chievement Stndrd 91031 pply geometric resoning in solving problems Copy correctly Up to 3% of workbook Copying or scnning from ES workbooks is subject to the NZ Copyright ct which limits copying to 3% of this workbook. Mthemtics nd Sttistics Eternlly ssessed (4 credits) Essentil em notes ngle properties of lines nd polygons When using geometric resoning to solve problems, the geometricl properties of ngles, lines nd polygons need to be known. Some importnt rules follow (with their bbrevited form shown in brckets). ngles lines nd points Sum of ngles on stright line is 180 Sum of ngles t point is 360 Verticlly opposite ngles re equl ( s on line 180 ) ( s t point 360 ) (vert opp s ) b b d c b + b 180 + b + c + d 360 b In reverse to prove tht line is stright, show tht djcent ngles on the line dd to 180. ngles nd prllel lines When line crosses pir of prllel lines, the corresponding ngles formed re equl When line crosses pir of prllel lines, the lternte ngles formed re equl When line crosses pir of prllel lines, the co-interior ngles formed dd to 180 (corr s // lines ) (lt s // lines ) (co-int s // lines supp) b b b b b + b 180 In reverse, to prove tht lines re prllel, show tht corresponding ngles re equl, lternte ngles re equl, or co-interior ngles re supplementry (dd to 180 ). Polygons polygon is mny-sided closed figure whose sides re No of sides Nme of polygon stright-line segments. 5 pentgon Some polygons hve specil nmes ccording to the number 6 hegon of sides they hve (see tble). 8 octgon regulr polygon hs ll sides of equl length nd ll ngles 9 nongon of equl size. 10 decgon 12 dodecgon
56 chievement Stndrd 91031 (Mthemtics nd Sttistics ) Tringles tringle is 3-sided polygon. In ny tringle: the longest side is lwys opposite the lrgest ngle; nd the sum of the lengths of ny two sides must be greter thn the length of the third side. The sum of the interior ngles of tringle is 180 ( sum 180 ) The size of the eterior ngle of tringle is equl to the sum of the sizes of the two interior opposite ngles. (et ) n equilterl tringle is regulr tringle with ll sides equl nd with ech interior ngle of size 60 (int eq 60 ) n isosceles tringle hs one is of symmetry. 2 sides re equl length nd 2 ngles re equl size. (bse s isos ) sclene tringle hs ll ngles nd sides of different size nd length. b c b e 60 60 60 b b c + b + c 180 e + b b Qudrilterls qudrilterl is 4-sided polygon. The sum of the interior ngles of polygon is 360 ( sum qud 360 ). Qudrilterls with specil properties (such s right ngles, prllel lines, symmetry, etc.) include the trpezium, the prllelogrm, the rhombus, the kite, the rectngle nd the squre. The properties of these specil qudrilterls need to be known. Emple Q. Find with resons the size of the ngle mrked in the digrm, where C is prllelogrm length length E 110. C // (property of prllelogrm) EC 110 (corr s // lines ) E length C length (property of prllelogrm) EC is isosceles (E C since E ) 180 110 2 35 (bse s isos ) 110 C ngles of polygons In generl, the sum of the interior ngles of polygon is: Sum (n 2) 180 The eterior ngles of ny polygon combine to mke full turn. Emple where n number of sides Sum of eterior ngles of polygon 360 The sum of the interior ngles of n octgon is 45 (8 2) 180 1080 [using (n 2) 180 ] Thus, ech interior ngle of regulr octgon is 135 1 080 8 135 [ech interior ngle size is equl] Ech eterior ngle of regulr octgon is 360 8 45 [or 180 135 45 ] ES Publictions (NZ) Ltd ISN 978-0-947504-34-2 Copying or scnning from ES workbooks is limited to 3% under the NZ Copyright ct.
