UNCORRECTED. 9Geometry in the plane and proof

Size: px
Start display at page:

Download "UNCORRECTED. 9Geometry in the plane and proof"

Transcription

1 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 Ph

2 268 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 Ph O O 1

3 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 = 180 (supplementry ngles) = 180 (supplementry ngles) xmple 1 1 = 3 Find the vlues of x nd y in the digrm. Solution O 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 = 180 y = Uncorrected 3rd smple pges mridge University Press vns, et l Ph

4 270 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 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 l 3 Uncorrected 3rd smple pges mridge University Press vns, et l Ph

5 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 Ph

6 272 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 Ph

7 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 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 Ph

8 274 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 Ph

9 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 Ph

10 276 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 ( ) = 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 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 ch of the interior ngles is = 150. G Uncorrected 3rd smple pges mridge University Press vns, et l Ph

11 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 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 Uncorrected 3rd smple pges mridge University Press vns, et l Ph SMPL PGS SF

12 278 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 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 Ph

13 9 ongruence nd proofs 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 Ph F G β J R Q Q R Q H I

14 280 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 Ph Z O

15 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 Ph

16 282 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 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 Ph

17 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 Ph

18 284 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 = 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 Ph

19 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 = c2 = c2 2 ( + )( + ) = c = 2 + c = c 2 onverse of Pythgors theorem Let e tringle with side lengths, nd c. If = c 2, then is right ngle. Proof ssume hs side lengths =, = nd c = such tht = 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 = 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 = x 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 m. Z W X Z Uncorrected 3rd smple pges mridge University Press vns, et l Ph

20 286 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 x + x 2 + h 2 = 121 Sustituting for x 2 + h 2 from (1) gives x + 81 = 121 Sustituting in (1), we hve x = 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 = c 2. If = 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 Ph

21 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 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 = 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 = cm Find the length of XY. Y 8 cm SMPL PGS 2 SF F Uncorrected 3rd smple pges mridge University Press vns, et l Ph

22 288 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 = 2P 2 + 2P For prllelogrm, prove tht = Rtios This section is revision of work of previous yers. xmple 12 ivide 300 in the rtio 3 : 2. Solution one prt = = 60 two prts = 60 2 = 120 three prts = 60 3 = 180 xmple 13 ivide 3000 in the rtio 3 : 2 : 1. Solution one prt = = 500 UNORRT Skillsheet xmple 12 two prts = = 1000 three prts = = 1500 xercise 9 1 SMPL PGS ivide 9000 in the rtio 2 : 7. xmple 13 2 ivide in the rtio 2 : 2 : x y F U SF 3 Given tht x : 6 = 9 : 15, find x. Uncorrected 3rd smple pges mridge University Press vns, et l Ph

23 9 9 Rtios 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 : , 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 : 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 = 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 Ph

24 290 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) (4, 12) (2, 6) (10, 6) (5, 3) (4, 1) (8, 2) 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) (1, 3) (2, 3) 3 (3, 3) (6, 3) 2 1 (1, 1) (2, 1) 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 Ph

25 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 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. = = cm 3 cm cm 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 Ph

26 292 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 = 0.8 = = 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 () cm x = x = = X = Y x x + 6 = x = 3(x + 6) 6.25 cm cm x cm 3 cm 2.5 cm 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 Ph

27 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 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 Ph

28 294 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 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? m SMPL PGS 20 m SF Uncorrected 3rd smple pges mridge University Press vns, et l Ph

29 9F 9F Introduction to similrity 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 Ph

30 296 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 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 Ph

31 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 Ph

32 298 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 Ph

33 9G 9H res, volumes nd similrity 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 = 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 Ph

34 300 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 Ph

35 9H res, volumes nd similrity 301 The rtio of their res is re of ( 5 ) 2 re of = = re of = Similrity nd volume = 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 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 = = 6 cm 3 F 2.5 cm Volume of F G H = = cm 3 Rtio of volumes = = = 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 Ph

36 302 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 = = cm Section summry The rtio of their volumes is Volume of V ( 5 ) 3 Volume of V = = Volume of V = = 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 Ph

37 9H 9H res, volumes nd similrity 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 = 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 Ph

38 304 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 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 Ph

39 9H 9H res, volumes nd similrity 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 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 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 Ph

40 306 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 Ph

41 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 = 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 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 Ph

42 308 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 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 Ph P R S p O q Q SF F

SAMPLE. Ratios and similarity. 9.1 Ratios This section is revision of work of previous years. Several examples are presented.

SAMPLE. Ratios and similarity. 9.1 Ratios This section is revision of work of previous years. Several examples are presented. Ojectives H P T R 9 Rtios nd similrity To divide quntity in given rtio To determine the rtio in which quntity hs een divided To pply the trnsformtions which re expnsions from the origin To define similrity

More information

STRAND J: TRANSFORMATIONS, VECTORS and MATRICES

STRAND J: TRANSFORMATIONS, VECTORS and MATRICES Mthemtics SKE: STRN J STRN J: TRNSFORMTIONS, VETORS nd MTRIES J3 Vectors Text ontents Section J3.1 Vectors nd Sclrs * J3. Vectors nd Geometry Mthemtics SKE: STRN J J3 Vectors J3.1 Vectors nd Sclrs Vectors

More information

GEOMETRICAL PROPERTIES OF ANGLES AND CIRCLES, ANGLES PROPERTIES OF TRIANGLES, QUADRILATERALS AND POLYGONS:

