Level 1: single concept, simple calculation, easier than HKDSE types Level 2: one or two concept, some calculations, similar to HKDSE easy types Level 3: involving high level, logical and abstract thinking skills, or with complicated calculations, similar to HKDSE difficult types ** Please answer any 30 questions ** Level 1 1. In the figure, ACD is a straight line. Find x. A. 40 B. 55 C. 85 D. 110 The answer is B. 140 = 30 + 2x 110 = 2x x = 55 2. In the figure, AE and BD intersect at C. Find z. A. 30 B. 50 C. 80 D. 100 The answer is C. AED = 30 In CDE, z = AED + 50 = 30 + 50 = 80 3. 1
In the figure, CBA and CDE are straight lines. It is given that AB = BD and BD // AF. Find y. A. 35 B. 50 C. 55 D. 70 The answer is A. AB = BD BAD = y DAE = BDA = y In ACE, 60 + ( CAD + DAE) = 130 60 + y + y = 130 2y = 70 y = 35 4. If each interior angle of a regular polygon is greater than its exterior angle by 135, find the number of sides of the regular polygon. A. 8 B. 12 C. 16 D. 20 The answer is C. Let n be the number of sides of the regular polygon. Sum of interior angles = (n 2) 180 All the interior angles of a regular polygon are equal. ( n 2) 180 Each interior angle = n Sum of exterior angles = 360 All the exterior angles of a regular polygon are equal. 360 Each exterior angle = n ( n 2) 180 n 360 n = 135 (n 2) 180 360 = n 135 n 180 360 360 = n 135 2
180n 135n = 720 45n = 720 n = 16 The number of sides of the regular polygon is 16. 5. S P 50 X R 60 Q In the figure, PQRS is a parallelogram. SN = RQ and PN intersects SR at X. Find NXR. A. 35 B. 55 C. 60 D. 85 The answer is D. SP = RQ (opp. sides of //gram) and SN = RQ (given) SP = SN SNP = SPN (base s, isos. ) PSR = PQR = 60 In SPN, PSN = PSR + NSR = 60 + 50 = 110 (opp. s of //gram) SNP + SPN + PSN = 180 ( sum of ) 2 SNP + 110 = 180 SNP = 35 In SNX, NXR = SNX + NSX (ext. of ) = 35 + 50 = 85 N 6. 3
A 9 B D E F 7 C In the figure, ABCD is a parallelogram. The angle bisectors of BAD and ABC meet DC at F and E respectively. If AB = 9 and EF = 7, find the perimeter of ABCD. A. 28 B. 30 C. 32 D. 34 The answer is D. DAF = BAF (given) BAF = DFA (alt. s, AB // DC) DAF = DFA AD = DF (sides opp. eq. s) CBE = ABE (given) ABE = CEB (alt. s, AB // DC) CBE = CEB CB = CE (sides opp. eq. s) AD = CB (opp. sides of //gram) DF = AD = CB = CE EF = DF + CE DC = DF + CE AB (opp. sides of //gram) 7 = 2DF 9 DF = 8 AD = BC = DF = 8 Perimeter of ABCD = AD + BC + AB + DC = 8 + 8 + 9 + 9 = 34 7. 4
A 6 Q D P S B R 8 9.6 C In the figure, P and Q are the mid-points of AB and AD respectively. BRC and DSC are straight lines. PQ = 6, RS = 8 and PQ // RS. If RC = 9.6, find BR. A. 3.2 B. 4.8 C. 6.4 D. 7.2 The answer is B. AP = PB and AQ = QD (given) 1 PQ = 2 BD BD = 2 6 = 12 PQ // RS (given) BD // RS CRS ~ CBD (AAA) 9.6 CB CR CB RS = BD 8 =12 9.6 12 CB = 8 = 14.4 BR = CB CR = 14.4 9.6 = 4.8 and PQ // BD (corr. sides, ~ s) (mid-pt. theorem) 8. In the figure, ADB and AEC are straight lines. If AB = BC, find x. 50 A. 50 B. 55 C. 60 D. 65 5
