Post on 05-Jul-2019
Pāngarau, Kaupae 1, 200890153 Te whakamahi whakaaro āhuahanga whaitake
hei whakaoti rapanga
Whiwhinga: Rua9.30 i te ata Rāhina 24 Whiringa-ā-rangi 2008
Tirohia mehemea e ōrite ana te Tau Ākonga ā-Motu kei tō pepa whakauru ki te tau kei runga ake nei.
Me whakautu e koe ngā pātai KATOA kei roto i te pukapuka nei.
Whakaaturia ō mahinga KATOA.
Ki te hiahia koe ki ētahi atu wāhi hei tuhituhi whakautu, whakamahia ngā whārangi kei muri i te pukapuka nei, ka āta tohu ai i ngā tau pātai.
Tirohia mehemea kei roto nei ngā whārangi 2–11 e raupapa tika ana, ā, kāore hoki he whārangi wātea.
HOATU TE PUKAPUKA NEI KI TE KAIWHAKAHAERE HEI TE MUTUNGA O TE WHAKAMĀTAUTAU.
Mā te Kaimāka anake Paearu Paetae
Paetae PaetaeKaiaka
Paetae Kairangi
Te whakamahi whakaaro āhuahanga whaitake hei whakaoti rapanga.
Te whakamahi, te tuhi hoki i ngā whakaaro āhuahanga whaitake hei whakaoti rapanga.
Te whakaoti i tētahi rapanga āhuahanga kua whakaroatia.
Whakakaotanga o te tairanga mahinga
901535
9 0 1 5 3 M1
For Supervisor’s use only
© Mana tohu Mātauranga o Aotearoa, 2008Pūmau te mana. Kia kaua rawa he wāhi o tēnei tuhinga e tāruatia ki te kore te whakaaetanga a te Mana Tohu Mātauranga o Aotearoa.
See back cover for an Englishtranslation of this cover
Kia 25 meneti tāu e whakautu ana i ngā pātai o tēnei pukaiti.
Mō te whiwhinga i te paetae kaiaka, homai ngā take mō ō whakautu.
PĀTAI TUATAHI
E whakaaturia ana te kaupae a Bob i te hoahoa.
He rārangi tika a ABEF.He tuaka hangarite a XY.Ko te koki ABC = 74°.
Tātaitia te rahi o te koki AED, homai hoki he take mō ō whakautu.
Tātaitai Take
Koki AED = °
PĀTAI TUARUA
E whakaaturia ana te arawhata a Bob i te hoahoa.
Ko te koki IHF = 119°.
HF = GF.
Tātaitia te rahi o te koki HFG, homai hoki he take mō ō whakautu.
Tātaitai Take
Koki HFG = °
74°
A B
C
X
YD
E Ftuaka hangarite
EHARA i tehoahoa āwhata
119°HI G
FEHARA i te
hoahoa āwhata
2
Pāngarau 90153, 2008
Mā te kaimākaanahe
You are advised to spend 25 minutes answering the questions in this booklet.
To achieve merit you must give reasons for your answers.
QUESTION ONE
Bob’s trestle is shown in the diagram.
ABEF is a straight line.XY is an axis of symmetry.Angle ABC = 74°.
Calculate the size of angle AED, giving reasons for your answers.
Calculations Reasons
Angle AED = °
QUESTION TWO
Bob’s step-ladder is shown in the diagram.
Angle IHF = 119°.
HF = GF.
Calculate the size of angle HFG, giving reasons for your answers.
Calculations Reasons
Angle HFG = °
74°
A B
C
X
YD
E Faxis of symmetry
Diagram is NOT to scale
119°HI G
FDiagram is
NOT to scale
3
Mathematics 90153, 2008
Assessor’suse only
PĀTAI TUATORU
E whakaaturia ana tētahi pihanga1 tapawhā hāngai, he mea ōpure, a JKLM.
Ko tētahi o ngā wāhi kōata o te pihanga, a PQRST, he taparima rite.
Tātaitia te rahi o te koki LRS, homai hoki he take mō ō whakautu.
Tātaitai Take
Koki LRS = °
PĀTAI TUAWHĀ
I te hoahoa, he porowhita haurua a ABCD, ko te pū ko O.
He tapawhā whakarara rite a EBFO.