GEOMETRICAL PROPERTIES OF ANGLES AND CIRCLES, ANGLES PROPERTIES OF TRIANGLES, QUADRILATERALS AND POLYGONS: GEOMETRICL PROPERTIES OF NGLES ND CIRCLES, NGLES PROPERTIES OF TRINGLES, QUDRILTERLS ND POLYGONS: 1.1 TYPES OF NGLES: CUTE NGLE RIGHT NGLE OTUSE NGLE STRIGHT NGLE REFLEX NGLE 40 0 4 0 90 0 156 0 180 0

More information

2 Calculate the size of each angle marked by a letter in these triangles.

2 Calculate the size of each angle marked by a letter in these triangles. Cmridge Essentils Mthemtics Support 8 GM1.1 GM1.1 1 Clculte the size of ech ngle mrked y letter. c 2 Clculte the size of ech ngle mrked y letter in these tringles. c d 3 Clculte the size of ech ngle mrked

More information

Shape and measurement

Shape and measurement C H A P T E R 5 Shpe nd mesurement Wht is Pythgors theorem? How do we use Pythgors theorem? How do we find the perimeter of shpe? How do we find the re of shpe? How do we find the volume of shpe? How do

More information

Similarity and Congruence

Similarity and Congruence Similrity nd ongruence urriculum Redy MMG: 201, 220, 221, 243, 244 www.mthletics.com SIMILRITY N ONGRUN If two shpes re congruent, it mens thy re equl in every wy ll their corresponding sides nd ngles

More information

Geometry AP Book 8, Part 2: Unit 3

Geometry AP Book 8, Part 2: Unit 3 Geometry ook 8, rt 2: Unit 3 IMRTNT NTE: For mny questions in this unit, there re multiple correct nswers, e.g. line segment cn e written s, RST is the sme s TSR, etc. Where pproprite, techers should e

More information

8Similarity UNCORRECTED PAGE PROOFS. 8.1 Kick off with CAS 8.2 Similar objects 8.3 Linear scale factors. 8.4 Area and volume scale factors 8.

8Similarity UNCORRECTED PAGE PROOFS. 8.1 Kick off with CAS 8.2 Similar objects 8.3 Linear scale factors. 8.4 Area and volume scale factors 8. 8.1 Kick off with S 8. Similr ojects 8. Liner scle fctors 8Similrity 8. re nd volume scle fctors 8. Review U N O R R E TE D P G E PR O O FS 8.1 Kick off with S Plese refer to the Resources t in the Prelims

More information

MEP Practice Book ES3. 1. Calculate the size of the angles marked with a letter in each diagram. None to scale

MEP Practice Book ES3. 1. Calculate the size of the angles marked with a letter in each diagram. None to scale ME rctice ook ES3 3 ngle Geometr 3.3 ngle Geometr 1. lculte the size of the ngles mrked with letter in ech digrm. None to scle () 70 () 20 54 65 25 c 36 (d) (e) (f) 56 62 d e 60 40 70 70 f 30 g (g) (h)

More information

Angles and triangles. Chapter Introduction Angular measurement Minutes and seconds

Angles and triangles. Chapter Introduction Angular measurement Minutes and seconds hpter 0 ngles nd tringles 0.1 Introduction stright line which crosses two prllel lines is clled trnsversl (see MN in igure 0.1). Trigonometry is suject tht involves the mesurement of sides nd ngles of

More information

Use the diagram to identify each angle pair as a linear pair, vertical angles, or neither.

Use the diagram to identify each angle pair as a linear pair, vertical angles, or neither. inl xm Review hpter 1 6 & hpter 9 Nme Use the points nd lines in the digrm to identify the following. 1) Three colliner points in Plne M. [],, H [],, [],, [],, [],, M [] H,, M 2) Three noncolliner points

More information

8Similarity ONLINE PAGE PROOFS. 8.1 Kick off with CAS 8.2 Similar objects 8.3 Linear scale factors. 8.4 Area and volume scale factors 8.

8Similarity ONLINE PAGE PROOFS. 8.1 Kick off with CAS 8.2 Similar objects 8.3 Linear scale factors. 8.4 Area and volume scale factors 8. 8.1 Kick off with S 8. Similr ojects 8. Liner scle fctors 8Similrity 8.4 re nd volume scle fctors 8. Review Plese refer to the Resources t in the Prelims section of your eookplus for comprehensive step-y-step

More information

GEOMETRY OF THE CIRCLE TANGENTS & SECANTS

GEOMETRY OF THE CIRCLE TANGENTS & SECANTS Geometry Of The ircle Tngents & Secnts GEOMETRY OF THE IRLE TNGENTS & SENTS www.mthletics.com.u Tngents TNGENTS nd N Secnts SENTS Tngents nd secnts re lines tht strt outside circle. Tngent touches the

More information

Answers for Lesson 3-1, pp Exercises

Answers for Lesson 3-1, pp Exercises Answers for Lesson -, pp. Eercises * ) PQ * ) PS * ) PS * ) PS * ) SR * ) QR * ) QR * ) QR. nd with trnsversl ; lt. int. '. nd with trnsversl ; lt. int. '. nd with trnsversl ; sme-side int. '. nd with

More information

R(3, 8) P( 3, 0) Q( 2, 2) S(5, 3) Q(2, 32) P(0, 8) Higher Mathematics Objective Test Practice Book. 1 The diagram shows a sketch of part of

