M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT Applications of Geometrical Properties 1)! 2U84-2i Find the area of the rhombus ABCD given AB = 10 cm and EB = 8 cm. 10 cm A B 8 cm E D C † « 96 cm2 » 2)! 2U84-8i W Y Z V 3)! NOT TO SCALE X In the triangle WXV, YZ = 9 cm, VX = 12 cm, WX = 8 cm and YZ || VX. Prove that WZY is similar to WXV and find the length of WZ.† « 6 cm » 2U84-9iv A NOT TO SCALE C B D F 4)! E In the figure, AB = AC, BD || FE, BF || DE and CAB = 54. Find the size of FED giving reasons.† « 63 » 2U85-7i P Q T NOT TO SCALE S R PQRS is a square with PQ = 1 unit. Find the perimeter of PTRS.† « 4 2 units » 5)! 2U85-7ii †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 154 M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT A Z D X Y NOT TO SCALE C B ABC and ABD are two triangles, X, Y and Z are points such that XY CB and YZ BD. Prove that XY : YZ = CB : BD.† « Proof » 6)! 2U86-5i List three properties of a rhombus.† « All sides are equal. Opposite sides are parallel. Diagonals bisect each other at right angles. The diagonals bisect the angles through which they pass. » 7)! 2U86-5ii A, B, C are collinear points. BD || AE, BA || DE, BC = BD and BCD = 58. Reproduce this diagram on your answer sheet and find the size of DEA. C 58 B A 8)! NOT TO SCALE D E † « 116 » 2U86-5iii In the triangle PSU, QR || SU, SP || TR, ST = 7·5 cm, PQ = 10 cm, PR = 12 cm and UT = 15 cm. Find the length of SQ giving reasons. P Q NOT TO SCALE R S T U « 20 cm » 9)! 2U86-5iv GL is a median in HFG and HJ || FK. a. Draw a neat sketch of this diagram on your answer sheet. b. Prove, giving reasons, that KL = LJ. †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 155 M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT F NOT TO SCALE K L J H 10)! G « Proof » 2U87-5i In the diagram AB || CD and GH AB. If y = 25 find the size of GMH. Hence or otherwise find the size of MFD.† y G A H M C 11)! B F « GMH = 65, MFD = 115 » 2U87-5ii PQRS is a trapezium with PQ || SR. Diagonals PR and SQ intersect at T. P Q NOT TO SCALE T 12)! D S R a. Reproduce this diagram on your answer sheet. b. Prove, giving reasons, that PQT ||| RST. c. Hence, find PQ, given that SR = 36 cm, PT = 5 cm and RT = 15 cm.† « a) Proof b) 12 cm » 2U87-5iii In the diagram below, UXY = UYX and XZ = YZ. U X Y Z a. b. c. V W Copy this diagram on your answer sheet. Prove that UVY UWX, giving reasons. Hence prove that VZW is isosceles.† « Proof » 13)! 2U88-6i †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 156 M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT F 14)! I E G J H In the figures FG = 10 cm, EG = 15 cm, EF = 12·5 cm, IJ = 4 cm, HJ = 6 cm and HI = 5 cm. a. Draw a neat sketch and mark on it all the given information. b. Show that EFG ||| HIJ giving reasons.† « Proof » 2U88-6ii L M P 15)! NOT TO SCALE NOT TO SCALE K N The figure above shows quadrilateral KLMN with diagonals KM and LN intersecting at P. a. Reproduce this diagram on your answer sheet. b. If the diagonals KM an LN bisect each other at right angles, prove that KLMN is a rhombus.† « Proof » 2U89-3b L NOT TO SCALE N K M In the diagram above, KN = NM, KL = LM, KNM = 110 and NKL = 45. i. Reproduce a neat sketch and mark on it all the given information. ii. Find the size of MKN and KLM, giving reasons.† L N 45 110 M ii) MKN = 35, KLM = 20 » « i) K 16)! 2U89-6c †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 157 M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT P Q R U T 17)! NOT TO SCALE S In the diagram above, SQ PQ, RU SQ and PS || QR. i. Prove that RQU ||| PSQ. ii. If RU = x units, QR = y units and PS is four times the length of RU, find the length of PQ in terms of x and y.† 4x2 « i) Proof ii) » y 2U90-2c L NOT TO SCALE N M K KLM is an isosceles triangle with KL = LM, LKM = 80, LN bisects KLM and KMN = 20. i. On your answer sheet, draw a neat sketch of the diagram above, showing all the given information. ii. Find the size of LMN, giving reasons for your answer. iii. Find the size of LNM, giving reasons.† L N 80 20 ii) LMN = 60 iii) LNM = 110 » « i) K 18)! M 2U90-5d PQRS is a quadrilateral with PR = QS, PQ PS and SR PS. i. On your answer sheet, draw a neat sketch and mark on it all the given information. ii. Prove that QPS and RSP are congruent. iii. Hence prove that PQRS is a parallelogram.