Te Pāngarau me te Tauanga, Kaupae 1, 2018 · New council rules state that a slide must have: • an angle (UTX) of less than 60° with the water AND • a ladder height of less than

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© Mana Tohu Mātauranga o Aotearoa, 2018. Pūmau te mana. Kia kaua rawa he wāhi o tēnei tuhinga e whakahuatia ki te kore te whakaaetanga tuatahi a te Mana Tohu Mātauranga o Aotearoa.

MĀ TE KAIMĀKA ANAKE

TAPEKE

Te Pāngarau me te Tauanga, Kaupae 1, 201891031M Te whakahāngai whakaaro āhuahanga

whaitake hei whakaoti rapanga

9.30 i te ata Rātū 20 Whiringa-ā-rangi 2018 Whiwhinga: Whā

Paetae Kaiaka KairangiTe whakahāngai whakaaro āhuahanga whaitake hei whakaoti rapanga.

Te whakahāngai whakaaro āhuahanga whaitake mā te whakaaro whaipānga hei whakaoti rapanga.

Te whakahāngai whakaaro āhuahanga whaitake mā te whakaaro waitara hōhonu hei whakaoti rapanga.

Tirohia mēnā e rite ana te Tau Ākonga ā-Motu (NSN) kei runga i tō puka whakauru ki te tau kei runga i tēnei whārangi.

Me whakamātau koe i ngā tūmahi KATOA kei roto i tēnei pukapuka.

Whakaaturia ngā mahinga KATOA.

Mēnā ka hiahia whārangi atu anō koe mō ō tuhinga, whakamahia te (ngā) whārangi wātea kei muri o tēnei pukapuka, ka āta tohu ai i te tau tūmahi.

Tirohia mēnā e tika ana te raupapatanga o ngā whārangi 2 – 23 kei roto i tēnei pukapuka, ā, kāore tētahi o aua whārangi i te takoto kau.

ME HOATU RAWA KOE I TĒNEI PUKAPUKA KI TE KAIWHAKAHAERE Ā TE MUTUNGA O TE WHAKAMĀTAUTAU.

Te Pāngarau me te Tauanga 91031M, 2018

MĀ TE KAIMĀKA

ANAKE

NGĀ PAPATĀKARO

TŪMAHI TUATAHI

(a) Ewhakaaturiaanairarohewāhangaangapikipikipapatakāro.

HewhakararateAHmeteGJ.

HehuapaeteAGmeteHJ.

Koki JGD = 51°

H

K

x

51°

A G D

J

Diagram is NOT to scale

(i) Tātaihiaterahi,x,otekokiJHK.

Whakamahia te whakaaro āhuahanga mārama hei parahau i tāu tuhinga.

KĀORE i tuhi ā-āwhatatia tēnei hoahoa

2

PLAYGROUNDS

QUESTION ONE

(a) Partofaplaygroundclimbingframeisshownbelow.

AHandGJareparallel.

AGandHJarehorizontal.

AngleJGD=51°

H

K

x

51°

A G D

J


(i) Calculatethesize,x,ofangleJHK.

Justify your answer with clear geometric reasoning.

3

Mathematics and Statistics 91031, 2018

ASSESSOR’S USE ONLY


MĀ TE KAIMĀKA

ANAKE

(ii) Ewhakaaturiaanatētahiatuwāhangaotētahiangapikipiki.

C

y

z

D B


Tuhia te koki z e ai ki y.



4

(ii) Anotherpartofaclimbingframeisshownbelow.

C

y

z

D B


Writetheanglezintermsofy.


5




MĀ TE KAIMĀKA

ANAKE

(b) Kuahangaiaheretiretikitētahipunakaukaumaiitētahiangatapatorumetearawhatapoutū.

HehuapaeaTYmete8mteroa.

He3mteroaoTX.

He2.5mteteiteioXU.

HepoutokopoutūaXUmeYW.

T Y

W

U

2.5 m

ladderheight

8 m3 m

X


Ekīanangāturehouotekauniherakoteretiretimemātua:

• whai koki (UTX) iti iho i te 60°kitewai,METE

• whaiarawhataeitiihoteteiteiite5mita.

MerapumēnāeūanatēneiretiretikiēneitureeRUAotekaunihera.

Whakaaturia ō mahinga katoa, ā, kia mārama te tuku i tō whakataunga.


teitei arawhata

6

(b) Aslideintoapoolismadefromatriangularframewithaverticalladder.

