
π π’π§. )
πΊπ¨ = π(π
ππ π’π§π)
πΊπ¨ =ππ
π π’π§π or π
π
π π’π§π
b. Joshua makes a scale drawing of the cube using a scale factor of π. Find the surface area of the cube that
Joshua drew.
π
πππ.β π = π ππ.; The edge lengths of Joshuaβs drawing would be π inches.
πΊπ¨ = π(π ππ. )π
πΊπ¨ = π(π πππ)
πΊπ¨ = ππ πππ
(3-Dimensional Form)
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Lesson 21: Surface Area
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c. What is the ratio of the surface area of the scale drawing to the surface area of the actual cube, and how does
the value of the ratio compare to the scale factor?
ππ Γ· πππ
ππ Γ·πππ
ππ βπππ
π β π = ππ. The ratios of the surface area of the scale drawing to the surface area of the actual cube is ππ:π.
The value of the ratio is ππ. The scale factor of the drawing is π, and the value of the ratio of the surface area
of the drawing to the surface area of the actual cube is ππ or ππ.
3. Find the surface area of each of the following right prisms using the formula πΊπ¨ = π³π¨ + ππ©.
a.
πΊπ¨ = π³π¨ + ππ©
π³π¨ = π· β π
π³π¨ = (πππ
π π¦π¦+ ππ π¦π¦+ π
π
π π¦π¦) β ππ π¦π¦
π³π¨ = ππ π¦π¦ β ππ π¦π¦
π³π¨ = πππ π¦π¦π
π© =ππππ πΊπ¨ = πππ π¦π¦π + π(
πππ π¦π¦π)
π© =ππβ (π
ππ π¦π¦) β (ππ π¦π¦) πΊπ¨ = πππ π¦π¦π + ππ π¦π¦π
π© =ππβ (ππ + π) π¦π¦π πΊπ¨ = πππ π¦π¦π
π© =ππβ ππ π¦π¦π
π© =πππ π¦π¦π
The surface area of the prism is πππ π¦π¦π.
ππππ mm
NYS COMMON CORE MATHEMATICS CURRICULUM 7β’3 Lesson 21
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b.
πΊπ¨ = π³π¨+ ππ©
π³π¨ = π· β π π© =ππππ
π³π¨ = (ππππππ. +π
ππππ. +π ππ. ) β π ππ π© =
ππβ π
πππππ.β π
ππππ.
π³π¨ = (πππππ
ππ. +πππππ. +π ππ. ) β π ππ π© =
ππβπππππ
ππ.βππππ.
π³π¨ = (πππππ
ππ. +πππππ
ππ. +πππππ
ππ. ) β π ππ. π© =π,ππππππ
ππ2
π³π¨ = (πππππ
ππ. ) β π ππ. π© = ππππππ2
π³π¨ =ππ,πππππ
πππ ππ© = π β πππππππ
π³π¨ = πππππ πππ ππ© = ππ
πππππ
πΊπ¨ = π³π¨+ ππ©
πΊπ¨ = πππππ π’π§π + ππ
ππ π’π§π
πΊπ¨ = ππππππ π’π§π
The surface area of the prism is ππππππ π’π§π.
c.
πΊπ¨ = π³π¨+ ππ©
π³π¨ = π· β π
π³π¨ = (π
π π’π§.+
π
π π’π§. +
π
ππ’π§. +
π
π π’π§. +
π
ππ’π§. +
π
π π’π§. ) β π π’π§.
π³π¨ = (πππ π’π§. ) β π π’π§.
π³π¨ = π π’π§π + πππ π’π§π π© = π¨πππππππππ + ππ¨ππππππππ
π³π¨ = πππ π’π§π π© = (
ππ π’π§.β
ππ
π’π§. ) + π βππ(ππ
π’π§.βππ π’π§. )
π© = (πππ π’π§π) + (
πππ π’π§π)
πΊπ¨ = πππ πππ + π(
ππ
πππ) π© =πππ
π’π§π +πππ π’π§π
πΊπ¨ = πππ πππ +
ππ πππ π© =
πππ π’π§π +
πππ
π’π§π
πΊπ¨ = πππ
πππ +ππ
πππ π© =πππ π’π§π
πΊπ¨ = πππ
πππ π© =ππ π’π§π
The surface area of the prism is πππ
π’π§π.
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Lesson 21: Surface Area
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d.