R(3, 8) P( 3, 0) Q( 2, 2) S(5, 3) Q(2, 32) P(0, 8) Higher Mathematics Objective Test Practice Book. 1 The diagram shows a sketch of part of Higher Mthemtics Ojective Test Prctice ook The digrm shows sketch of prt of the grph of f ( ). The digrm shows sketch of the cuic f ( ). R(, 8) f ( ) f ( ) P(, ) Q(, ) S(, ) Wht re the domin nd rnge of

More information

Triangles The following examples explore aspects of triangles:

Triangles The following examples explore aspects of triangles: Tringles The following exmples explore spects of tringles: xmple 1: ltitude of right ngled tringle + xmple : tringle ltitude of the symmetricl ltitude of n isosceles x x - 4 +x xmple 3: ltitude of the

More information

Section 1.3 Triangles

Section 1.3 Triangles Se 1.3 Tringles 21 Setion 1.3 Tringles LELING TRINGLE The line segments tht form tringle re lled the sides of the tringle. Eh pir of sides forms n ngle, lled n interior ngle, nd eh tringle hs three interior

More information

What s in Chapter 13?

What s in Chapter 13? Are nd volume 13 Wht s in Chpter 13? 13 01 re 13 0 Are of circle 13 03 res of trpeziums, kites nd rhomuses 13 04 surfce re of rectngulr prism 13 05 surfce re of tringulr prism 13 06 surfce re of cylinder

More information

Alg 3 Ch 7.2, 8 1. C 2) If A = 30, and C = 45, a = 1 find b and c A

Alg 3 Ch 7.2, 8 1. C 2) If A = 30, and C = 45, a = 1 find b and c A lg 3 h 7.2, 8 1 7.2 Right Tringle Trig ) Use of clcultor sin 10 = sin x =.4741 c ) rete right tringles π 1) If = nd = 25, find 6 c 2) If = 30, nd = 45, = 1 find nd c 3) If in right, with right ngle t,

More information

2) Three noncollinear points in Plane M. [A] A, D, E [B] A, B, E [C] A, B, D [D] A, E, H [E] A, H, M [F] H, A, B

2) Three noncollinear points in Plane M. [A] A, D, E [B] A, B, E [C] A, B, D [D] A, E, H [E] A, H, M [F] H, A, B Review Use the points nd lines in the digrm to identify the following. 1) Three colliner points in Plne M. [],, H [],, [],, [],, [],, M [] H,, M 2) Three noncolliner points in Plne M. [],, [],, [],, [],,

More information

Alg. Sheet (1) Department : Math Form : 3 rd prep. Sheet

Alg. Sheet (1) Department : Math Form : 3 rd prep. Sheet Ciro Governorte Nozh Directorte of Eduction Nozh Lnguge Schools Ismili Rod Deprtment : Mth Form : rd prep. Sheet Alg. Sheet () [] Find the vlues of nd in ech of the following if : ) (, ) ( -5, 9 ) ) (,

More information

Bridging the gap: GCSE AS Level

Bridging the gap: GCSE AS Level Bridging the gp: GCSE AS Level CONTENTS Chpter Removing rckets pge Chpter Liner equtions Chpter Simultneous equtions 8 Chpter Fctors 0 Chpter Chnge the suject of the formul Chpter 6 Solving qudrtic equtions

More information

Stage 11 Prompt Sheet

Stage 11 Prompt Sheet Stge 11 rompt Sheet 11/1 Simplify surds is NOT surd ecuse it is exctly is surd ecuse the nswer is not exct surd is n irrtionl numer To simplify surds look for squre numer fctors 7 = = 11/ Mnipulte expressions

More information

15 - TRIGONOMETRY Page 1 ( Answers at the end of all questions )

15 - TRIGONOMETRY Page 1 ( Answers at the end of all questions ) - TRIGONOMETRY Pge P ( ) In tringle PQR, R =. If tn b c = 0, 0, then Q nd tn re the roots of the eqution = b c c = b b = c b = c [ AIEEE 00 ] ( ) In tringle ABC, let C =. If r is the inrdius nd R is the

More information

Comparing the Pre-image and Image of a Dilation

Comparing the Pre-image and Image of a Dilation hpter Summry Key Terms Postultes nd Theorems similr tringles (.1) inluded ngle (.2) inluded side (.2) geometri men (.) indiret mesurement (.6) ngle-ngle Similrity Theorem (.2) Side-Side-Side Similrity

More information

Coimisiún na Scrúduithe Stáit State Examinations Commission

Coimisiún na Scrúduithe Stáit State Examinations Commission M 30 Coimisiún n Scrúduithe Stáit Stte Exmintions Commission LEAVING CERTIFICATE EXAMINATION, 005 MATHEMATICS HIGHER LEVEL PAPER ( 300 mrks ) MONDAY, 3 JUNE MORNING, 9:30 to :00 Attempt FIVE questions

More information

1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE

1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE ELEMENTARY ALGEBRA nd GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the exmples, work the prolems, then check your nswers t the end of ech topic. If you don t get the nswer given, check

More information

Minnesota State University, Mankato 44 th Annual High School Mathematics Contest April 12, 2017

Minnesota State University, Mankato 44 th Annual High School Mathematics Contest April 12, 2017 Minnesot Stte University, Mnkto 44 th Annul High School Mthemtics Contest April, 07. A 5 ft. ldder is plced ginst verticl wll of uilding. The foot of the ldder rests on the floor nd is 7 ft. from the wll.

