9Geometry in the plne nd proof Ojectives To consider necessry nd sufficient conditions for two lines to e prllel. To determine the ngle sum of polygon. To define congruence of two figures. To determine when two tringles re congruent. To write geometric proofs. To use Pythgors theorem nd its converse. To define similrity of two figures. To determine when two tringles re similr. To determine nd pply similrity fctors for res nd volumes. To use vectors to prove geometric results. To investigte properties of the golden rtio. There re three min resons for the study of geometry t school. UNORRT The first reson is tht the properties of figures in two nd three dimensions re helpful in other res of mthemtics. The second reson is tht the suject provides good setting to show how lrge ody of results my e deduced from smll numer of ssumptions. The third reson is tht it gives you, the student, the opportunity to prctise writing coherent, logicl mthemticl rguments. SMPL PGS In this chpter nd the next, we use some of the proof techniques introduced in the previous chpter. Review of geometry from Yers 9 nd 10 is included, ut in such wy tht you cn see the uilding of the results. hpter 9 Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
268 hpter 9: Geometry in the plne nd proof 9 Points, lines nd ngles In this section we do not pretend to e fully rigorous, ut im to mke you wre tht ssumptions re eing mde nd tht we se the proofs of the results on these ssumptions. The ssumptions do seem ovious to us, ut there re wys of mking the study of geometry even more rigorous. However, whtever we do, we will need to ccept set of results s our strting point. Points, lines nd plnes We egin with few sic concepts. No forml definitions re given. Point Line Plne In geometry, point is used to indicte position. In the physicl world, we my illustrte the ide of line s tightly stretched wire or fold in piece of pper. line hs no width nd is infinite in length. plne hs no thickness nd it extends infinitely in ll directions. We mke the following ssumptions out points nd lines: Given point nd line, the point my or my not lie on the line. Two distinct points re contined in exctly one line. Two distinct lines do not hve more thn one point in common. ngles ry is portion of line consisting of point O nd ll the points on one side of O. n ngle is the figure formed y two distinct rys which hve common endpoint O. The common endpoint is clled the vertex of the ngle. If the two rys re prt of one stright line, the ngle is clled stright ngle nd mesures 180. right ngle is n ngle of 90. n cute ngle is n ngle which is less thn 90. n otuse ngle is n ngle which is greter thn 90 nd less thn 180. UNORRT Supplementry ngles re two ngles whose sum is 180. omplementry ngles re two ngles whose sum is 90. Nming ngles The convention for nming n ngle is to fully descrie the rys of the ngle nd the endpoint where the rys meet. SMPL PGS The mrked ngle is denoted y. When there is no chnce of miguity, it cn e written s. Sometimes n ngle cn simply e numered s shown, nd in proof we refer to the ngle s 1. Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400 O O 1
9 Points, lines nd ngles 269 The importnt thing is tht the writing of your rgument must e cler nd unmiguous. With complicted digrms, the nottion is sfest. Theorem If two stright lines intersect, then the opposite ngles re equl in pirs. Such ngles re sid to e verticlly opposite. Proof using ngle nmes O nd O re supplementry. Tht is, O + O = 180. lso, O nd O re supplementry. Tht is, O + O = 180. Hence O = O. The proof cn lso e presented with the lelling technique. Proof using numer lels 1 + 2 = 180 (supplementry ngles) 2 + 3 = 180 (supplementry ngles) xmple 1 1 = 3 Find the vlues of x nd y in the digrm. Solution O 1 4 2 3 x (x + 5) y UNORRT x + (x + 5) = 90 2x = 85 x = 42.5 y + (x + 5) = 180 (complementry ngles) (supplementry ngles) SMPL PGS y + 47.5 = 180 y = 132.5 Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
270 hpter 9: Geometry in the plne nd proof Prllel lines Given two distinct lines l 1 nd l 2 in the plne, either the lines intersect in single point or the lines hve no point in common. In the ltter cse, the lines re sid to e prllel. We cn write this s l 1 l 2. Here is nother importnt ssumption. Plyfir s xiom Given ny point P not on line l, there is only one line through P prllel to l. From this we hve the following results for three distinct lines l 1, l 2 nd l 3 in the plne: If l 1 l 2 nd l 2 l 3, then l 1 l 3. If l 1 l 2 nd l 3 intersects l 1, then l 3 lso intersects l 2. We prove the first of these nd leve the other s n exercise. The proof is y contrdiction. Proof Let l 1, l 2 nd l 3 e three distinct lines in the plne such tht l 1 l 2 nd l 2 l 3. Now suppose tht l 1 is not prllel to l 3. Then l 1 nd l 3 meet t point P. ut y Plyfir s xiom, there is only one line prllel to l 2 pssing through P. Therefore l 1 = l 3. ut this gives contrdiction, s l 1 nd l 3 re distinct y ssumption. orresponding, lternte nd co-interior ngles The following types of pirs of ngles ply n importnt role in considering prllel lines. In the digrm, the lines l 1 nd l 2 re crossed y trnsversl l 3. orresponding ngles: ngles 1 nd 5 ngles 2 nd 6 ngles 3 nd 7 ngles 4 nd 8 lternte ngles: ngles 3 nd 5 ngles 4 nd 6 o-interior ngles: ngles 3 nd 6 ngles 4 nd 5 l 2 l 1 5 6 8 7 UNORRT The following result is esy to prove, nd you should complete it s n exercise. Theorem When two lines re crossed y trnsversl, ny one of the following three conditions implies the other two: SMPL PGS pir of lternte ngles re equl pir of corresponding ngles re equl pir of co-interior ngles re supplementry. 4 1 3 2 l 3 Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9 Points, lines nd ngles 271 The next result is importnt s it gives us the ility to estlish properties of the ngles ssocited with prllel lines crossed y trnsversl, nd it lso gives us n esily pplied method for proving tht two lines re prllel. Theorem If two prllel lines re crossed y trnsversl, then lternte ngles re equl. onversely, if two lines crossed y trnsversl form n equl pir of lternte ngles, then the two lines re prllel. xmple 2 Find the vlues of the pronumerls. Note: The rrows indicte tht the two lines re prllel. Solution = 65 (corresponding) d = 65 (lternte with ) = 115 (co-interior with d) e = 115 (corresponding with ) c = 115 (verticlly opposite e) xmple 3 For shown in the digrm, the line XZ is drwn through vertex prllel to. Use this construction to prove tht the sum of the interior ngles of tringle is stright ngle (180 ). X e 65 d c UNORRT Solution = X = Z (lternte ngles) (lternte ngles) SMPL PGS X + Z + is stright ngle. Therefore + + = 180. Z Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