pply geometric resoning in solving problems 57 NCE prctice Questions ngle properties of lines nd polygons 1. clothes drying rck hs two horizontl levels on which the clothes cn be hung s shown by lines E nd HI on the digrm longside. E is prllel to HI nd prllel to the ground JN. The rck is symmetricl round the line CL. J H K F C E 24 G I L M N igrm is NOT to scle Yer 2015 ns. p. 126 C CF ngle KCL 24. Find the size of ngle CF. Justify your nswer with cler geometric resoning. b. Find the size of ngle GC. Justify your nswer with cler geometric resoning. 2. elow ucklnd s Sky Tower is cr prk mde of rmps. The rmps re t 2 ngle. Yer 2016 ns. p. 126 There re verticl pillrs regulrly plced long the rmps for strength. ll pillrs re prllel to ech other. LM is horizontl.. Clculte the size of ngle in the digrm longside. Justify your nswer with cler geometric resoning. y N H pillr rmp rmp 2 L pillr igrm is NOT to scle M ES Publictions (NZ) Ltd ISN 978-0-947504-34-2 Copying or scnning from ES workbooks is limited to 3% under the NZ Copyright ct.
58 chievement Stndrd 91031 (Mthemtics nd Sttistics ) b. Clculte the size of ngle y in the digrm bove. Justify your nswer with cler geometric resoning. c. Prt of the rmp hd etr scffolding dded for support, s shown in the digrm longside. The lines SK nd YT re prllel. ngle WSY is 174. The lines WS nd PY re both horizontl. Clculte the size of ngle in the digrm. Justify your nswer with cler geometric resoning. W 174 P S rmp scffold K Y rmp igrm is NOT to scle scffold 2 T d. From the side, the crprk looks like the digrm below. igrm is NOT to scle E 176 J C F G N H 2 2 I K L M ngle EGJ is 176. IK nd LM re horizontl. Prove tht the lines nd C re prllel. Justify your nswer with cler geometric resoning. ES Publictions (NZ) Ltd ISN 978-0-947504-34-2 Copying or scnning from ES workbooks is limited to 3% under the NZ Copyright ct.
pply geometric resoning in solving problems 59 3. hmed drws pentgon nd drws the eterior ngles (shown dotted). 57 122 Yer 2014 ns. p. 126 hmed works out the size of ngle c s shown below. c 360 57 147 140 122 c 106 c. Give the geometric reson for his clcultion. 147 140 b. Eplin the mening of the negtive mesurement of the ngle c, nd stte when this will occur. 4. In this pentgon, ngle EGR + ngle GR 180.. How does this prove tht lines GE nd R re prllel? R Yer 2014 ns. p. 126 P w b. Prove tht the ngle PE (lso lbelled w bove) is given by the eqution: ngle PE ngle PR + ngle PEG. Eplin your geometric resoning clerly. G E 5.. Find the size of ngle e. Eplin your method clerly, nd give geometric resons for ech step. igrm is NOT to scle r e Yer 2013 ns. p. 126 n g m 70 30 ES Publictions (NZ) Ltd ISN 978-0-947504-34-2 Copying or scnning from ES workbooks is limited to 3% under the NZ Copyright ct.
60 chievement Stndrd 91031 (Mthemtics nd Sttistics ) b. Find n epression for the size of ngle r, in terms of ngles nd y. Eplin your method clerly, nd give geometric resons for ech step. igrm is NOT to scle r e n g m y Yer 2014 ns. p.127 6. hmed knows tht ny qudrilterl will tessellte. This mens tht it cn be used to cover surfce leving no gps. n emple is given here: In this digrm, si copies of the sme shpe hve been tessellted. The si shpes hve been coloured differently so tht you cn see how the tesselltion tkes plce. Eplin, using geometric resoning, why it is true tht ny qudrilterl will tessellte. You my wish to refer to the lbelled digrm longside or to the tesselltion bove. Q U ES Publictions (NZ) Ltd ISN 978-0-947504-34-2 Copying or scnning from ES workbooks is limited to 3% under the NZ Copyright ct.
pply geometric resoning in solving problems 61 7. CE is regulr pentgon with centre O.. Find the vlue of nd eplin your nswer. Yer 2015 ns. p. 127 O E C y b. Find the vlue of y. Justify your nswer with cler geometric resoning. 8. simplified digrm of the position of the legs of ucklnd s Sky Tower is shown below s regulr octgon. Point O is t the centre of the octgon. Yer 2016 ns. p. 127 T O K y Y S Show tht ngle y is hlf the size of ngle. Justify your nswer with cler geometric resoning. ES Publictions (NZ) Ltd ISN 978-0-947504-34-2 Copying or scnning from ES workbooks is limited to 3% under the NZ Copyright ct.