More information

4 VECTORS. 4.0 Introduction. Objectives. Activity 1

4 VECTORS. 4.0 Introduction. Objectives. Activity 1 4 VECTRS Chpter 4 Vectors jectives fter studying this chpter you should understnd the difference etween vectors nd sclrs; e le to find the mgnitude nd direction of vector; e le to dd vectors, nd multiply

More information

Things to Memorize: A Partial List. January 27, 2017

Things to Memorize: A Partial List. January 27, 2017 Things to Memorize: A Prtil List Jnury 27, 2017 Chpter 2 Vectors - Bsic Fcts A vector hs mgnitude (lso clled size/length/norm) nd direction. It does not hve fixed position, so the sme vector cn e moved

More information

GM1 Consolidation Worksheet

GM1 Consolidation Worksheet Cmridge Essentils Mthemtis Core 8 GM1 Consolidtion Worksheet GM1 Consolidtion Worksheet 1 Clulte the size of eh ngle mrked y letter. Give resons for your nswers. or exmple, ngles on stright line dd up

More information

+ R 2 where R 1. MULTIPLE CHOICE QUESTIONS (MCQ's) (Each question carries one mark)

+ R 2 where R 1. MULTIPLE CHOICE QUESTIONS (MCQ's) (Each question carries one mark) 2. C h p t e r t G l n c e is the set of ll points in plne which re t constnt distnce from fixed point clled centre nd constnt distnce is known s rdius of circle. A tngent t ny point of circle is perpendiculr

More information

Answers to Exercises. c 2 2ab b 2 2ab a 2 c 2 a 2 b 2

Answers to Exercises. c 2 2ab b 2 2ab a 2 c 2 a 2 b 2 Answers to Eercises CHAPTER 9 CHAPTER LESSON 9. CHAPTER 9 CHAPTER. c 9. cm. cm. b 5. cm. d 0 cm 5. s cm. c 8.5 cm 7. b cm 8.. cm 9. 0 cm 0. s.5 cm. r cm. 7 ft. 5 m.. cm 5.,, 5. 8 m 7. The re of the lrge

More information

Chapter 12. Lesson Geometry Worked-Out Solution Key. Prerequisite Skills (p. 790) A 5 } perimeter Guided Practice (pp.

Chapter 12. Lesson Geometry Worked-Out Solution Key. Prerequisite Skills (p. 790) A 5 } perimeter Guided Practice (pp. Chpter 1 Prerequisite Skills (p. 790) 1. The re of regulr polygon is given by the formul A 5 1 p P, where is the pothem nd P is the perimeter.. Two polygons re similr if their corresponding ngles re congruent

More information

Date Lesson Text TOPIC Homework. Solving for Obtuse Angles QUIZ ( ) More Trig Word Problems QUIZ ( )

Date Lesson Text TOPIC Homework. Solving for Obtuse Angles QUIZ ( ) More Trig Word Problems QUIZ ( ) UNIT 5 TRIGONOMETRI RTIOS Dte Lesson Text TOPI Homework pr. 4 5.1 (48) Trigonometry Review WS 5.1 # 3 5, 9 11, (1, 13)doso pr. 6 5. (49) Relted ngles omplete lesson shell & WS 5. pr. 30 5.3 (50) 5.3 5.4

More information

Intermediate Math Circles Wednesday, November 14, 2018 Finite Automata II. Nickolas Rollick a b b. a b 4

Intermediate Math Circles Wednesday, November 14, 2018 Finite Automata II. Nickolas Rollick a b b. a b 4 Intermedite Mth Circles Wednesdy, Novemer 14, 2018 Finite Automt II Nickols Rollick nrollick@uwterloo.c Regulr Lnguges Lst time, we were introduced to the ide of DFA (deterministic finite utomton), one

More information

On the diagram below the displacement is represented by the directed line segment OA.

On the diagram below the displacement is represented by the directed line segment OA. Vectors Sclrs nd Vectors A vector is quntity tht hs mgnitude nd direction. One exmple of vector is velocity. The velocity of n oject is determined y the mgnitude(speed) nd direction of trvel. Other exmples

More information

MTH 4-16a Trigonometry

MTH 4-16a Trigonometry MTH 4-16 Trigonometry Level 4 [UNIT 5 REVISION SECTION ] I cn identify the opposite, djcent nd hypotenuse sides on right-ngled tringle. Identify the opposite, djcent nd hypotenuse in the following right-ngled

More information

CONIC SECTIONS. Chapter 11

CONIC SECTIONS. Chapter 11 CONIC SECTIONS Chpter. Overview.. Sections of cone Let l e fied verticl line nd m e nother line intersecting it t fied point V nd inclined to it t n ngle α (Fig..). Fig.. Suppose we rotte the line m round

More information

THE KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE NO CALCULATORS 90 MINUTES

THE KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE NO CALCULATORS 90 MINUTES THE 08 09 KENNESW STTE UNIVERSITY HIGH SHOOL MTHEMTIS OMPETITION PRT I MULTIPLE HOIE For ech of the following questions, crefully blcken the pproprite box on the nswer sheet with # pencil. o not fold,

More information

Individual Contest. English Version. Time limit: 90 minutes. Instructions:

Individual Contest. English Version. Time limit: 90 minutes. Instructions: Elementry Mthemtics Interntionl Contest Instructions: Individul Contest Time limit: 90 minutes Do not turn to the first pge until you re told to do so. Write down your nme, your contestnt numer nd your

More information

JEE Advnced Mths Assignment Onl One Correct Answer Tpe. The locus of the orthocenter of the tringle formed the lines (+P) P + P(+P) = 0, (+q) q+q(+q) = 0 nd = 0, where p q, is () hperol prol n ellipse

More information

8.6 The Hyperbola. and F 2. is a constant. P F 2. P =k The two fixed points, F 1. , are called the foci of the hyperbola. The line segments F 1

8.6 The Hyperbola. and F 2. is a constant. P F 2. P =k The two fixed points, F 1. , are called the foci of the hyperbola. The line segments F 1 8. The Hperol Some ships nvigte using rdio nvigtion sstem clled LORAN, which is n cronm for LOng RAnge Nvigtion. A ship receives rdio signls from pirs of trnsmitting sttions tht send signls t the sme time.