272 hpter 9: Geometry in the plne nd proof 9 Section summry Pirs of ngles complementry ( nd ) supplementry (c nd d) verticlly opposite (e nd f ) lternte (c nd e) corresponding (c nd f ) co-interior (d nd e) c d Prllel lines If two prllel lines re crossed y trnsversl, then: lternte ngles re equl corresponding ngles re equl co-interior ngles re supplementry. If two lines crossed y trnsversl form n equl pir of lternte ngles, then the two lines re prllel. xercise 9 1 onsider the digrm shown. Stte whether ech of the following ngles is cute, otuse, right or stright: i ii HF iii iv F Stte which ngle is: i corresponding to ii lternte to F iii verticlly opposite F iv co-interior to F c Stte which ngles re: UNORRT xmple 1, 2 2 i complementry to ii supplementry to lculte the vlues of the unknowns for ech of the following. Give resons. 115 f e H G F x (x + 10) y SMPL PGS SF Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9 9 Points, lines nd ngles 273 xmple 3 3 c e F 60 K 70 e d α H J M β c f L 87 Side of is extended to point X nd line is drwn prllel to side. Prove tht the sum of two interior ngles of tringle is equl to the opposite exterior ngle. Hint: Using the digrm, this mens showing tht + = X. G d f β α J α 120 F θ 108 4 Recll tht prllelogrm is qudrilterl whose opposite sides re prllel. prllelogrm is shown on the right. Let = α. Find the sizes of nd in terms of α. Hence find the size of in terms of α. UNORRT 5 Prove the converse of the result in Question 4. Tht is, prove tht if the opposite ngles of qudrilterl re equl, then the qudrilterl is prllelogrm. 6 Prove tht is perpendiculr to. β F K α X α α β β SMPL PGS 7 The lines PQ nd RS re prllel. trnsversl meets PQ t X nd RS t Y. Lines X nd Y re isectors of the ngles PXY nd XYS. Prove tht X is prllel to Y. SF F Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
274 hpter 9: Geometry in the plne nd proof 9 8 For the digrm on the right, show tht α + β = 90. α X β 9 For ech of the following, use construction line to find the ngle mrked θ: 42 θ 65 9 Tringles nd polygons We first define polygons. θ P Q 35 F O G line segment is portion of line consisting of two distinct points nd nd ll the points etween them. If distinct points 1, 2,..., n in the plne re connected in order y the line segments 1 2, 2 3,..., n 1, then the figure formed is polygon. The points 1, 2,..., n re the vertices of the polygon, nd the line segments 1 2, 2 3,..., n 1 re its sides. Types of polygons simple polygon is polygon such tht no two sides hve point in common except vertex. convex polygon is polygon tht contins ech line segment connecting ny pir of points on its oundry. UNORRT For exmple, the left-hnd figure is convex, while the right-hnd figure is not. convex polygon non-convex polygon Note: In this chpter we will lwys ssume tht the polygons eing considered re convex. SMPL PGS regulr polygon is polygon in which ll the ngles re equl nd ll the sides re equl. Nmes of polygons tringle (3 sides) hexgon (6 sides) nongon (9 sides) qudrilterl (4 sides) pentgon (5 sides) heptgon (7 sides) octgon (8 sides) decgon (10 sides) dodecgon (12 sides) F Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
Tringles 9 Tringles nd polygons 275 tringle is figure formed y three line segments determined y set of three points not on one line. If the three points re, nd, then the figure is clled tringle nd commonly written. The points, nd re clled the vertices of the tringle. Tringle inequlity n importnt property of tringle is tht ny side is shorter thn the sum of the other two. In : < + c, < c + nd c < +. Note: For lelled s shown, we hve c < < if nd only if < <. The following two results hve een proved in xmple 3 nd in Question 3 of xercise 9. ngles of tringle The sum of the three interior ngles of tringle is 180. The sum of two interior ngles of tringle is equl to the opposite exterior ngle. lssifiction of tringles quilterl tringle Isosceles tringle Sclene tringle Importnt lines in tringle Medin ltitude xmple 4 tringle in which ll three sides re equl tringle in which two sides re equl tringle in which ll three sides re unequl medin of tringle is line segment from vertex to the midpoint of the opposite side. n ltitude of tringle is line segment from vertex to the opposite side (possily extended) which forms right ngle where it meets the opposite side. UNORRT The sides of tringle re 6 x, 4x + 1 nd 2x + 3. Find the vlue of x for which the tringle is isosceles, nd show tht if it is isosceles, then it is equilterl. Solution 6 x = 4x + 1 5x = 5 x = 1 xplntion We wnt to show tht if ny two side lengths re equl, then the third length is the sme. SMPL PGS When x = 1, we hve 6 x = 5, 4x + 1 = 5 nd 2x + 3 = 5. Hence the tringle is equilterl with ech side of length 5 units. It is enough to show tht the three lines y = 6 x, y = 4x + 1 nd y = 2x + 3 intersect in common point. c Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