πΊπ¨ = π³π¨+ ππ©
π³π¨ = π· β π
π³π¨ = (ππ ππ¦+ ππ ππ¦ + π. π ππ¦ + π.π ππ¦) β ππ
πππ¦
π³π¨ = (ππ ππ¦+ ππ. π ππ¦) β ππ
π ππ¦
π³π¨ = (ππ. π)ππ¦ β πππππ¦
π³π¨ = (ππ. π ππ¦π + ππ. π ππ¦π)
π³π¨ = ππ. π ππ¦π
πΊπ¨ = π³π¨ + ππ© π© =ππ(ππ ππ¦ β π ππ¦ ) +
ππ (ππ ππ¦ β ππ ππ¦ )
πΊπ¨ = ππ. π πππ + π(ππ ππ¦π) π© =ππ(ππ ππ¦π + πππ ππ¦π)
πΊπ¨ = ππ. π ππ¦π + πππ ππ¦π π© =ππ(πππ ππ¦π)
πΊπ¨ = πππ. π ππ¦π π© = ππ ππ¦π
The surface area of the prism is πππ. π ππ¦π.
4. A cube has a volume of ππ π¦π. What is the cubeβs surface area?
A cubeβs length, width, and height must be equal. ππ = π β π β π = ππ, so the length, width, and height of the cube
are all π π¦.
πΊπ¨ = πππ
πΊπ¨ = π(π π¦)π
πΊπ¨ = π(ππ π¦π)
πΊπ¨ = ππ π¦π
5. The height of a right rectangular prism is πππ ππ. The length and width of the prismβs base are π ππ. and π
ππ ππ. Use
the formula πΊπ¨ = π³π¨ + ππ© to find the surface area of the right rectangular prism.
πΊπ¨ = π³π¨+ ππ©
π³π¨ = π· β π π© = ππ
π³π¨ = (π ππ.+π ππ.+πππ ππ.+π
ππ ππ. ) β π
ππ ππ. π© = π ππ.β π
ππ ππ.
π³π¨ = (π ππ.+ π ππ. +π ππ. ) β πππ ππ. πΊπ¨ = π³π¨+ ππ π© = π πππ
π³π¨ = π ππ.β πππ ππ. πΊπ¨ = ππ
ππ
πππ + π(π πππ)
π³π¨ = ππ πππ + πππ
πππ πΊπ¨ = ππππ
πππ + π πππ
π³π¨ = ππππ πππ πΊπ¨ = ππ
ππ πππ
The surface area of the right rectangular prism is ππππ πππ.
NYS COMMON CORE MATHEMATICS CURRICULUM 7β’3 Lesson 21
Lesson 21: Surface Area
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6. The surface area of a right rectangular prism is ππππ π’π§π. The dimensions of its base are π π’π§ and π π’π§ Use the
formula πΊπ¨ = π³π¨+ ππ© and π³π¨ = π·π to find the unknown height π of the prism.
πΊπ¨ = π³π¨+ πB
πΊπ¨ = π· β π + ππ©
ππππ π’π§π = ππ π’π§.β (π) + π(ππ π’π§π)
ππππ π’π§π = ππ π’π§.β (π) + ππ π’π§π
ππππ π’π§π β ππ π’π§π = ππ π’π§.β (π) + ππ π’π§π β ππ π’π§π
ππππ π’π§π = ππ π’π§.β (π) + π π’π§π
ππππ π’π§π β
πππ π’π§.
= ππ π’π§ βπ
ππ π’π§.β (π)
ππ
π π’π§π β
π
ππ π’π§= π β π
π
π π’π§. = π
π =ππ π’π§. or π
ππ π’π§.
The height of the prism is πππ π’π§.
7. A given right triangular prism has an equilateral triangular base. The height of that equilateral triangle is
approximately π. π ππ¦. The distance between the bases is π ππ¦. The surface area of the prism is πππππ ππ¦π. Find
the approximate lengths of the sides of the base.
πΊπ¨ = π³π¨+ ππ© Let π represent the number of centimeters in each side of the equilateral triangle.
π³π¨ = π· β π π© =ππππ πππ
ππ πππ = π³π¨ + ππ©
π³π¨ = π(π ππ¦) β π ππ¦ π© =ππβ (π ππ¦) β π. π ππ¦ πππ
ππ ππ¦π = πππ ππ¦π + π(π. πππ ππ¦π)
π³π¨ = πππ ππ¦π π© = π. πππ ππ¦π πππππ ππ¦π = πππ ππ¦π + π.ππ ππ¦π
πππππ ππ¦π = ππ. ππ ππ¦π
πππππ ππ¦π = ππ
ππππ ππ¦π
πππ
π ππ¦π =
πππ
πππ ππ¦π
πππ
π ππ¦π β
ππ
πππ ππ¦=πππ
πππ ππ¦π β
ππ
πππ ππ¦
ππππ
πππππ¦ = π
π =ππππ
πππππ¦
π β π. π ππ¦
The lengths of the sides of the equilateral triangles are approximately π. π ππ¦ each.