62 chievement Stndrd 91031 (Mthemtics nd Sttistics ) Essentil em notes Similrity oubling ll sides of polygon cretes similr polygon, with the sme ngle sizes. If ny two polygons re similr, then: ngles in corresponding positions re equl corresponding sides re in proportion. If two shpes re similr, then one shpe cn be considered to be n enlrgement of the other. The scle fctor k for this enlrgement cn be found by dividing n enlrged length by the corresponding originl length. Ech side length in the enlrged shpe is k times the length of the corresponding side in the originl shpe. If the side lengths of shpe increse by fctor of k, then the re increses by fctor of k 2. In solids, if sides re incresed by fctor of k, then the volume increses by fctor of k 3. Emple Two similr qudrilterls re shown. y 12 cm 16.25 cm 15 cm 17 cm Q. Find the length of the side mrked.. Scle fctor 15 or 1.25 12 [dividing corresponding side lengths] 17 1.25 [multiplying corresponding side length by scle fctor] 21.25 cm ES Publictions (NZ) Ltd ISN 978-0-947504-34-2 Copying or scnning from ES workbooks is limited to 3% under the NZ Copyright ct.
pply geometric resoning in solving problems 63 Q. Find the length of the side mrked y.. y 16.25 1.25 [dividing corresponding side length by scle fctor] y 13 cm Q. If the re of the smller qudrilterl is 174 cm 2, find the re of the lrger qudrilterl.. re increses by fctor of 1.25 2 [squre of scle fctor] re 174 1.25 2 271.875 cm 2 Similr tringles If two tringles hve the sme ngle sizes then they re similr nd their sides will be in proportion. In reverse, if corresponding sides re in proportion, then corresponding ngle sizes will be the sme. Emple Q. Find in the digrm longside.. In the digrm, E C (corr s // lines ) nd E C (corr s // lines ) tringles E nd C re similr [sme ngles] E C E C [since sides re in proportion] 12 8 14 [since C 8 + 6 14] 14 96 [cross-multiplying] 6.9 (1 dp) 8 E 6 12 C Similr figures my overlp ech other, so identify them crefully. Emple Q. Find the length of the sides mrked nd y.. E is similr to C since E C (corr s // lines ) E C (corr s // lines ) E C, so 10 8 4 E 14 y C C C E 4 24 10 10 96 9.6 To find y: E C E [corresponding sides re in proportion] [substituting side lengths] [cross-multiplying] [corresponding sides re in proportion] 8 9.6 4 [substituting side lengths] 4 8 9.6 [cross-multiplying] 19.2 [dividing by 4] So y 19.2 8 [ y + 8] y 11.2 ES Publictions (NZ) Ltd ISN 978-0-947504-34-2 Copying or scnning from ES workbooks is limited to 3% under the NZ Copyright ct.
64 chievement Stndrd 91031 (Mthemtics nd Sttistics ) NCE prctice Questions Similrity Yer 2014 ns. p. 127 1. In the digrm longside,, C,, nd E lie on the circumference of circle, nd is NOT the centre of the circle.. Use geometric resoning to eplin why tringles C nd E re similr. C E b. Prove tht C E. Justify your nswer with cler geometric resoning. Yer 2012 ns. p. 127 2. Find the length of the line MN. You must show your working nd give resons. 4 cm J K 6 cm L igrm is NOT to scle M 12 cm N ES Publictions (NZ) Ltd ISN 978-0-947504-34-2 Copying or scnning from ES workbooks is limited to 3% under the NZ Copyright ct.