276 hpter 9: Geometry in the plne nd proof xmple 5 Find the vlues of,, c nd d, giving resons. Solution = 50 = 60 (verticlly oposite ngles) (supplementry ngles) c = 180 (50 + 60) = 70 d = 60 (ngle sum of tringle) (corresponding ngles FG) ngle sum of polygon If polygon hs n sides, then we cn drw n 3 digonls from vertex. In this wy, we cn divide the polygon into n 2 tringles, ech with n ngle sum of 180. We hve drwn hexgon to illustrte this, ut we could hve used ny polygon. ngle sum of polygon 50 120 c d F The sum of the interior ngles of n n-sided polygon is (n 2)180. (n 2) ch interior ngle of regulr n-sided polygon hs size 180. n xmple 6 regulr dodecgon is shown to the right. UNORRT Find the sum of the interior ngles of dodecgon. Find the size of ech interior ngle of regulr dodecgon. Solution SMPL PGS The ngle sum of polygon with n sides is (n 2)180. Therefore the ngle sum of dodecgon is 1800. ch of the interior ngles is 1800 12 = 150. G Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9 9 Tringles nd polygons 277 xmple 4 xmple 5 Section summry Polygons The sum of the interior ngles of n n-sided polygon is (n 2)180. In regulr polygon, ll the ngles re equl nd ll the sides re equl. (n 2) ch interior ngle of regulr n-sided polygon hs size 180. n Tringles n equilterl tringle is tringle in which ll three sides re equl. n isosceles tringle is tringle in which two sides re equl. sclene tringle is tringle in which ll three sides re unequl. The sum of the three interior ngles of tringle is 180. The sum of two interior ngles of tringle is equl to the opposite exterior ngle. In : < + c, < c + nd c < + c < < if nd only if < < xercise 9 1 Is it possile for tringle to hve sides of lengths: c 12 cm, 9 cm, 20 cm 5 cm, 5 cm, 5 cm 2 escrie ech of the tringles in Question 1. d c 24 cm, 24 cm, 40 cm 12 cm, 9 cm, 2 cm? 3 If tringle hs sides 10 cm nd 20 cm, wht cn e sid out the third side? 4 The sides of tringle re 2n 1, n + 5 nd 3n 8. 5 6 Find the vlue(s) of n for which the tringle is isosceles. Is there vlue of n which mkes the tringle equilterl? UNORRT The sides of tringle re 2n 1, n + 7 nd 3n 9. Prove tht if the tringle is isosceles, then it is equilterl. lculte the vlue of the unknowns for ech of the following. Give resons. F α 60 α θ β γ γ β 70 θ 64 60 40 Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400 SMPL PGS SF
278 hpter 9: Geometry in the plne nd proof 9 xmple 6 7 c 65 γ d X α 65 β α α Z e g 120 α β r q 52 α m n p 68 f 125 α Y 45 g c e Find the interior-ngle sum nd the size of ech ngle of regulr polygon with: 6 sides 12 sides 8 In the decgon shown on the right, ech side hs een extended to form n exterior ngle. xplin why the sum of the interior ngles plus the sum of the exterior ngles is 1800. Hence find the sum of the decgon s 10 exterior ngles. 9 Prove tht the sum of the exterior ngles of ny polygon is 360. c f 20 sides UNORRT 10 If the sum of the interior ngles of polygon is four times the sum of the exterior ngles, how mny sides does the polygon hve? 11 ssume tht the sum of the interior ngles of polygon is k times the sum of the exterior ngles (where k N). Prove tht the polygon hs 2(k + 1) sides. SMPL PGS SF F Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9 ongruence nd proofs 279 9 ongruence nd proofs Two plne figures re clled congruent if one figure cn e moved on top of the other figure, y sequence of trnsltions, rottions nd reflections, so tht they coincide exctly. ongruent figures hve exctly the sme shpe nd size. For exmple, the two figures shown re congruent. We cn write: pentgon pentgon FGHI J When two figures re congruent, we cn find trnsformtion tht pirs up every prt of one figure with the corresponding prt of the other, so tht: pired ngles hve the sme size pired line segments hve the sme length pired regions hve the sme re. ongruent tringles We used the four stndrd congruence tests for tringles in hpter 3. The SSS congruence test If the three sides of one tringle re respectively equl to the three sides of nother tringle, then the two tringles re congruent. The SS congruence test If two sides nd the included ngle of one tringle re respectively equl to two sides nd the included ngle of nother tringle, then the two tringles re congruent. The S congruence test If two ngles nd one side of one tringle re respectively equl to two ngles nd the mtching side of nother tringle, then the two tringles re congruent. α P PQR R α P PQR UNORRT α β P Q α PQR SMPL PGS The RHS congruence test If the hypotenuse nd one side of one right-ngled tringle re respectively equl to the hypotenuse nd one side of nother right-ngled tringle, then the two tringles re congruent. R P PQR Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400 F G β J R Q Q R Q H I
280 hpter 9: Geometry in the plne nd proof lssifiction of qudrilterls Trpezium Prllelogrm Rhomus Rectngle Squre Proofs using congruence xmple 7 qudrilterl with t lest one pir of opposite sides prllel qudrilterl with oth pirs of opposite sides prllel prllelogrm with pir of djcent sides equl qudrilterl in which ll ngles re right ngles qudrilterl tht is oth rectngle nd rhomus Let nd XYZ e such tht = YXZ, = XY nd = XZ. If P nd Q re the midpoints of nd YZ respectively, prove tht P = XQ. Solution P From the given conditions, we hve XYZ (SS). Therefore P = XYQ nd = YZ. X Thus P = Y Q, s P nd Q re the midpoints of nd YZ respectively. Hence P XYQ (SS) nd so P = XQ. xmple 8 Prove tht, in prllelogrm, the digonls isect ech other. Y Prove tht if the digonls of qudrilterl isect ech other, then the qudrilterl is prllelogrm. Solution UNORRT Note tht opposite sides of prllelogrm re equl. (See Question 8 of xercise 9.) In tringles O nd O: O = O O = O O = O = O O Hence O = O nd O = O. (lternte ngles ) (lternte ngles ) (verticlly opposite) (opposite sides of prllelogrm re equl) (S) Q SMPL PGS Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400 Z O
9 ongruence nd proofs 281 O = O (digonls isect ech other) O = O O = O O = O O O O O (digonls isect ech other) (verticlly opposite) (verticlly opposite) (SS) (SS) Therefore O = O nd so, since lternte ngles re equl. Similrly, we hve. Hence is prllelogrm. xmple 9 Prove tht the tringle formed y joining the midpoints of the three sides of n isosceles tringle (with the midpoints s the vertices of the new tringle) is lso isosceles. Solution ssume is isosceles with = nd =. (See Question 3 of xercise 9.) Then we hve =, where nd re the midpoints of nd respectively. We lso hve F = F, where F is the midpoint of. Therefore F F (SS). Hence F = F nd so F is isosceles. Section summry ongruent figures hve exctly the sme shpe nd size. If tringle is congruent to tringle XYZ, this cn e written s XYZ. Two tringles re congruent provided ny one of the following four conditions holds: SSS the three sides of one tringle re equl to the three sides of the other tringle UNORRT SS two sides nd the included ngle of one tringle re equl to two sides nd the included ngle of the other tringle S two ngles nd one side of one tringle re equl to two ngles nd the mtching side of the other tringle SMPL PGS RHS the hypotenuse nd one side of right-ngled tringle re equl to the hypotenuse nd one side of nother right-ngled tringle. O F Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