He whakarara a BF ki CD.
Ko te koki EBF = 59°.
Tātaitia te rahi o te koki COD.
Tuhia tētahi pūtake āhuahanga mō ia wāhanga e whai haere ana ki tō whakautu.
Koki COD = °
L S T M
K Q J
R P
EHARA i tehoahoa āwhata
A O D
E
F
B
C59°
EHARA i tehoahoa āwhata
4
Pāngarau 90153, 2008
Mā te kaimākaanahe
1mataaho
5
Mathematics 90153, 2008
Assessor’suse only
QUESTION THREE
The diagram shows a rectangular stained-glass window, JKLM.
One of the pieces of glass in the window, PQRST, is a regular pentagon.
Calculate the size of angle LRS, giving reasons for your answers.
Calculations Reasons
Angle LRS = °
QUESTION FOUR
In the diagram, ABCD is a semicircle, centre, O.
EBFO is a rhombus.
BF is parallel to CD.
Angle EBF = 59°.
Calculate the size of angle COD.
You must give a geometric reason for each step leading to your answer.
Angle COD = °
L S T M
K Q J
R P
Diagram is NOT to scale
A O D
E
F
B
C59°
Diagram is NOT to scale
PĀTAI TUARIMA
I te hoahoa nei, kei tētahi porowhita a P, Q, R me S e takoto ana, ko O te pū.
He pātapa a TR ki te porowhita kei R.Ko te koki PQR = x°.Ko te koki SRT = y°.
Tātaitia te rahi o te koki SPO, e ai ki x me y.
Tuhia tētahi pūtake āhuahanga mō ia wāhanga e whai haere ana ki tō whakautu.
Koki SPO = °
S
P
O
Q
RT
EHARA i tehoahoa āwhata
6
Pāngarau 90153, 2008
Mā te kaimākaanahe
QUESTION FIVE
In the diagram P, Q, R and S lie on a circle, centre, O.
TR is a tangent to the circle at R.Angle PQR = x°.Angle SRT = y°.
Find the size of angle SPO, in terms of x and y.
You must give a geometric reason for each step leading to your answer.
Angle SPO = °
S
P
O
Q
RT
Diagram is NOT to scale
7
Mathematics 90153, 2008
Assessor’suse only
PĀTAI TUAONO
He tapatoru a ABE me ACF.Ko te koki BCD = 28°.Ko te koki DEF = 36°.
Tātaitia te rahi o te koki FDE.
Tuhia tētahi pūtake āhuahanga mō ia wāhanga e whai haere ana ki tō whakautu.
Koki FDE = °
A
B
C
D
E
F
EHARA i tehoahoa āwhata
8
Pāngarau 90153, 2008
Mā te kaimākaanahe
9
Mathematics 90153, 2008
Assessor’suse only
QUESTION SIX
ABE and ACF are two triangles.Angle BCD = 28°.Angle DEF = 36°.
Find the size of angle FDE.
You must give a geometric reason for each step leading to your answer.
Angle FDE = °
A
B
C
D
E
F
Diagram is NOT to scale
10
Pāngarau 90153, 2008
Mā te kaimākaanahe
Tau pātai
He puka tāpiri tēnei hei whakaoti i ō whakautu mē e hiahiatia ana. Āta tohua te tau o te pātai.
11
Mathematics 90153, 2008
Assessor’suse only
Question number
Extra paper for continuation of answers if required.Clearly number the question.
Level 1 Mathematics, 200890153 Use geometric reasoning to solve problems
Credits: Two9.30 am Monday 24 November 2008
Check that the National Student Number (NSN) on your admission slip is the same as the number at the top of this page.
You should answer All the questions in this booklet.
You should show ALL working.
If you need more space for any answer, use the page(s) provided at the back of this booklet and clearly number the question.
Check that this booklet has pages 2–11 in the correct order and that none of these pages is blank.
YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION.
For Assessor’s use only Achievement Criteria
Achievement Achievement with Merit
Achievement with Excellence
Use geometric reasoning to solve problems.
Use, and state, geometric reasons in solving problems.
Solve an extended geometrical problem.
Overall Level of Performance
© New Zealand Qualifications Authority, 2008All rights reserved. No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.
90
15
3M
English translation of the wording on the front cover