More information

Geometry: similarity and mensuration

Geometry: similarity and mensuration Geometry: similrity nd mensurtion 8 VCEcoverge Are of study Units & Geometry nd trigonometry In this ch chpter 8A Properties of ngles, tringles nd polygons 8B Are nd perimeter 8C Totl surfce re 8D Volume

More information

3.1 Review of Sine, Cosine and Tangent for Right Angles

3.1 Review of Sine, Cosine and Tangent for Right Angles Foundtions of Mth 11 Section 3.1 Review of Sine, osine nd Tngent for Right Tringles 125 3.1 Review of Sine, osine nd Tngent for Right ngles The word trigonometry is derived from the Greek words trigon,

More information

Answers: ( HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 15 December 2017

Answers: ( HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 15 December 2017 Answers: (0- HKMO Het Events) reted y: Mr. Frncis Hung Lst updted: 5 Decemer 07 - Individul - Group Individul Events 6 80 0 4 5 5 0 6 4 7 8 5 9 9 0 9 609 4 808 5 0 6 6 7 6 8 0 9 67 0 0 I Simplify 94 0.

More information

Geometry of the Circle - Chords and Angles. Geometry of the Circle. Chord and Angles. Curriculum Ready ACMMG: 272.

Geometry of the Circle - Chords and Angles. Geometry of the Circle. Chord and Angles. Curriculum Ready ACMMG: 272. Geometry of the irle - hords nd ngles Geometry of the irle hord nd ngles urriulum Redy MMG: 272 www.mthletis.om hords nd ngles HRS N NGLES The irle is si shpe nd so it n e found lmost nywhere. This setion

More information

Definition :- A shape has a line of symmetry if, when folded over the line. 1 line of symmetry 2 lines of symmetry

Definition :- A shape has a line of symmetry if, when folded over the line. 1 line of symmetry 2 lines of symmetry Symmetry Lines of Symmetry Definition :- A shpe hs line of symmetry if, when folded over the line the hlves of the shpe mtch up exctly. Some shpes hve more thn one line of symmetry : line of symmetry lines

More information

7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus

7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus 7.1 Integrl s Net Chnge nd 7. Ares in the Plne Clculus 7.1 INTEGRAL AS NET CHANGE Notecrds from 7.1: Displcement vs Totl Distnce, Integrl s Net Chnge We hve lredy seen how the position of n oject cn e

More information

Section 13.1 Right Triangles

Section 13.1 Right Triangles Section 13.1 Right Tringles Ojectives: 1. To find vlues of trigonometric functions for cute ngles. 2. To solve tringles involving right ngles. Review - - 1. SOH sin = Reciprocl csc = 2. H cos = Reciprocl

More information

Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University

Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions

More information

MEP Practice Book ES19

MEP Practice Book ES19 19 Vectors M rctice ook S19 19.1 Vectors nd Sclrs 1. Which of the following re vectors nd which re sclrs? Speed ccelertion Mss Velocity (e) Weight (f) Time 2. Use the points in the grid elow to find the

More information

Parallel Projection Theorem (Midpoint Connector Theorem):

Parallel Projection Theorem (Midpoint Connector Theorem): rllel rojection Theorem (Midpoint onnector Theorem): The segment joining the midpoints of two sides of tringle is prllel to the third side nd hs length one-hlf the third side. onversely, If line isects

More information

Problem Set 9. Figure 1: Diagram. This picture is a rough sketch of the 4 parabolas that give us the area that we need to find. The equations are:

Problem Set 9. Figure 1: Diagram. This picture is a rough sketch of the 4 parabolas that give us the area that we need to find. The equations are: (x + y ) = y + (x + y ) = x + Problem Set 9 Discussion: Nov., Nov. 8, Nov. (on probbility nd binomil coefficients) The nme fter the problem is the designted writer of the solution of tht problem. (No one

More information

Pythagoras Theorem. Pythagoras

Pythagoras Theorem. Pythagoras 11 Pythgors Theorem Pythgors Theorem. onverse of Pythgors theorem. Pythgoren triplets Proof of Pythgors theorem nd its converse Problems nd riders bsed on Pythgors theorem. This unit fcilittes you in,

More information

/ 3, then (A) 3(a 2 m 2 + b 2 ) = 4c 2 (B) 3(a 2 + b 2 m 2 ) = 4c 2 (C) a 2 m 2 + b 2 = 4c 2 (D) a 2 + b 2 m 2 = 4c 2

/ 3, then (A) 3(a 2 m 2 + b 2 ) = 4c 2 (B) 3(a 2 + b 2 m 2 ) = 4c 2 (C) a 2 m 2 + b 2 = 4c 2 (D) a 2 + b 2 m 2 = 4c 2 SET I. If the locus of the point of intersection of perpendiculr tngents to the ellipse x circle with centre t (0, 0), then the rdius of the circle would e + / ( ) is. There re exctl two points on the

More information

Trigonometric Functions

Trigonometric Functions Exercise. Degrees nd Rdins Chpter Trigonometric Functions EXERCISE. Degrees nd Rdins 4. Since 45 corresponds to rdin mesure of π/4 rd, we hve: 90 = 45 corresponds to π/4 or π/ rd. 5 = 7 45 corresponds

More information

P 1 (x 1, y 1 ) is given by,.