The answer is D. In ABC, AB = BC (given) BAC = BCA = x (base s, isos. ) AD = DB (given) AE = EC (given) DE // BC (mid-pt. theorem) AED = ACB = x (corr. s, DE // BC) In ADE, ADE + AED + DAE = 180 ( sum of ) 50 + x + x = 180 2x = 130 x = 65 9. In the figure, OA // CB and COA = 134. Find OAB. A. 46 B. 67 C. 80 D. 113 The answer is B. Reflex COA + COA = 360 ( s at a pt.) Reflex COA + 134 = 360 Reflex COA = 226 Reflex COA = 2 CBA ( at centre twice at circumference) 226 = 2 CBA CBA = 113 OAB + CBA = 180 (int. s, OA // CB) OAB + 113 = 180 OAB = 67 10. 6
In the figure, AD // BC and AEB = 60. Find ACB. A. 30 B. 45 C. 60 D. 70 The answer is A. Let ACB = x. DAE = ACB = x (alt. s, AD // BC) ADB = ACB = x ( s in the same segment) In ADE, AEB = ADB + DAE (ext. of ) 60 = 2x x = 30 ACB = 30 11. A B O 25 C In the figure, O is the centre. AB : = AOB : BOC (arcs prop. to s at centre) 7 3 BC = AB 2, and ACO = 25, find BOC. A. 77 B. 78 C. 79 D. 80 The answer is B. AO = OC (radii) OAC = OCA = 25 (base s, isos. ) OAC + OCA + AOC = 180 ( sum of ) BC 25 + 25 + AOC = 180 AOC = 130
Let AOB = x, then BOC = 3 x + x = 130 2 5 x = 130 2 x = 52 BOC = 130 52 = 3 x 2. 78 12. A 35 O B C In the figure, O is the centre. It is given that BAC = 35, find BOC. A. 35 B. 70 C. 80 D. 105 The answer is B. BOC = 2 BAC ( at centre twice at circumference) BOC = 2 35 = 70 13. A E O 84 29 B C D In the figure, O is the centre and AB // CD. Find BAD. A. 33.5 B. 39.5 C. 46.5 D. 57.5 The answer is A. 8
A O 29 B E 84 C D Join AC. BAD = BCD ( s in the same segment) ABC = BCD (alt. s, AB // CD) In ABD, ABE + EBD + BDA + BAD = 180 ( sum of ) ABE + 29 + 84 + BAD = 180 2 BAD = 67 BAD = 33.5 14. In the figure, O is the centre. If the sum of ABC and AOC is 258, find x. A. 16 B. 18 C. 20 D. 22 The answer is C. AOC = 2 ABC ( at centre twice at circumference) AOC = 2 (4x + 6 ) = 8x + 12 8x + 12 + 4x + 6 = 258 12x + 18 = 258 x = 20 15. 9
A 74 C O E b 39 B D In the figure, O is the centre. CEB is the diameter and AED is a straight line. Find b. A. 55 B. 58 C. 60 D. 62 The answer is A. CBD = CAD = 74 Since CEB is a diameter, CDB = 90 CDB = CDE + EDB = 90 39 + EDB = 90 EDB = 51 In EBD, ( s in the same segment) ( in semi-circle) EBD + BED + EDB = 180 ( sum of ) 74 + b + 51 = 180 b = 55 16. E A x B O 40 D C F In the figure, O is the centre, DE and DF are tangents to the circle at points A and C respectively. If BDF = 40, then x = A. 120. B. 130. C. 140. D. 150. The answer is B. ADO = CDO (tangent properties) 10