282 hpter 9: Geometry in the plne nd proof 9 xercise 9 1 In ech prt, find pirs of congruent tringles. Stte the congruence tests used. c 2 cm 2 cm 38 40 6 cm 7 cm 45 8 cm 8 cm 5 cm 12 cm 5 cm 13 cm 2 Nme the congruent tringles nd stte the congruence test used: α α 2 cm 40 7 cm c d e α α 45 8 cm UNORRT xmple 7 3 f α SMPL PGS Prove tht if is isosceles with =, then =. α 4 cm 3 cm 4 Prove tht if is such tht =, then is isosceles. (This is the converse of Question 3.) SF Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9 9 ongruence nd proofs 283 xmple 8 5 For the qudrilterl shown, prove tht. 6 FG is regulr hexgon. Find the vlues of,, c nd d. Prove tht nd G. 7 is regulr pentgon. 8 Find the vlues of,, c, d, e nd f. Prove tht nd. Proofs involving prllelogrms Prove ech of the following: α β G e f c d d c In prllelogrm, opposite sides re equl nd opposite ngles re equl. If ech side of qudrilterl is equl to the opposite side, then the qudrilterl is prllelogrm. F F β α c If ech ngle of qudrilterl is equl to the opposite ngle, then the qudrilterl is prllelogrm. d If one side of qudrilterl is equl nd prllel to the opposite side, then the qudrilterl is prllelogrm. UNORRT 9 Let e prllelogrm nd let P nd Q e the midpoints of nd respectively. Prove tht PQ is prllelogrm. 10 Let PQRS e prllelogrm whose digonls meet t O. Let X, Y, Z nd W e the midpoints of PO, QO, RO nd SO respectively. Prove tht XYZW is prllelogrm. SMPL PGS 11 Proofs involving rhomuses Prove ech of the following: The digonls of rhomus isect ech other t right ngles. The digonls of rhomus isect the vertex ngles through which they pss. c If the digonls of qudrilterl isect ech other t right ngles, then the qudrilterl is rhomus. SF F Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
284 hpter 9: Geometry in the plne nd proof 9 xmple 9 12 Proofs involving rectngles Prove ech of the following: 13 The digonls of rectngle re equl nd isect ech other. prllelogrm with one right ngle is rectngle. c If the digonls of qudrilterl re equl nd isect ech other, then the qudrilterl is rectngle. is pentgon in which ll the sides re equl nd digonl is equl to digonl. Prove tht =. 14 is equilterl nd its sides re extended to points X, Y nd Z so tht Y, Z nd X re ll equl in length to the sides of. Prove tht XYZ is lso equilterl. 15 is qudrilterl in which = nd =. The digonl is extended to point K. Prove tht K = K. 16 Prove tht if the ngle of tringle is equl to the sum of the other two ngles, then the length of side is equl to twice the length of the line segment joining with the midpoint of. 17 Prove tht if NO is the se of isosceles tringle MNO nd if the perpendiculr from N to MO meets MO t, then ngle NO is equl to hlf of ngle NMO. 18 If medin of tringle is drwn, prove tht the perpendiculrs from the other vertices upon this medin re equl. (The medin my e extended.) 9 Pythgors theorem Pythgors theorem Let e tringle with side lengths, nd c. If is right ngle, then 2 + 2 = c 2 UNORRT Pythgors theorem cn e illustrted y the digrm shown here. The sum of the res of the two smller squres is equl to the re of the squre on the longest side (hypotenuse). There re mny different proofs of Pythgors theorem. One ws given t the strt of hpter 8. Here we give nother proof, due to Jmes. Grfield, the 20th President of the United Sttes. re = c 2 c c re = 2 re = 2 SMPL PGS F Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9 Pythgors theorem 285 Proof The proof is sed on the digrm shown on the right. re of trpezium XYZW = 1 2 ( + )( + ) re of WX + re of YZ + re of WZ Thus Hence 1 = 1 2 + 1 2 + 1 2 c2 = + 1 2 c2 2 ( + )( + ) = + 1 2 c2 2 + 2 + 2 = 2 + c 2 2 + 2 = c 2 onverse of Pythgors theorem Let e tringle with side lengths, nd c. If 2 + 2 = c 2, then is right ngle. Proof ssume hs side lengths =, = nd c = such tht 2 + 2 = c 2. onstruct second tringle XYZ with YZ = nd ZX = such tht XZY is right ngle. y Pythgors theorem, the length of the hypotenuse of XYZ is 2 + 2 = c 2 = c Therefore XYZ (SSS). Hence is right ngle. xmple 10 The digonl of soccer field is 130 m nd the length of the long side of the field is 100 m. Find the length of the short side, correct to the nerest centimetre. UNORRT Solution Let x e the length of the short side. Then x 2 + 100 2 = 130 2 x 2 = 130 2 100 2 x = 6900 Y Y X c 100 m c c 130 m SMPL PGS orrect to the nerest centimetre, the length of the short side is 83.07 m. Z W X Z Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
286 hpter 9: Geometry in the plne nd proof 9 xmple 11 onsider with = 9 cm, = 11 cm nd = 10 cm. Find the length of the ltitude of on. Solution Let N e the ltitude on s shown, with N = h cm. Let N = x cm. Then N = (10 x) cm. In N: In N: x 2 + h 2 = 81 (1) (10 x) 2 + h 2 = 121 (2) xpnding in eqution (2) gives 100 20x + x 2 + h 2 = 121 Sustituting for x 2 + h 2 from (1) gives 100 20x + 81 = 121 Sustituting in (1), we hve x = 3 9 + h 2 = 81 h 2 = 72 h = 6 2 The length of ltitude N is 6 2 cm. Section summry Pythgors theorem nd its converse Let e tringle with side lengths, nd c. If is right ngle, then 2 + 2 = c 2. If 2 + 2 = c 2, then is right ngle. UNORRT xercise 9 1 n 18 m ldder is 7 m wy from the ottom of verticl wll. How fr up the wll does it rech? SMPL PGS xmple 10 2 Find the length of the digonl of rectngle with dimensions 40 m y 9 m. 3 In circle of centre O, chord is of length 4 cm. The rdius of the circle is 14 cm. Find the distnce of the chord from O. 9 x h N 10 c 11 SF Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9 9 Pythgors theorem 287 xmple 11 4 squre hs n re of 169 cm 2. Wht is the length of the digonl? 