P 1 (x 1, y 1 ) is given by,. MA00 Clculus nd Bsic Liner Alger I Chpter Coordinte Geometr nd Conic Sections Review In the rectngulr/crtesin coordintes sstem, we descrie the loction of points using coordintes. P (, ) P(, ) O The distnce

More information

( β ) touches the x-axis if = 1

( β ) touches the x-axis if = 1 Generl Certificte of Eduction (dv. Level) Emintion, ugust Comined Mthemtics I - Prt B Model nswers. () Let f k k, where k is rel constnt. i. Epress f in the form( ) Find the turning point of f without

More information

1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE

1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE ELEMENTARY ALGEBRA nd GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the exmples, work the prolems, then check your nswers t the end of ech topic. If you don t get the nswer given, check

More information

USA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year

USA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year 1/1/21. Fill in the circles in the picture t right with the digits 1-8, one digit in ech circle with no digit repeted, so tht no two circles tht re connected by line segment contin consecutive digits.

More information

6.2 The Pythagorean Theorems

6.2 The Pythagorean Theorems PythgorenTheorems20052006.nb 1 6.2 The Pythgoren Theorems One of the best known theorems in geometry (nd ll of mthemtics for tht mtter) is the Pythgoren Theorem. You hve probbly lredy worked with this

More information

Andrew Ryba Math Intel Research Final Paper 6/7/09 (revision 6/17/09)

Andrew Ryba Math Intel Research Final Paper 6/7/09 (revision 6/17/09) Andrew Ryb Mth ntel Reserch Finl Pper 6/7/09 (revision 6/17/09) Euler's formul tells us tht for every tringle, the squre of the distnce between its circumcenter nd incenter is R 2-2rR, where R is the circumrdius

More information

Polynomials and Division Theory

Polynomials and Division Theory Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the

More information

THE NUMBER CONCEPT IN GREEK MATHEMATICS SPRING 2009

THE NUMBER CONCEPT IN GREEK MATHEMATICS SPRING 2009 THE NUMBER CONCEPT IN GREEK MATHEMATICS SPRING 2009 0.1. VII, Definition 1. A unit is tht by virtue of which ech of the things tht exist is clled one. 0.2. VII, Definition 2. A number is multitude composed

More information

I1.1 Pythagoras' Theorem. I1.2 Further Work With Pythagoras' Theorem. I1.3 Sine, Cosine and Tangent. I1.4 Finding Lengths in Right Angled Triangles

I1.1 Pythagoras' Theorem. I1.2 Further Work With Pythagoras' Theorem. I1.3 Sine, Cosine and Tangent. I1.4 Finding Lengths in Right Angled Triangles UNIT I1 Pythgors' Theorem nd Trigonometric Rtios: Tet STRAND I: Geometry nd Trigonometry I1 Pythgors' Theorem nd Trigonometric Rtios Tet Contents Section I1.1 Pythgors' Theorem I1. Further Work With Pythgors'

More information

青藜苑教育 The digrm shows the position of ferry siling between Folkestone nd lis. The ferry is t X. X 4km The pos

青藜苑教育 The digrm shows the position of ferry siling between Folkestone nd lis. The ferry is t X. X 4km The pos 青藜苑教育 www.thetopedu.com 010-6895997 1301951457 Revision Topic 9: Pythgors Theorem Pythgors Theorem Pythgors Theorem llows you to work out the length of sides in right-ngled tringle. c The side opposite

More information

MATHEMATICS AND STATISTICS 1.6

MATHEMATICS AND STATISTICS 1.6 MTHMTIS N STTISTIS 1.6 pply geometri resoning in solving prolems ternlly ssessed 4 redits S 91031 inding unknown ngles When finding the size of unknown ngles in figure, t lest two steps of resoning will

More information

A study of Pythagoras Theorem

A study of Pythagoras Theorem CHAPTER 19 A study of Pythgors Theorem Reson is immortl, ll else mortl. Pythgors, Diogenes Lertius (Lives of Eminent Philosophers) Pythgors Theorem is proly the est-known mthemticl theorem. Even most nonmthemticins

More information

S56 (5.3) Vectors.notebook January 29, 2016

S56 (5.3) Vectors.notebook January 29, 2016 Dily Prctice 15.1.16 Q1. The roots of the eqution (x 1)(x + k) = 4 re equl. Find the vlues of k. Q2. Find the rte of chnge of 剹 x when x = 1 / 8 Tody we will e lerning out vectors. Q3. Find the eqution

More information

Level I MAML Olympiad 2001 Page 1 of 6 (A) 90 (B) 92 (C) 94 (D) 96 (E) 98 (A) 48 (B) 54 (C) 60 (D) 66 (E) 72 (A) 9 (B) 13 (C) 17 (D) 25 (E) 38