nswers nd eplntions chievement Stndrd 91027 (Mthemtics nd Sttistics 1.2) pply lgebric procedures in solving problems 1.2 lgebric terms nd epressions p. 3 1.. 2m 4m 5 2 m +3 4m using common + 3 5 15 denomintor 10 m +12m 15 22 m (M) 15 b. 3 4 + 2 3 3 3 + 2 4 12 dding frctions 9 + 8 12 17 12 () 2.. (2 3 ) 2 2 2 3 2 inde lws 4 6 () b. 2 8 2 () dividing numertor nd denomintor by 4 4 9 8 3.. 4 2 4 2 9 n 2 4 inde lws 9 n 4 equting powers n 5 () solving b. 4 2 b 4 b 3 4 ( 2 4 ) (b b 3 ) grouping terms with sme bse dding powers of terms 4 6 b 4 () with sme bse 4. 3n 2 12n + 6 + n epnding 3n 2 11n + 6 simplifying 2n 2 12n + n 6 + n 2 + 3 epnding 3n 2 11n 3 simplifying Compring nd it cn be seen tht cn be written s (3n 2 11n 3) + 9 since 3 + 9 6 So + 9 (lterntively, subtrcting the epressions for nd gives 9, so + 9) (E) 5. The lowest common multiple of the denomintors of the frctions is 5, so 2 3 + + 5 5 2 (3 + ) + 5 5 equivlent frctions with common denomintor 10 5 2 3 + 5 simplifying 2 + 3 + 10 5 () dding numertors 6.. 2( 1) 3( + 2) 2 2 3 6 epnding 8 () simplifying b. ( + 4)( 2) ( 2) + 4( 2) epnding 2 2 + 4 8 epnding 2 + 2 8 () simplifying c. 2(3 5) 4(3 5) 6 2 10 12 + 20 epnding brckets 6 2 22 + 20 () d. ( + 2)(4 5) (4 5) + 2(4 5) 4 2 5 + 8 10 4 2 + 3 10 () simplifying e. ( + 6) 2 ( + 6)( + 6) 2 + 6 + 6 + 36 epnding 2 + 12 + 36 () simplifying f. (2 7) 2 (2 7)(2 7) 2(2 7) 7(2 7) 4 2 14 14 + 49 epnding 4 2 28 + 49 () simplifying 7.. 2 8 + 12 ( 6)( 2) () fctors of 12 tht dd to 8 re 6, 2 b. Require fctor pir of 60 tht dds to 7: these re 12 nd 5, so 2 7 60 ( 12)( + 5) () highest common fctor of b 2 nd 2 b is b c. Multiplying the coefficient of 2 by the constnt gives 3 6 18 Fctors of 18 tht dd to 11 re 9 nd 2 3 2 11 + 6 3 2 9 2 + 6 splitting middle term 3( 3) 2( 3) fctorising in pirs (3 2)( 3) ( 3) is common fctor lterntively, use tril nd improvement methods (product of first terms in brckets is 3 2 nd product of second terms in brckets is 6 djust possible pirs to get the correct middle term of 11). () d. 4 2 1 (2) 2 1 2 epressing s difference of squres (2 + 1)(2 1) () 2 b 2 ( + b)( b) 8.. re of rectngle bse height Epressing the formul for the re in fctorised form: 2 2 ( + 1)( 2)
112 nswers nd eplntions nswers 9.. So if one side hs length ( + 1) metres, then Length of the other side ( 2) metres () b. Since side lengths re positive b. + 1 > 0 nd 2 > 0 So > 1 nd > 2 So must be more thn 2 metres (M) m 2 m m(m 1) m 2 (M) fctorising 1 (m + 1)(m 1) m (E) (m 1) m + 1 cncelling (m 1) 2 3 28 ( 7)( + 4) + 4 ( + 4) 7 (M) fctorising dividing top nd bottom by ( + 4) 10.. 