5 Find the re of squre with digonl of length: 10 cm 8 cm 6 is squre of side length 2 cm. If is point on extended nd =, find the length of. 7 In squre of side length 2 cm, the midpoints of ech side re joined to form new squre. Find the re of the new squre. 8 onsider with = 7 cm, = 6 cm nd = 5 cm. Find the length of N, the ltitude on. 9 Which of the following re the three side lengths of right-ngled tringle? 5 cm, 6 cm, 7 cm 3.9 cm, 3.6 cm, 1.5 cm c 2.4 cm, 2.4 cm, 4 cm d 82 cm, 18 cm, 80 cm 10 Prove tht tringle with sides lengths x 2 1, 2x nd x 2 + 1 is right-ngled tringle. 11 onsider such tht = 20 cm, = 15 cm nd the ltitude N hs length 12 cm. Prove tht is right-ngled tringle. 12 Find the length of n ltitude in n equilterl tringle with side length 16 cm. 13 Three semicircles re drwn on the sides of this right-ngled tringle. Let 1, 2 nd 3 e the res of these semicircles. Prove tht 3 = 1 + 2. 3 c 1 UNORRT 14 Rectngle hs = 6 cm nd = 8 cm. Line segments Y nd X re drwn such tht X points X nd Y lie on nd X = Y = 90. 6 cm Find the length of XY. Y 8 cm SMPL PGS 2 SF F Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
288 hpter 9: Geometry in the plne nd proof 9 15 Find the vlues of x nd y. 16 If P is point in rectngle such tht P = 3 cm, P = 4 cm nd P = 5 cm, find the length of P. 17 Let Q e n ltitude of, where Q lies etween nd. Let P e the midpoint of. Prove tht 2 + 2 = 2P 2 + 2P 2. 18 For prllelogrm, prove tht 2 2 + 2 2 = 2 + 2. 9 Rtios This section is revision of work of previous yers. xmple 12 ivide 300 in the rtio 3 : 2. Solution one prt = 300 5 = 60 two prts = 60 2 = 120 three prts = 60 3 = 180 xmple 13 ivide 3000 in the rtio 3 : 2 : 1. Solution one prt = 3000 6 = 500 UNORRT Skillsheet xmple 12 two prts = 500 2 = 1000 three prts = 500 3 = 1500 xercise 9 1 SMPL PGS ivide 9000 in the rtio 2 : 7. xmple 13 2 ivide 15 000 in the rtio 2 : 2 : 1. 4 6 3 x y F U SF 3 Given tht x : 6 = 9 : 15, find x. Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9 9 Rtios 289 4 The rtio of the numers of ornge flowers to pink flowers in grden is 6 : 11. There re 144 ornge flowers. How mny pink flowers re there? 5 Given tht 15 : 2 = x : 3, find x. 6 The ngles of tringle re in the rtio 6 : 5 : 7. Find the sizes of the three ngles. 7 Three men X, Y nd Z shre n mount of money in the rtio 2 : 3 : 7. If Y receives $2 more thn X, how much does Z receive? 8 n lloy consists of copper, zinc nd tin in the rtio 1 : 3 : 4 (y weight). If there is 10 g of copper in the lloy, find the weights of zinc nd tin. 9 The rtio of red eds to white eds to green eds in g is 7 : 2 : 1. If there re 56 red eds, how mny white eds nd how mny green eds re there? 10 On mp, the length of rod is represented y 45 mm. If the scle is 1 : 125 000, find the ctul length of the rod. 11 Five thousnd two hundred dollrs ws divided etween mother nd dughter in the rtio 8 : 5. Find the difference etween the sums they received. 12 Points,, nd re plced in tht order on line so tht = 2 =. xpress s frction of. 13 If the rdius of circle is incresed y two units, find the rtio of the new circumference to the new dimeter. 14 In clss of 30 students, the rtio of oys to girls is 2 : 3. If six oys join the clss, find the new rtio of oys to girls in the clss. 15 If : = 3 : 4 nd : ( + c) = 2 : 5, find the rtio : c. 16 The scle of mp is 1 : 250 000. Find the distnce, in kilometres, etween two towns which re 3.5 cm prt on the mp. UNORRT 17 Prove tht if c d = c d, then = c d. SMPL PGS 18 Prove tht if x = y = c z = 2 3, then + + c x + y + z = 2 3. 19 Prove tht if x y = m n, then x + y x y = m + n m n. SF F Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
290 hpter 9: Geometry in the plne nd proof 9F Introduction to similrity The two tringles nd shown in the digrm re similr. Note: O = 2O, O = 2O, O = 2O Tringle cn e considered s the imge of tringle under mpping of the plne in which the coordintes re multiplied y 2. This mpping is clled n expnsion from the origin of fctor 2. From now on we will cll this fctor the similrity fctor. The rule for this mpping cn e written in trnsformtion nottion s (x, y) (2x, 2y). 12 10 8 6 4 2 0 (4, 12) (2, 6) (10, 6) (5, 3) (4, 1) (8, 2) 2 4 6 8 10 12 There is lso mpping from to, which is n expnsion from the origin of fctor 1 2. The rule for this is (x, y) ( 1 2 x, 1 2 y). Two figures re clled similr if we cn enlrge one figure so tht its enlrgement is congruent to the other figure. Mtching lengths of similr figures re in the sme rtio. Mtching ngles of similr figures re equl. For exmple, the rectngle with side lengths 1 nd 2 is similr to the rectngle with side lengths 3 nd 6. Here the similrity fctor is 3 nd the rule for the mpping is (x, y) (3x, 3y). Notes: ny two circles re similr. ny two squres re similr. ny two equilterl tringles re similr. Similr tringles If tringle is similr to tringle, we cn write this s 9 (3, 9) (6, 9) 8 7 6 5 4 (1, 3) (2, 3) 3 (3, 3) (6, 3) 2 1 (1, 1) (2, 1) 0 1 2 3 4 5 6 UNORRT The tringles re nmed so tht ngles of equl mgnitude hold the sme position. Tht is, mtches to, mtches to nd mtches to. So we hve = = = k where k is the similrity fctor. SMPL PGS Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9F Introduction to similrity 291 There re four stndrd tests for two tringles to e similr. The similrity test If two ngles of one tringle re respectively equl to two ngles of nother tringle, then the two tringles re similr. 100 100 45 35 The SS similrity test If the rtios of two pirs of mtching sides re equl nd the included ngles re equl, then the two tringles re similr. = 45 The SSS similrity test If the sides of one tringle cn e mtched up with the sides of nother tringle so tht the rtio of mtching lengths is constnt, then the two tringles re similr. = = 1 2 2 cm 3 cm 1 2 1 cm 45 45 5 cm 35 3 cm 6 cm The RHS similrity test If the rtio of the hypotenuses of two right-ngled tringles equls the rtio of nother pir of sides, then the two tringles re similr. UNORRT 7 cm 3 cm SMPL PGS 3 cm 2 7 cm 2 Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