Level I MAML Olympiad 2001 Page 1 of 6 (A) 90 (B) 92 (C) 94 (D) 96 (E) 98 (A) 48 (B) 54 (C) 60 (D) 66 (E) 72 (A) 9 (B) 13 (C) 17 (D) 25 (E) 38 Level I MAML Olympid 00 Pge of 6. Si students in smll clss took n em on the scheduled dte. The verge of their grdes ws 75. The seventh student in the clss ws ill tht dy nd took the em lte. When her score

More information

Form 5 HKCEE 1990 Mathematics II (a 2n ) 3 = A. f(1) B. f(n) A. a 6n B. a 8n C. D. E. 2 D. 1 E. n. 1 in. If 2 = 10 p, 3 = 10 q, express log 6

Form 5 HKCEE 1990 Mathematics II (a 2n ) 3 = A. f(1) B. f(n) A. a 6n B. a 8n C. D. E. 2 D. 1 E. n. 1 in. If 2 = 10 p, 3 = 10 q, express log 6 Form HK 9 Mthemtics II.. ( n ) =. 6n. 8n. n 6n 8n... +. 6.. f(). f(n). n n If = 0 p, = 0 q, epress log 6 in terms of p nd q.. p q. pq. p q pq p + q Let > b > 0. If nd b re respectivel the st nd nd terms

More information

Review of Gaussian Quadrature method

Review of Gaussian Quadrature method Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge

More information

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.) MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give

More information

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz

More information

Lesson-5 ELLIPSE 2 1 = 0

Lesson-5 ELLIPSE 2 1 = 0 Lesson-5 ELLIPSE. An ellipse is the locus of point which moves in plne such tht its distnce from fied point (known s the focus) is e (< ), times its distnce from fied stright line (known s the directri).

More information

KEY CONCEPTS. satisfies the differential equation da. = 0. Note : If F (x) is any integral of f (x) then, x a

KEY CONCEPTS. satisfies the differential equation da. = 0. Note : If F (x) is any integral of f (x) then, x a KEY CONCEPTS THINGS TO REMEMBER :. The re ounded y the curve y = f(), the -is nd the ordintes t = & = is given y, A = f () d = y d.. If the re is elow the is then A is negtive. The convention is to consider

More information

Trigonometry. VCEcoverage. Area of study. Units 3 & 4 Geometry and trigonometry

Trigonometry. VCEcoverage. Area of study. Units 3 & 4 Geometry and trigonometry Trigonometry 9 VEcoverge re of study Units & Geometry nd trigonometry In this ch chpter 9 Pythgors theorem 9 Pythgoren trids 9 Three-dimensionl Pythgors theorem 9D Trigonometric rtios 9E The sine rule

More information

Edexcel GCE Core Mathematics (C2) Required Knowledge Information Sheet. Daniel Hammocks

Edexcel GCE Core Mathematics (C2) Required Knowledge Information Sheet. Daniel Hammocks Edexcel GCE Core Mthemtics (C) Required Knowledge Informtion Sheet C Formule Given in Mthemticl Formule nd Sttisticl Tles Booklet Cosine Rule o = + c c cosine (A) Binomil Series o ( + ) n = n + n 1 n 1

More information

Chapter 9 Definite Integrals

Chapter 9 Definite Integrals Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished

More information

PYTHAGORAS THEOREM WHAT S IN CHAPTER 1? IN THIS CHAPTER YOU WILL:

PYTHAGORAS THEOREM WHAT S IN CHAPTER 1? IN THIS CHAPTER YOU WILL: PYTHAGORAS THEOREM 1 WHAT S IN CHAPTER 1? 1 01 Squres, squre roots nd surds 1 02 Pythgors theorem 1 03 Finding the hypotenuse 1 04 Finding shorter side 1 05 Mixed prolems 1 06 Testing for right-ngled tringles

More information

set is not closed under matrix [ multiplication, ] and does not form a group.

set is not closed under matrix [ multiplication, ] and does not form a group. Prolem 2.3: Which of the following collections of 2 2 mtrices with rel entries form groups under [ mtrix ] multipliction? i) Those of the form for which c d 2 Answer: The set of such mtrices is not closed

More information

2. VECTORS AND MATRICES IN 3 DIMENSIONS

2. VECTORS AND MATRICES IN 3 DIMENSIONS 2 VECTORS AND MATRICES IN 3 DIMENSIONS 21 Extending the Theory of 2-dimensionl Vectors x A point in 3-dimensionl spce cn e represented y column vector of the form y z z-xis y-xis z x y x-xis Most of the

More information

13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS

13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS 33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in

More information

3 Angle Geometry. 3.1 Measuring Angles. 1. Using a protractor, measure the marked angles.

3 Angle Geometry. 3.1 Measuring Angles. 1. Using a protractor, measure the marked angles. 3 ngle Geometry MEP Prtie ook S3 3.1 Mesuring ngles 1. Using protrtor, mesure the mrked ngles. () () (d) (e) (f) 2. Drw ngles with the following sizes. () 22 () 75 120 (d) 90 (e) 153 (f) 45 (g) 180 (h)

More information

SUMMER KNOWHOW STUDY AND LEARNING CENTRE

SUMMER KNOWHOW STUDY AND LEARNING CENTRE SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18

More information

Lesson Notes: Week 40-Vectors

Lesson Notes: Week 40-Vectors Lesson Notes: Week 40-Vectors Vectors nd Sclrs vector is quntity tht hs size (mgnitude) nd direction. Exmples of vectors re displcement nd velocity. sclr is quntity tht hs size but no direction. Exmples

More information

MEP Scheme of Work: YEAR 8A

MEP Scheme of Work: YEAR 8A MATHEMATICAL DIAGRAMS. Milege Chrts Extrcting distnces from milege chrt The distnces etween sttions on the GNER rilwy re 4 York shown opposite. 83670 Doncster () Complete the top row. 68 47 Peterorough

More information

What else can you do?