= 40 DAO = 90 x = AOB = DAO + ADO (ext. of ) =130 17. A O 106 B C D In the figure, O is the centre of the small circle, ABD is a straight line. ABC = 106, find BDC. A. 32 B. 53 C. 74 D. 95 The answer is A. Reflex angle AOC = 2 ABC ( at centre twice at circumference) = 212 AOC = 360 reflex angle AOC = 148 BDC = 180 AOC (opp. s, cyclic quad.) = 32 18. C 28 B A 43 T In the figure, TA is the tangent to the circle ABC. If ACB = 28 o and 11
ATB = 43 o, find ABC. A. 54 o B. 60 o C. 65 o D. 71 o The answer is D. BAT = ACB ( in alt. segment) = 28 o ABC = BAT + BTA (ext. of ) = 28 o + 43 o = 71 o 19. S B A 38 P T C In the figure, PAB is a straight line and ST is the tangent to the circle at A. If PAT = 38, then ACB = A. 32. B. 38. C. 46. D. 57. The answer is B. SAB = PAT (vert. opp. s) = 38 o ACB = SAB ( in alt. segment) = 38 o 20. B C O F A x D 140 E In the figure, the two circles ABCD and DCFE intersect at two points C and D. O is the centre of the circle DCFE and BCF is a straight line. If DOF = 140, find x. 12
A. 65 B. 70 C. 75 D. 80 The answer is B. B C O F A x D 140 E Join CD. 1 DCF = 2 DOF 1 = 2 (140 o ) = 70 o ( at centre twice at circumference) x = DCF (ext., cyclic quad.) = 70 o 13
Level 2 21. In the figure, CB is the angle bisector of ABE, AB // CD and EB = ED. Find EBD. A. 30 B. 40 C. 50 D. 60 The answer is B. DCB = 30 CB is the angle bisector of ABE. CBE = ABC = 30 EB = ED EBD = EDB (30 + CBE + EBD) + ( EDB + 40 ) = 180 30 + 30 + EBD + EBD + 40 = 180 2 EBD = 80 EBD = 40 22. In the figure, CDEF is a square, AF = AB = BF = FC and FBC = 55. Find FAE. A. 15 B. 20 C. 25 D. 30 14
The answer is B. AF = AB = BF AFB is an equilateral triangle. AFB = 60 FC = BF FCB = 55 In BFC, 55 + FCB + BFC = 180 55 + 55 + BFC = 180 BFC = 70 CDEF is a square. FC = FE and CFE = 90. AFB + BFC + CFE + AFE = 360 60 + 70 + 90 + AFE = 360 AFE = 140 FE = FC and FC = AF. FE = AF FEA = FAE In AFE, FAE + AFE + FEA = 180 FAE + 140 + FAE = 180 2 FAE = 40 FAE = 20 23. In the figure, ABCGH and CDEFG are two regular pentagons. Find AGD. A. 36 B. 72 C. 90 D. 108 The answer is D. Sum of interior angles of a regular pentagon = (5 2) 180 = 540 15
All the interior angles of a regular polygon are equal. 540 AHG = HGC = CGF = 5 ABCGH is a regular pentagon. AH = HG HAG = HGA In AHG, AHG + HAG + HGA = 180 108 + HGA + HGA = 180 2 HGA = 72 HGA = 36 Similarly, DGC = 36. AGC AGD = HGC HGA = 108 36 = 72 = AGC + DGC = 72 + 36 = 108 = 108 24. In the figure, ABCDE is a regular pentagon and PQR is an equilateral triangle. Find x + y. A. 176 B. 182 C. 188 D. 192 The answer is D. PQR is an equilateral triangle. PQR = 60 16
