pāngarau, kaupae 1, 2008 - nzqa.govt.nz · kia 25 meneti tāu e whakautu ana i ngā pātai o...

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.


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


A O D

E

F

B

C59°


4



1mataaho

5



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.


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


A O D

E

F

B

C59°


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.


Koki SPO = °

S

P

O

Q

RT


6



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.


Angle SPO = °

S

P

O

Q

RT


7



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.


Koki FDE = °

A

B

C

D

E

F


8



9



QUESTION SIX

ABE and ACF are two triangles.Angle BCD = 28°.Angle DEF = 36°.

Find the size of angle FDE.


Angle FDE = °

A

B

C

D

E

F


10



Tau pātai

He puka tāpiri tēnei hei whakaoti i ō whakautu mē e hiahiatia ana. Āta tohua te tau o te pātai.

11



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

pāngarau, kaupae 1, 2008 - nzqa.govt.nz · kia 25 meneti tāu e whakautu ana i ngā pātai o...

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