TYishorizontaland8mlong.

TXis3mlong.

XUis2.5mhigh.

XUandYWarebothverticalsupports.

T Y

W

U

2.5 m

ladderheight

8 m3 m

X


Newcouncilrulesstatethataslidemusthave:

• anangle(UTX)oflessthan60° with the water AND

• aladderheightoflessthan5metres.

FindoutwhetherornotthisslidepassesBOTHofthesecouncilregulations.

Show your working and state your final conclusion clearly.

7




MĀ TE KAIMĀKA

ANAKE

(c) HemaheretāMadalynmōtētahipapatākarootetakiwākeirotokotētahiruakirikiri(koO),tētahitārere(koG),tētahitiemi1(koX)meteretireti(koF).

North

G

X O

F


He ōrite te tawhitiotetārere,tiemi,meteretiretimaiitepokapūoteruakirikiri.

He 130°teahungaoteretiretimaiiteruakirikiri.

He285°teahungaotetiemimaiiteruakirikiri.

He 350°teahungaotetāreremaiiteruakirikiri.

ItetūaMadalynitetiemimeteangaatukitetārere.

Heahatekokiitirawakahurihiaeiakiaangaatuiakiteretireti?

Whakaaturia ō mahinga katoa me te whakamahi i te whakaaro āhuahanga mārama hei parahau i tāu tuhinga.


Raki

1tīeke

8

(c) Madalynhasaplanofalocalplaygroundwhichhasasandpit(labelledO),swing(labelledG),seesaw(labelledX),andslide(labelledF).

North

G

X O

F


Theswing,seesaw,andslideareallthesame distancefromthecentreofthesandpit.

Theslideisonabearingof130°fromthesandpit.

Theseesawisonabearingof285°fromthesandpit.

Theswingisonabearingof350°fromthesandpit.

Madalynwasstandingattheseesawandfacingtheswing.

Whatisthesmallestangleshewouldturntofacetheslide?

Show your working and justify your answer with clear geometric reasoning.

9




MĀ TE KAIMĀKA

ANAKE

TŪMAHI TUARUA

(a) Hemeahangaheangapikipikimaiiteporohitahauruamengātapatoru.

KuahangariteteangapikipikihurinoaiteFE.

Koki CAD = 33°

KKADtewhitiangaoteporohitahaurua.

x

y

AE

B

33°

F

L

C

D


(i) Tātaihiaterahi,x,otekokiALD.


(ii) Tātaihiaterahi,y,otekokiBAC.



10

QUESTION TWO

(a) Aclimbingframeismadefromasemi-circleandtriangles.

TheclimbingframeissymmetricalaboutFE.

AngleCAD=33°

ADisthediameterofthesemi-circle.

x

y

AE

B

33°

F

L

C

D


(i) Calculatethesize,x,oftheangleALD.


(ii) Calculatethesize,y,ofangleBAC.


11




MĀ TE KAIMĀKA

ANAKE

(b) Keitehoahoatiahetāparepikipikiporohitaanō.

KoOtepūoteporohita.

HewhakararangārārangiOCmeteBD.

OC = BD

A

B

CO

D


HāponotiakaōriteteroaoterārangitorotikaACkiteroaoterārangitorotikaOD.



12

(b) Anothercircularclimbingframeisbeingdesigned.

PointOisthecentreofthecircle.

LinesOCandBDareparallel.

OC = BD

A

B

CO

D


ProvethatthelengthofthestraightlineACequalsthelengthofthestraightlineOD.


13




MĀ TE KAIMĀKA

ANAKE

(c) Keitehiahiatētahikaihoahoakitewaihangaitētahiangametētahitaparimariteitiihoirotoitētahitaparimaritenuiake.

KotepūwāhiOko: • tepūongātaparimaerua • 0.6mitamaiiteakituotetaparimaitiake.

Kongāpoumaiingāakituotetaparimaitiakekitetahaotetaparimanuiakehe: • 2 mita te roa • kiakokihāngaikingātahaotetaparimanuiake.

F

O

L

KC

B

2 m 0.6 m


TātaihiateroaoFL,tētahitahaotetaparimaritenuiake.

Āta whakaaturia ō mahinga.


14

(c) Adesignerwantstocreateaframewithasmallregularpentagoninsidealargerregularpentagon.