Sum of interior angles of a regular pentagon = (5 2) 180 = 540 All interior angles of a regular polygon are equal. 540 AED = 5 = 108 SQTE is a quadrilateral. x + SQT + y + TES = (4 2) 180 x + 60 + y + 108 = 360 x + y = 192 25. P 35 X Q Y 80 S R In the figure, PQRS and PXRY are parallelograms. SYXQ is a straight line. RQX is 5 greater than SPY. Find SPY. A. 30 B. 35 C. 45 D. 65 The answer is A. XPY = XRY = 80 (opp. s of //gram) In XPY, PYX = 180 XPY PXY ( sum of ) = 180 80 35 = 65 PSY = RQX (alt. s, PS // QR) In PSY, PSY + SPY = PYX (ext. of ) RQX + SPY = 65 ( SPY + 5 ) + SPY = 65 (given) 2 SPY + 5 = 65 SPY = 30 26. 17
P 108 8 cm Q S R In the figure, PQRS is a rhombus. PQR = 108 and PR = 8 cm. Find the area of PQRS, correct to 3 significant figures. A. 23.2 cm 2 B. 44.0 cm 2 C. 46.5 cm 2 D. 88.1 cm 2 The answer is A. Refer to the notations in the figure. P Q N S R PQN = RQN (property of rhombus) 1 108 = 2 = 54 PN = NR 1 8 = 2 cm = 4 cm In PNQ, (property of rhombus) PNQ = 90 (property of rhombus) PN tan PQN = NQ 4 NQ = tan 54 cm Area of PQRS = 4 area of PQN 1 4 4 4 = 2 tan54 cm 2 = 23.2 cm 2, cor. to 3 sig. fig. 27. 18
In the figure, AFED is a rectangle and ABCO is a square. AE and FD intersect at O. It is given that OCD = 72. Find x. A. 9 B. 18 C. 27 D. 36 The answer is B. OA = OD (property of rectangle) and OA = OC (by definition) OC = OD ODC = OCD = 72 (base s, isos. ) In OCD, ODC + OCD + COD = 180 ( sum of ) 72 + 72 + COD = 180 COD = 36 AOC = 90 (property of square) AOD = AOC + COD = 90 + 36 = 126 In OAD, OAD = ODA (base s, isos. ) OAD + ODA + AOD = 180 ( sum of ) 2 OAD + 126 = 180 OAD = 27 OAC = 45 (property of square) x + OAD x + 27 = 45 x = 18 = OAC 28. 19
In the figure, E and F are the mid-points of AB and DC respectively. AC intersects DE and FB at G and H respectively. It is given that DG = GE and DE // FB. Find FH : HB. A. 1 : 2 B. 1 : 3 C. 1 : 4 D. 1 : 5 The answer is C. CF = FD and DE // FB (given) CH = HG (intercept theorem) 1 FH = 2 DG (mid-pt. theorem) DG = 2FH AE = EB and DE // FB (given) AG = GH (intercept theorem) 1 GE = 2 HB (mid-pt. theorem) DG = GE 2FH = FH HB 1 2 HB 1 = 4 i.e. FH : HB = 1 : 4 (given) 29. In the figure, AC is a diagonal of rectangle ABCD. E is a point on AC such that DE AC. If AC = 4 cm and DC = 2 cm, then DE = A. 3 cm. B. 12 cm. C. 1 cm. D. 2 cm. 20
The answer is A. ADC = 90 (property of rectangle) In ACD, cos ACD 2 = 4 ACD = 60 In DCE, DC = AC DE sin DCE = DC DE sin 60 = 2 cm 3 2 DE = 2 cm = 3 cm 30. In the figure, OE = OF = 5 cm and OC = 13 cm. Find the length of AB. A. 12 cm B. 13 cm C. 24 cm D. 26 cm The answer is C. In OCF, OF 2 + CF 2 = OC 2 (Pyth. theorem) CF = = 2 OC OF 2 2 2 13 5 cm = 12 cm 21