† P Q « i) S 19)! 2U91-4c †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 158 R ii) iii) Proof » M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT In the diagram given below, ABC is a right angle triangle with BAC = 90, CQ = CR, PB = RB and ACB = 40. A P i. ii. 20)! Q 40° B R C Copy this diagram onto your answer booklet. Write down the size of PRQ. (No reasons are required in your solution).† « 45 » 2U91-7a P Q V 21)! T S R PQRS is a parallelogram. TQ bisects PQR and VS bisects PSR. i. Copy this diagram onto your answer booklet. ii. State why PQR = PSR. iii. Prove that PVS and RTQ are congruent. iv. Hence find the length of TV if PR = 20 cm and TR = 8 cm.† « ii) Opposite 's in a parallelogram iii) Proof iv) 4 cm » 2U92-5a N L 123 K 2 22)! NOT TO SCALE M NOT TO SCALE J In the diagram above JKLM is a quadrilateral and LMN is a triangle. JM || LN, JK = KL, JM = ML = MN, KLM = 123, JKL = 2 and JML = . i. Copy this diagram onto your answer sheet. ii. Show that JML = 38 giving reasons. iii. Determine the size of LNM giving reasons.† « ii) Proof iii) 38 » 2U92-7a P NOT TO Q SCALE N M Z S R In the given diagram PQ || RS. MQ bisects PQR, NR bisects QRS and MQ = NR. †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 159 M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT i. ii. iii. iv. Copy this diagram onto your answer sheet and mark on it all the given information. Explain how you know that MQZ = NRZ. Prove that QMZ RNZ. Hence prove that the intervals QR and MN bisect each other.† P Q N M Z R « i) 23)! S ii) iii) iv) Proof » 2U93-4a In the diagram CA = AD = DB and EBD = 20. Copy this diagram onto your answer sheet. C NOT TO SCALE D i. ii. 20 E A Show ADC = 40, giving reasons. Hence find the size of CAE, giving reasons.† B « i) Proof ii) 60 » 24)! 2U93-5c In the diagram CT bisects ACB, AE is perpendicular to CT and M is the midpoint of AB. AE produced meets BC at the point P. C P E i. ii. iii. iv. A T M B Copy this diagram onto your answer sheet and mark in all the given information. Prove that ACE is congruent to PCE. Explain why AE = EP. Hence prove that EM is parallel to PB.† C P E A T M B « i) ii) Proof iii) Corresponding sides in congruent 's iv) Proof » 25)! 2U94-3b In the diagram AE || BD, AC || ED, AED = 130 and ABC = 90. †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 160 M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT A B C NOT TO SCALE 130 E i. ii. 26)! D Copy this diagram onto your answer sheet. Find the size of BAC giving reasons.† « 40 » 2U94-7c In the figure triangles ACB and APO are equilateral. A P NOT TO SCALE O i. ii. iii. iv. B C Copy this diagram onto your answer sheet and include all the given information. Explain why BAO = PAC. Prove AOB APC. Hence prove OB = CP.† A P O 27)! C ii) Each angle is equal to 60 OAC iii) iv) Proof » « i) B 2U95-1d In the diagram AB || CE, ABF = 75 and BFE = 35. A C NOT TO SCALE B 75 F 35 D E Find the size of giving reasons.† « 40 » 28)! 2U95-5d E 2y F y L 96 G H †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 161 M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT The diagram shows a rhombus EFGH. A line EL is drawn through E so that HEL = 2 FEL. i. Copy the diagram onto your answer page. ii. FGH = 96, find the size of ELF giving reasons.† « 106 » 29)! 2U95-9b In the diagram ABCD is a square. AB is produced to E so that AB = BE and BC is produced to F so that BC = CF. A B E D i. ii. iii. NOT TO SCALE C F Copy the diagram onto your answer page. Prove AED BFA. Hence prove AED = BFA.† « Proof » 30)! 2U96-1f D A x NOT TO SCALE 30 35 B 50 C E Find the value of x.† « 115 » 31)! 2U96-2b In the diagram, AB || CD, AD = CD and BAC = 120. Copy the diagram onto your answer sheet. A B 120 C i. ii. 32)! D NOT TO SCALE Explain why ACD = 60. Show that ADC is equilateral, giving reasons.† « Proof » 2U96-9a ABC is right-angled at A and AD is drawn perpendicular to BC. AB = 15 cm and AD = 12 cm. Copy the given diagram onto your answer sheet. †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 162 M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT B NOT TO SCALE 15 cm 12 cm D A i. ii. iii. C Show that BD = 9 cm. Prove that ABC is similar to DBA. Hence find the length of AC.