292 hpter 9: Geometry in the plne nd proof xmple 14 Give the reson for tringle eing similr to tringle. Find the vlue of x. Solution Tringle is similr to tringle y SS, since nd xmple 15 5 6.25 = 0.8 = 3 3.75 = 20 = 5 cm x cm Give the reson for tringle eing similr to tringle XY. Find the vlue of x. Solution orresponding ngles re of equl mgnitude (). 20 3 cm x 3.013 = 5 6.25 x = 5 6.25 3.013 = 2.4104 X = Y x x + 6 = 3 5.5 5.5x = 3(x + 6) 6.25 cm 3.013 cm x cm 3 cm 2.5 cm 20 3.75 cm UNORRT Section summry 2.5x = 18 x = 7.2 SMPL PGS Two figures re similr if we cn enlrge one figure so tht its enlrgement is congruent to the other figure. Mtching lengths of similr figures re in the sme rtio. Mtching ngles of similr figures re equl. 6 cm Y X Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9F 9F Introduction to similrity 293 xmple 14 Two tringles re similr provided ny one of the following four conditions holds: two ngles of one tringle re equl to two ngles of the other tringle SS the rtios of two pirs of mtching sides re equl nd the included ngles re equl SSS the sides of one tringle cn e mtched up with the sides of the other tringle so tht the rtio of mtching lengths is constnt RHS the rtio of the hypotenuses of two right-ngled tringles equls the rtio of nother pir of sides. xercise 9F 1 Give resons why ech of the following pirs of tringles re similr nd find the vlue of x in ech cse: c d 4 cm 13 cm 56 82 9 cm 5 cm 14 cm 12 cm 6 cm 10 cm x cm x cm x cm 4 cm 6 cm UNORRT SMPL PGS Q 2 cm 8 cm x cm P SF 10 cm R Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
294 hpter 9: Geometry in the plne nd proof 9F xmple 15 2 Give resons why ech of the following pirs of tringles re similr nd find the vlue of x in ech cse: c x cm 6 cm 12 cm P 2 cm x cm P Q 8 cm 16 cm Q 8 cm 3 Given tht = 14, = 12, = 15 nd = 4, find, nd. d x cm x cm 2 cm 1.5 cm 3 cm 2 cm 10 cm 14 12 2 cm 4 tree csts shdow of 33 m nd t the sme time stick 30 cm long csts shdow of 224 cm. How tll is the tree? UNORRT 5 20 m high neon sign is supported y 40 m steel cle s shown. n nt crwls long the cle strting t. How high is the nt when it is 15 m from? 40 m 6 hill hs grdient of 1 in 20, i.e. for every 20 m horizontlly there is 1 m increse in height. If you go 300 m horizontlly, how high up will you e? 15 4 0.3 m SMPL PGS 20 m SF Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9F 9F Introduction to similrity 295 7 mn stnds t nd looks t point Y cross the river. He gets friend to plce stone t X so tht the three points, X nd Y re colliner (tht is, they ll lie on single line). He then mesures, X nd X to e 15 m, 30 m nd 45 m respectively. Find Y, the distnce cross the river. 15 m 45 m X 30 m Y 8 Find the height, h m, of tree tht csts shdow 32 m long t the sme time tht verticl stright stick 2 m long csts shdow 6.2 m long. 9 plnk is plced stright up stirs tht re 20 cm wide nd 12 cm deep. Find x, where x cm is the width of the widest rectngulr ox of height 8 cm tht cn e plced on stir under the plnk. 10 The sloping edge of technicl drwing tle is 1 m from front to ck. lculte the height ove the ground of point, which is 30 cm from the front edge. 11 Two similr rods 1.3 m long hve to e hinged together to support tle 1.5 m wide. The rods hve een fixed to the floor 0.8 m prt. Find the position of the hinge y finding the vlue of x. 80 cm 30 cm x m plnkx cm 8 cm 20 cm 1 m 1.5 m 0.8 m 12 cm 92 cm (1.3 x) m UNORRT 12 mn whose eyes re 1.7 m from the ground, when stnding 3.5 m in front of wll 3 m high, cn just see the top of tower tht is 100 m wy from the wll. Find the height of the tower. SMPL PGS 13 mn is 8 m up 10 m ldder, the top of which lens ginst verticl wll nd touches it t height of 9 m ove the ground. Find the height of the mn ove the ground. SF Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
296 hpter 9: Geometry in the plne nd proof 9F 14 spotlight is t height of 0.6 m ove ground level. verticl post 1.1 m high stnds 3 m wy nd 5 m further wy there is verticl wll. How high up the wll does the shdow rech? 15 onsider the digrm on the right. Prove tht. Find x. c Use Pythgors theorem to find y nd z. d Verify tht y : z = :. 16 Find. spotlight 5 0.6 m verticl post 1.1 m 3 m 5 m 17 mn who is 1.8 m tll csts shdow of 0.76 m in length. If t the sme time telephone pole csts 3 m shdow, find the height of the pole. 18 In the digrm shown, RT = 4 cm nd S T = 10 cm. Find the length NT. UNORRT 19 is tringulr frme with = 14 m, = 10 m nd = 7 m. point P on, 1.5 m from, is linked y rod to point Q on, 3 m from. lculte the length PQ. 10 S SMPL PGS 20 Using this digrm, find, x nd y. 7 6 x z y 12 4 N R 4 y x wll 2 T SF F Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9G Proofs involving similrity 297 9G Proofs involving similrity xmple 16 The ltitudes nd F of intersect t H. Prove tht F = Solution = F = F F F = xmple 17 ( nd F re ltitudes) (common) () is prllelogrm with cute. M is perpendiculr to extended, nd N is perpendiculr to extended. Prove tht Solution M = N M = N = (lternte ngles ) (lternte ngles ) N UNORRT Hence M = N N = M = 90 (given) M N M N = SMPL PGS ut = nd =, giving M N = Hence M = N. H F M Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
298 hpter 9: Geometry in the plne nd proof 9G Skillsheet xmple 16 xmple 17 xmple 18 is trpezium with digonls intersecting t O. line through O, prllel to the se, meets t X. Prove tht X = X. Solution Thus nd OX OX X = OX X = OX ivide (1) y (2): X X = X = X xercise 9G 1 2 () () (1) (2) Let M e the midpoint of line segment. ssume tht X nd MY re equilterl tringles on opposite sides of nd tht XY cuts t Z. Prove tht XZ YZ nd hence prove tht Z = 2Z. is rectngle. ssume tht P, Q nd R re points on, nd respectively such tht PQR is right ngle. Prove tht Q Q = P R. 3 is digonl of regulr pentgon. Find the sizes of nd., nd re digonls of regulr pentgon, with nd meeting t X. Prove tht () 2 = X. 4 hs right ngle t, nd is the ltitude to. Prove tht =. UNORRT Prove tht () 2 =. c Prove tht () 2 =. xmple 18 5 is trpezium with one of the prllel sides. The digonls meet t O. OX is the perpendiculr from O to, nd XO extended meets t Y. Prove tht OX OY = O O =. SMPL PGS 6 P is the point on side of such tht P : = 1 : 3, nd Q is the point on such tht Q : = 1 : 3. The line segments Q nd P intersect t X. Prove tht X : Q = 3 : 5. O X SF F Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9G 9H res, volumes nd similrity 299 7 P nd Q re points on sides nd respectively of such tht PQ. The medin meets PQ t M. Prove tht PM = MQ. 8 is stright line nd = =. n equilterl tringle P is drwn with se. Prove tht (P) 2 =. 