What else can you do? Wht else cn you do? ngle sums The size of specil ngle types lernt erlier cn e used to find unknown ngles. tht form stright line dd to 180c. lculte the size of + M, if L is stright line M + L = 180c( stright

More information

PROPERTIES OF AREAS In general, and for an irregular shape, the definition of the centroid at position ( x, y) is given by

PROPERTIES OF AREAS In general, and for an irregular shape, the definition of the centroid at position ( x, y) is given by PROPERTES OF RES Centroid The concept of the centroid is prol lred fmilir to ou For plne shpe with n ovious geometric centre, (rectngle, circle) the centroid is t the centre f n re hs n is of smmetr, the

More information

Chapter. Trigonometry. Contents: A Using scale diagrams

Chapter. Trigonometry. Contents: A Using scale diagrams hpter 12 Trigonometry ontents: Using scle digrms Lelling tringles The trigonometric rtios Trigonometric prolem solving erings 3-dimensionl prolem solving 268 TRIGOOMTRY (hpter 12) OPIG PROLM indi s office

More information

Area and Perimeter. Area and Perimeter. Curriculum Ready.

Area and Perimeter. Area and Perimeter. Curriculum Ready. Are nd Perimeter Curriculum Redy www.mthletics.com This ooklet shows how to clculte the re nd perimeter of common plne shpes. Footll fields use rectngles, circles, qudrnts nd minor segments with specific

More information

, MATHS H.O.D.: SUHAG R.KARIYA, BHOPAL, CONIC SECTION PART 8 OF

, MATHS H.O.D.: SUHAG R.KARIYA, BHOPAL, CONIC SECTION PART 8 OF DOWNLOAD FREE FROM www.tekoclsses.com, PH.: 0 903 903 7779, 98930 5888 Some questions (Assertion Reson tpe) re given elow. Ech question contins Sttement (Assertion) nd Sttement (Reson). Ech question hs

More information

ICSE Board Class IX Mathematics Paper 4 Solution

ICSE Board Class IX Mathematics Paper 4 Solution ICSE Bord Clss IX Mthemtics Pper Solution SECTION A (0 Mrks) Q.. () Consider x y 6 5 5 x y 6 5 5 0 6 0 6 x y 6 50 8 5 6 7 6 x y 6 7 6 x y 6 x 7,y (b) Dimensions of the brick: Length (l) = 0 cm, bredth

More information

List all of the possible rational roots of each equation. Then find all solutions (both real and imaginary) of the equation. 1.

List all of the possible rational roots of each equation. Then find all solutions (both real and imaginary) of the equation. 1. Mth Anlysis CP WS 4.X- Section 4.-4.4 Review Complete ech question without the use of grphing clcultor.. Compre the mening of the words: roots, zeros nd fctors.. Determine whether - is root of 0. Show

More information

Log1 Contest Round 3 Theta Individual. 4 points each 1 What is the sum of the first 5 Fibonacci numbers if the first two are 1, 1?

Log1 Contest Round 3 Theta Individual. 4 points each 1 What is the sum of the first 5 Fibonacci numbers if the first two are 1, 1? 008 009 Log1 Contest Round Thet Individul Nme: points ech 1 Wht is the sum of the first Fiboncci numbers if the first two re 1, 1? If two crds re drwn from stndrd crd deck, wht is the probbility of drwing

More information

Mathematics. Area under Curve.

Mathematics. Area under Curve. Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding

More information

1 (=0.5) I3 a 7 I4 a 15 I5 a (=0.5) c 4 N 10 1 (=0.5) N 6 A 52 S 2

1 (=0.5) I3 a 7 I4 a 15 I5 a (=0.5) c 4 N 10 1 (=0.5) N 6 A 52 S 2 Answers: (98-84 HKMO Finl Events) Creted by Mr. Frncis Hung Lst updted: December 05 Individul Events SI 900 I 0 I (=0.5) I 7 I4 5 I5 80 b 7 b b 5 b 6 b 8 b 4 c c 4 c 0 x (=0.5) c 4 N 0 d 9 d 5 d 5 y d

More information

HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE NO CALCULATORS 90 MINUTES (A) 3 (B) 5 (C) 9 (D) 15 (E) 25

HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE NO CALCULATORS 90 MINUTES (A) 3 (B) 5 (C) 9 (D) 15 (E) 25 HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE For ech of the following 5 questions, crefull lcken the pproprite o on the nswer sheet with # pencil. o not fold, end, or write str mrks on either

More information

PYTHAGORAS THEOREM,TRIGONOMETRY,BEARINGS AND THREE DIMENSIONAL PROBLEMS

PYTHAGORAS THEOREM,TRIGONOMETRY,BEARINGS AND THREE DIMENSIONAL PROBLEMS PYTHGORS THEOREM,TRIGONOMETRY,ERINGS ND THREE DIMENSIONL PROLEMS 1.1 PYTHGORS THEOREM: 1. The Pythgors Theorem sttes tht the squre of the hypotenuse is equl to the sum of the squres of the other two sides

More information