PointOis: • thecentreofbothpentagons • 0.6metresfromthevertexofthesmallerpentagon.

Polesfromtheverticesofthesmallerpentagontothesideofthelargerpentagonare: • 2metreslong • atrightanglestothesidesofthelargerpentagon.

F

O

L

KC

B

2 m 0.6 m


CalculatethelengthofFL,onesideofthelargerregularpentagon.

Show your working clearly.

15




MĀ TE KAIMĀKA

ANAKE

TŪMAHI TUATORU

(a) Kuahangaiahetāreremaiitētahipou2.3mteroa,kawhakatītahatiamaiitewhenua.

He0.5mmaiitepapatetūrutārere.

HerārangihuapaeaBC,1.8mteroa.

HepoutūaAB.

C

0.5 m

1.8 m

2.3 m

B

A

w


length of rope, v

(i) Tātaihiaterahi,w,otekokiCAB


(ii) Tātaihiateroaotetaura,v,epupurianaitetūrutārere.



roa o te taura,v

16

QUESTION THREE

(a) Aswingismadefromonepole2.3mlong,placedatanangleintheground.

Theswingseatis0.5mofftheground.

BCisahorizontallineoflength1.8m.

ABisvertical.

C

0.5 m

1.8 m

2.3 m

B

A

w


length of rope, v

(i) Calculatethesize,w,ofangleCAB.


(ii) Calculatethelengthoftherope,v,holdingtheswingseat.


17




MĀ TE KAIMĀKA

ANAKE

(b) Kuairitētahitarawhitiporohitamengāwaeaerereanairoto.

KoOtepūoteporohita.

Koki OAB = 50°

Koki ODB = 22°

A B D

O

G

E

JF

z

50° 22°


Tātaihiaterahi,z,otekokiEJO.



18

(b) Acircularhoopishungwithwiresrunningthroughit.

Oisthecentreofthecircularhoop.

AngleOAB=50°

AngleODB=22°

A B D

O

G

E

JF

z

50° 22°


Calculatethesize,z,ofangleEJO.


19




MĀ TE KAIMĀKA

ANAKE

(c) Heangataha-toruaABCE,ā,ihangaiaeaikingākōreroiraro:

• He2.5mteroaotepouABirungaakeitewhenua,ā,kaurukitewhenuaite85°.

• He3mteroaongāpouCBmeEBirungaakeitewhenua.

• Heōritetewehengaongāpouetoruite120°hurinoaitepokapūX(keirarotonuitepūwāhiB).

E

E

C

C85°

120°

120°

d

d

B

A

A

X

X

Diagrams are NOT to scale

Side view Top view

3 m

3 m

2.5 m

Tātaihiaad,tetawhitiiwaengaiaCmeEitepapa.



Tirohangakōtaha Tirohangamaiirunga

20

(c) ABCEisathree-sidedframeanditisbuiltasstatedbelow:

• PoleABis2.5mlongabovethegroundanditentersthegroundat85°.

• PolesCBandEBareboth3mlongabovetheground.

• Thethreepolesareequallyspacedoutat120°aboutthecentralpointX(whichisdirectlybelowpointB).

E

E

C

C85°

120°

120°

d

d

B

A

A

X

X

Diagrams are NOT to scale

Side view Top view

3 m

3 m

2.5 m

Calculated,thedistancebetweenCandEatgroundlevel.


21




MĀ TE KAIMĀKA

ANAKETAU TŪMAHI

He whārangi anō ki te hiahiatia.Tuhia te (ngā) tau tūmahi mēnā e tika ana.

22

23



QUESTION NUMBER

Extra paper if required.Write the question number(s) if applicable.

Level 1 Mathematics and Statistics, 201891031 Apply geometric reasoning in solving problems

9.30 a.m. Tuesday 20 November 2018 Credits: Four

Achievement Achievement with Merit Achievement with ExcellenceApply geometric reasoning in solving problems.

Apply geometric reasoning, using relational thinking, in solving problems.

Apply geometric reasoning, using extended abstract thinking, in solving problems.

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 attempt ALL the questions in this booklet.

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 – 23 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.

91

03

1M

English translation of the wording on the front cover

Documents

Te Pāngarau me te Tauanga, Kaupae 1, 2018 · New council rules state that a slide must have: • an angle (UTX) of less than 60° with the water AND • a ladder height of less than