9 is qudrilterl such tht = nd =. Prove tht isects. 10 hs right ngle t. The isector of meets t, nd is the 1 perpendiculr from to. Prove tht + 1 = 1. 11 Proportions in right-ngled tringle Prove tht, for right-ngled tringle, the ltitude on its hypotenuse forms two tringles which re similr to the originl tringle, nd hence to ech other. Prove Pythgors theorem y using prt (or y using similr tringles directly). 9H res, volumes nd similrity In this section we look t the res of similr shpes nd the volumes of similr solids. Similrity nd re If two shpes re similr nd the similrity fctor is k (tht is, if for ny length of one shpe, the corresponding length of the similr shpe is k), then re of similr shpe = k 2 re of originl shpe For exmple, if tringles nd re similr with = k, then re of = k 2 re of c h UNORRT c This cn e shown y oserving tht, since, we hve SMPL PGS re of = 1 2 h = 1 2 (k)(kh) = k 2 ( 1 2 h ) h F U = k 2 re of Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
300 hpter 9: Geometry in the plne nd proof Here re some more exmples of similr shpes nd the rtio of their res. Similr circles 3 cm 4 cm re = 9π cm 2 re = 16π cm 2 Similr rectngles 3 cm 2 cm 6 cm re = 6 cm 2 re = 24 cm 2 Similr tringles 5 cm 4 cm 3 cm 10 cm 8 cm re = 6 cm 2 re = 24 cm 2 xmple 19 Similrity fctor = 4 3 Rtio of res = 16π ( 4 9π = 3 4 cm Similrity fctor = 2 6 cm ) 2 Rtio of res = 24 6 = 4 = 22 Similrity fctor = 2 Rtio of res = 24 6 = 4 = 22 The two rectngles shown elow re similr. The re of rectngle is 20 cm 2. Find the re of rectngle. UNORRT 3 cm 5 cm SMPL PGS Solution The rtio of their side lengths is = 5 3. Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9H res, volumes nd similrity 301 The rtio of their res is re of ( 5 ) 2 re of = = 25 3 9. re of = 25 9 20 Similrity nd volume = 55 5 9 cm2 Two solids re considered to e similr if they hve the sme shpe nd the rtios of their corresponding liner dimensions re equl. For exmple, the two cuoids FGH nd F G H shown re similr, with similrity fctor 2.5. 3 cm F 2 cm G 1 cm H 7.5 cm If two solids re similr nd the similrity fctor is k, then volume of similr solid = k 3 volume of originl solid For exmple, for the two cuoids shown, we hve Volume of FGH = 2 1 3 = 6 cm 3 F 2.5 cm Volume of F G H = 5 2.5 7.5 = 93.75 cm 3 Rtio of volumes = 93.75 = 15.625 = 2.5 3 6 Here is nother exmple: V 3 cm 3 cm V 3 cm 5 cm 5 cm 5 cm 5 cm H UNORRT SMPL PGS Similrity fctor = 5 3 ( 5 3 Rtio of volumes = 3) G Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
302 hpter 9: Geometry in the plne nd proof 9H Skillsheet xmple 20 The two squre pyrmids re similr nd VO = 9 cm. Find the rtio of the lengths of their ses, nd hence find the height V O of pyrmid V. 9 cm V O 4 cm V O 5 cm The volume of V is 48 cm 3. Find the rtio of their volumes, nd hence find the volume of V. Solution The rtio of the length of their ses is = 5 4 V O = 5 4 9 = 11.25 cm Section summry The rtio of their volumes is Volume of V ( 5 ) 3 Volume of V = = 125 4 64 Volume of V = 125 64 48 = 93.75 cm 3 If two shpes re similr nd the similrity fctor is k (tht is, if for ny length of one shpe, the corresponding length of the similr shpe is k), then re of similr shpe = k 2 re of originl shpe If two solids re similr nd the similrity fctor is k, then xercise 9H 1 volume of similr solid = k 3 volume of originl solid These four rectngles re similr: UNORRT SMPL PGS Write down the rtio of the lengths of their ses. y counting rectngles, write down the rtio of their res. c Is there reltionship etween these two rtios? SF Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9H 9H res, volumes nd similrity 303 2 These four prllelogrms re similr: SF xmple 19 3 Write down the rtio of the lengths of their ses. y counting prllelogrms, write down the rtio of their res. c Is there reltionship etween these two rtios? The two rectngles shown re similr. The re of rectngle is 7 cm 2. Find the re of rectngle. 3 cm 4 Tringle is similr to tringle XYZ with XY = YZ = ZX = 2.1 5 cm The re of tringle XYZ is 20 cm 2. Find the re of tringle. 5 Tringles nd re equilterl tringles. Find the length of F. Find. c Find the rtio re of re of UNORRT 2 cm F 2 cm 2 cm 2 cm F cm SMPL PGS 6 The res of two similr tringles re 16 nd 25. Wht is the rtio of pir of corresponding sides? 7 The res of two similr tringles re 144 nd 81. If the se of the lrger tringle is 30, wht is the corresponding se of the smller tringle? Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
304 hpter 9: Geometry in the plne nd proof 9H xmple 20 8 These three solids re similr. Write down the rtio of the lengths of the ses. Write down the rtio of the lengths of the heights. c y counting cuoids equl in shpe nd size to cuoid, write down the rtio of the volumes. d Is there reltionship etween the nswers to, nd c? 9 These re two similr rectngulr locks. 10 Write down the rtio of their: 8 cm 4 cm 3 cm i longest edges ii depths iii heights. 12 cm y counting cues of side length 1 cm, write down the rtio of their volumes. c Is there ny reltionship etween the rtios in nd? These three solids re spheres. Write down the rtio of the rdii of the three spheres. The volume of sphere of rdius r is given y V = 4 3 πr3. xpress the volume of ech sphere s multiple of π. Hence write down the rtio of their volumes. c Is there ny reltionship etween the rtios found in nd? In ech of Questions 11 20, the ojects re mthemticlly similr. 11 The sides of two cues re in the rtio 2 : 1. Wht is the rtio of their volumes? 12 The rdii of two spheres re in the rtio 3 : 4. Wht is the rtio of their volumes? 13 Two regulr tetrhedrons hve volumes in the rtio 8 : 27. Wht is the rtio of their sides? 