† « i) ii) Proof iii) 20 cm » 33)! 2U97-3b A NOT TO SCALE B C D ABC is an equilateral triangle. BC is produced to D so that BC = CD. i. Copy the diagram onto your answer sheet and mark on it all given information. ii. Prove that BAD = 90.† A 60 60 « i) B 34)! 60 C D ii) Proof » 2U97-10a A 12 cm B NOT TO SCALE 9 cm M 35)! D C ABCD is a rectangle with AB = 12 cm, AD = 9 cm and AM is perpendicular to BD. i. Copy the diagram onto your answer sheet. ii. Find the length of BD. iii. Prove that ABM is similar to DBA. iv. Hence find the length of BM.† « ii) 15 cm iii) Proof iv) 9·6 cm » 2U98-1e T y 126 P Q R S In the diagram PQT = 126 and QTR = 90. Find the value of y.† « 144 » †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 163 M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT 36)! 2U98-5c M 37)! R P O PN is a diagonal of the rectangle MNOP. R is the point on PO and PQR = 90. i. Prove that PQR is similar to NMP. ii. Given MP = 5 cm, MN = 10 cm and QR = 2 cm, find the length of PQ.† « i) Proof ii) 4 cm » 2U99-1d Q 85 NOT TO P 120 SCALE S 38)! N Q 75 R T In the diagram, PQ || TR, PQR = 85, QPS = 120, PSR = 75 and SRT = . Copy the diagram onto your answer sheet. Find the value of .† « 15 » 2U99-4d W X NOT TO SCALE P Q Z 39)! Y WXYZ is a parallelogram. XP bisects WXY and ZQ bisects WZY. Copy the diagram onto your answer sheet. i. Explain why WXY = WZY. ii. Prove WXP is congruent to YZQ. iii. Hence find the length of PQ given WY = 20 cm and QY = 8 cm.† « i) ii) Proof iii) 4 cm » 2U99-7a NOT TO Y SCALE 8 cm P X 6 cm Q R S In the diagram, PQRS is a rectangle and SR = 3 PS. R, Q and Y are collinear points. XQ = 6 cm and YQ = 8 cm. i. Prove PXS is similar to QXY. ii. Hence find the length of PS. † « i) Proof ii) †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 164 8 cm » 3 M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT 40)! 2U00-1c X 53 NOT TO SCALE Y Z 108 41)! V U W The diagram shows XY parallel to UW, XYU = 53, UZV = 108 and ZVW = . Find the value of . Give reasons.† « = 161 » 2U00-6b D C A 42)! 43)! X B NOT TO SCALE Y In the diagram, ABCD is a parallelogram. X is a point on AB. DX and CB are both produced to Y. i. Copy this diagram onto your answer sheet. ii. Prove that ADX is similar to CYD. iii. Hence find the length of XY given AX = 8 cm, DC = 12 cm and DX = 10 cm.† « ii) Proof iii) 5 cm » 2U01-3c A C NOT TO x R SCALE 110 B S In the diagram, AC = BC, RCA and CBS are straight lines, ABS = 110 and BCR = x. Copy the diagram onto your writing sheet. Find the value of x giving reasons.† « x = 140 » 2U01-10a A NOT TO SCALE 1 D B 72 x 1 36 36 M x C In the diagram, ABC is an isosceles triangle where ABC = BCA = 72, AB = AC = 1 and BC = 2x. Angle BCA is bisected by CD and angle BAC is bisected by AM which is also the perpendicular bisector of BC. Copy the diagram onto your writing sheet. i. Show that AD = 2x. ii. Show that triangles ABC and CBD are similar. †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 165 M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT iii. iv. By using (ii), find the exact value of x. Hence find the exact value of sin 18.† « i) ii) Proof iii) x 44)! 2U02-4c B 45)! 1 5 1 5 iv) » 4 4 C A D H E NOT TO SCALE G F ABCDEFGH is regular octagon. i. Explain clearly why ABC is 135. ii. Calculate the size of GAH. iii. Using (i), or otherwise, calculate the size of CGF. iv. Hence, calculate the size of AGC.† « i) Proof ii) 225 iii) 675 iv) 45 » 2U02-9a ABCD is a rectangle and AE BD. AE = 5 cm and DE = 2 cm. A B i. ii. iii. E D C Copy the diagram and prove that triangles AED and BCD are similar. Hence, show that AD2 = BD·DE. Find the area of ABCD.† « i) ii) Proof ii) 725 cm2 » 46)! 2U03-3d In the diagram, PQRS is a parallelogram. QR is produced to U so that QR = RU. P Q S i. ii. 47)! T R NOT TO SCALE U Giving clear reasons, show that the triangles PST and URT are congruent. Hence, or otherwise, show that T is the midpoint of SR.† « Proof » 2U04-3a For what values of a, will ax2 + 5x + a be positive definitive?† †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 166 M_BANK\YR12-2U\APPLICATIONSOFGEOMETRICALPROPERTIES.CAT « a > †©CSSA OF NSW 1984 - 2006 ©EDUDATA SOFTWARE PTY LTD: DATA VER5.0 2006 167 5 » 2