10 2 y 5 2 y + 8y 2 5 2 y + 8y 2 () dding like terms b. i. 5 2 y + 8y 2 y(5 + 8y) () ii. 5 2 y + y 2 7 2 y y 2 2 2 y simplifying by dding like terms y(y 2) (M) y is the highest common fctor 3b 2 4 3 b +b 2 4b 2 4 3 b c. dding like terms 4b 2 4b 2 4b(b 2 ) fctorising 4b 2 (b 2 ) (/M) b cncelling common fctors in numertor nd denomintor d. To subtrct terms with frctions, require common denomintor 2 3 8 ( 3) chnging 2 to equivlent 4 4 4 frction (over 4) 8 ( 3) 4 subtrcting numertors 8 + 3 4 epnding brcket 7 + 3 4 (M) simplifying 11.. Chnging the order of line 1 s shown (moving middle number to n outer position) gives different nswer in line 4 Line 1: 3 5 7 1 Line 2: 3 + 5 8 5 + 7 12 7 + 1 8 Line 3: 8 + 12 20 12 + 8 20 Line 4: 20 + 20 40 ut chnging the order of line 1 s shown below (swpping the middle two numbers) gives the sme nswer. Line 1: 1 5 3 7 Line 2: 1 + 5 6 5 + 3 8 3 + 7 10 Line 3: 6 + 8 14 8 + 10 18 Line 4: 14 + 18 32 So, if the middle two numbers swp plces nd/or the outer two numbers swp plces, then the nswer in line 4 is the sme. (M) b. Let the numbers in the first line be, b, c, d Line 1: b c d Line 2: + b b + c c + d Line 3: + 2b + c b + 2c + d Line 4: + 3b + 3c + d So the finl totl in line 4 is ( + d) + 3(b + c). So, it cn be seen tht the position of the outer two numbers ( nd d) in Line 1 cn be swpped, nd/or the position of the inner two numbers (b nd c) in Line 1 cn be swpped without ffecting the totl in line 4. (E) c. From prt ii, if the numbers written in the tringle re, b, c, d, then the number t the bottom of the tringle is ( + d) + 3(b + c) This totl will be divisible by 3 since 3(b + c) is if ( + d) is divisible by 3 divisible by 3 So for the number t the bottom of the tringle to be divisible by 3, it is necessry tht the sum of the first nd lst numbers in Line 1 is divisible by 3. Let the consecutive numbers Json writes in Line 1 be n, (n + 1), (n + 2) nd (n + 3) The sum of the first nd lst numbers in Line 1 is n + n + 3 2n + 3 2n + 3 will be divisible by 3 if 2n is divisible by 3 2n is divisible by 3 if n is divisible by 3 So if the number t the bottom of the tringle is divisible by 3 then the first number Json writes must be divisible by 3. (E) 12.. T n 2 n + n substituting m n in formul n 2 So T n n nd is therefore the product of two numbers which re not one nd itself. T is therefore composite. (M) b. T (n 4)(n + 3) fctorising Since n > 5, it follows tht (n 4) > 1 subtrcting 4 from both sides (n + 3) > 8 dding 3 to both sides So T cn be written s the product of two numbers which re not one nd itself. So T is composite. (E) 1.2 Liner equtions nd inequtions 1.. 7 3 1 3 6 subtrcting 7 6 3 dividing by 3 2 () b. 6 + 27 15 epnding 6 12 subtrcting 15 2 () dividing by 6 c. 3 + 2 7 6 rerrnging 5 1 1 5 () d. 3 > 6 + 8 doubling both sides 5 3 > 8 subtrcting 6 5 > 11 dding 3 11 < (M) 5 dividing by 5 (reverse inequlity) 2.. 7 + 1 36 multiplying both sides by 4 7 35 subtrcting 1 5 () dividing by 7 b. 5(2m 3) 6(m 4) 10m 15 6m 24 epnding 4m 15 24 subtrcting 6m 4m 9 dding 15 p. 8 ES Publictions (NZ) Ltd ISN 978-0-947504-34-2 Copying or scnning from ES workbooks is limited to 3% under the NZ Copyright ct.