14 Two right cones hve volumes in the rtio 64 : 27. Wht is the rtio of: their heights their se rdii? 6 cm 2cm 4 1 2 cm UNORRT SMPL PGS 3cm 5cm SF 15 Two similr ottles re such tht one is twice s high s the other. Wht is the rtio of: their surfce res their cpcities? Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9H 9H res, volumes nd similrity 305 16 ch liner dimension of model cr is 1 of the corresponding cr dimension. Find 10 the rtio of: c the res of their windscreens the widths of the crs d the cpcities of their oots the numer of wheels they hve. 17 Three similr jugs hve heights 8 cm, 12 cm nd 16 cm. If the smllest jug holds 1 litre, find the cpcities of the other two. 2 18 Three similr drinking glsses hve heights 7.5 cm, 9 cm nd 10.5 cm. If the tllest glss holds 343 millilitres, find the cpcities of the other two. 19 toy mnufcturer produces model crs which re similr in every wy to the ctul crs. If the rtio of the door re of the model to the door re of the cr is 1 : 2500, find: the rtio of their lengths the rtio of the cpcities of their petrol tnks c the width of the model, if the ctul cr is 150 cm wide d the re of the rer window of the ctul cr if the re of the rer window of the model is 3 cm 2. 20 The rtio of the res of two similr lels on two similr jrs of coffee is 144 : 169. Find the rtio of: the heights of the two jrs their cpcities. 21 In the figure, if M is the midpoint of F nd K is the midpoint of, then how mny times lrger is the re of F thn the re of KM? If the re of F is 15, find the re of KM. 22 In the digrm, is equilterl, = F nd is the midpoint of. Find the rtio of the re of to the re of F. UNORRT SMPL PGS 23 The res of two similr tringles re 144 cm 2 nd 81 cm 2. If the length of one side of the first tringle is 6 cm, wht is the length of the corresponding side of the second? M F K F SF F Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
306 hpter 9: Geometry in the plne nd proof 9I Geometric proofs using vectors In this section we see how vectors cn e used s tool for proving geometric results. We require the following two definitions. olliner points oncurrent lines Three or more points re colliner if they ll lie on single line. Three or more lines re concurrent if they ll pss through single point. Here re some properties of vectors from hpter 6 tht will e useful: Prllel vectors For k R +, the vector k is in the sme direction s nd hs mgnitude k, nd the vector k is in the opposite direction to nd hs mgnitude k. Two non-zero vectors nd re prllel if nd only if = k for some k R \ {0}. If nd re prllel with t lest one point in common, then nd lie on the sme stright line. For exmple, if = k for some k R \ {0}, then, nd re colliner. Sclr product Two non-zero vectors nd re perpendiculr if nd only if = 0. = 2 Liner comintions of non-prllel vectors For two non-zero vectors nd tht re not prllel, if m + n = p + q, then m = p nd n = q. xmple 21 Three points P, Q nd R hve position vectors p, q nd k(2p + q) respectively, reltive to fixed origin O. The points O, P nd Q re not colliner. Find the vlue of k if: QR is prllel to p PR is prllel to q c P, Q nd R re colliner. Solution QR = QO + OR = q + k(2p + q) = 2kp + (k 1)q If QR is prllel to p, then there is some λ R \ {0} such tht 2kp + (k 1)q = λp This implies tht 2k = λ nd k 1 = 0 Hence k = 1. PR = PO + OR = p + k(2p + q) = (2k 1)p + kq If PR is prllel to q, then there is some m R \ {0} such tht (2k 1)p + kq = mq This implies tht 2k 1 = 0 nd k = m Hence k = 1 2. UNORRT SMPL PGS Note: Since points O, P nd Q re not colliner, the vectors p nd q re not prllel. Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
9I Geometric proofs using vectors 307 c If points P, Q nd R re colliner, then there exists n R \ {0} such tht n PQ = QR n( p + q) = 2kp + (k 1)q This implies tht n = 2k nd n = k 1 Therefore 3k 1 = 0 nd so k = 1 3. xmple 22 Suppose tht O is prllelogrm. Let = O nd c = O. xpress ech of the following in terms of nd c: i O ii Find in terms of nd c: i O 2 ii 2 c Hence, prove tht if the digonls of prllelogrm re of equl length, then the prllelogrm is rectngle. Solution i O = O + = O + O i = + c O 2 = O O = ( + c) ( + c) = + c + c + c c = 2 + 2 c + c 2 ii ii O = + = O O = c 2 = = ( c) ( c) = c c + c c = 2 2 c + c 2 c ssume tht the digonls of the prllelogrm O re of equl length. Then O =. This implies tht O 2 = 2 UNORRT 2 + 2 c + c 2 = 2 2 c + c 2 4 c = 0 SMPL PGS c = 0 We hve shown tht O O = 0. So O = 90. Hence the prllelogrm O is rectngle. Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400
308 hpter 9: Geometry in the plne nd proof 9I Skillsheet xmple 21 xercise 9I 1 In the digrm, OR = 4 OP, p = OP, q = OQ nd 5 PS : SQ = 1 : 4. xpress ech of the following in terms of p nd q: i OR ii RP iii PO iv PS v RS Wht cn e sid out line segments RS nd OQ? c Wht type of qudrilterl is ORSQ? d The re of tringle PRS is 5 cm 2. Wht is the re of ORSQ? 2 The position vectors of three points, nd reltive to n origin O re, nd k respectively. The point P lies on nd is such tht P = 2P. The point Q lies on nd is such tht Q = 6Q. 3 Find in terms of nd : i the position vector of P ii the position vector of Q Given tht OPQ is stright line, find: i the vlue of k ii the rtio OP PQ c The position vector of point R is 7. Show tht PR is prllel to. 3 The position vectors of two points nd reltive to n origin O re 3i + 3.5 j nd 6i 1.5 j respectively. i Given tht 1 O = O nd 1 =, write down the position vectors of 3 4 nd. ii Hence find. Given tht O nd intersect t X nd tht OX = po nd X = q, find the position vector of X in terms of: i p ii q c Hence determine the vlues of p nd q. 4 Points P nd Q hve position vectors p nd q, with reference to n origin O, nd M is the point on PQ such tht β PM = α MQ βp + αq Prove tht the position vector of M is given y m = α + β. Write the position vectors of P nd Q s p = k nd q = l, where k nd l re positive rel numers nd nd re unit vectors. UNORRT SMPL PGS i Prove tht the position vector of ny point on the internl isector of POQ hs the form λ( + ). ii If M is the point where the internl isector of POQ meets PQ, show tht α β = k l Uncorrected 3rd smple pges mridge University Press vns, et l. 2018 978-1-108-45165-9 Ph 03 8671 1400 P R S p O q Q SF F