OF CD (given) CF = FD (perpendicular from centre to chord bisects chord) CD = 2CF = 2 12 cm = 24 cm OE = OF (given) AB = CD (chords equidistant from centre are equal) = 24 cm 31. In the figure, BD is a diameter of the circle. ABD = 42 and CBD = 38. Find DEC. A. 76 B. 84 C. 86 D. 90 The answer is C. BCD = 90 ( in semi-circle) BCE + ACD = 90 BCE + 42 = 90 BCE = 48 ACD = ABD = 42 ( s in the same segment) In BCE, DEC = CBD + BCE (ext. of ) = 38 + 48 = 86 32. In the figure, O is the centre of the semi-circle ABCD. OB // DC and 22
AOE = 56. Find BAE. A. 28 B. 34 C. 56 D. 62 The answer is A. ADC = AOE = 56 (corr. s, OB // DC) ACD = 90 ( in semi-circle) In ACD, CAD + ACD + ADC = 180 ( sum of ) CAD + 90 + 56 = 180 CAD = 34 OA = OB (radii) OAB = OBA (base s, isos. ) In OAB, OAB + OBA + AOB = 180 ( sum of ) 2 OAB + 56 = 180 2 OAB = 124 OAB = 62 BAE = OAB CAD = 62 34 = 28 33. In the figure, AB = AC and BOC = 130. Find OCA. A. 32.5 B. 35 C. 65 D. 115 The answer is A. 23
Join OA. AB = AC (given) AOB = AOC (equal arcs, equal angles) AOC + AOB + BOC = 360 ( s at a pt.) 2 AOC + 130 = 360 2 AOC = 230 AOC = 115 OA = OC (radii) OAC = OCA (base s, isos. ) In AOC, OCA + OAC + AOC = 180 ( sum of ) 2 OCA + 115 = 180 2 OCA = 65 OCA = 32.5 34. In the figure, O is the centre. OC and AB intersect at a point D and AD = BD. Which of the following must be true? I. DC AB II. AC = BC III. OD = DC A. I only B. II only C. I and II only D. I, II and III The answer is C. I: AD = BD given 24
OD AB i.e. DC AB I must be true. line joining centre to mid-pt. of chord chord II: In CAD and CBD, AD = BD given CDA = CDB = 90 proved CD = CD common side CAD CBD SAS AC = BC corr. sides, s II must be true. III: There are not enough conditions to prove that OD = DC. III may not be true. Only I and II must be true. 35. In the figure, O is the centre. If BD = DC, AOB = 98 and BOC = 120, find ABD. A. 98 B. 101 C. 109 D. 128 The answer is B. Join OD. 25
BD = DC (given) BOD = DOC (equal arcs, equal angles) 1 BOD = 2 1 = 2 = 60 120 BOC In BOD, OB = OD (radii) OBD = ODB (base s, isos. ) OBD + ODB + BOD = 180 ( sum of ) 2 OBD + 60 = 180 OBD = 60 In AOB, OA = OB (radii) OAB = OBA (base s, isos. ) ABD OBA + OAB + AOB = 180 ( sum of ) 2 OBA + 98 = 180 OBA = 41 = OBA + OBD = 41 + 60 = 101 36. In the figure, ABP, DCP, BCQ and ADQ are straight lines. I is the in-centre of CDQ. BAD = 45 and IQD = 28. Find BPC. A. 28 B. 34 C. 40 D. 46 The answer is B. I is the in-centre of CDQ. 26
IQC = IQD = 28 In ABQ, PBC = BAD + CQD (ext. of ) = 45 + 28 + 28 = 101 CDA = PBC (ext., cyclic quad.) = 101 In APD, BPC + BAD + CDA = 180 ( sum of ) BPC + 45 + 101 = 180 BPC = 34 37. In the figure, DCQ is a straight line. ODC = 70 and BPC = 54. Find BCQ. A. 54 B. 74 C. 106 D. 108 The answer is B. Join AC. OC = OD (radii) OCD = ODC = 70 (base s, isos. ) In ODC, 27
DOC + OCD + ODC = 180 ( sum of ) DOC + 70 + 70 = 180 DOC = 40 DOC = 2 DAC ( at centre twice at circumference) 40 = 2 DAC DAC = 20 BAC = BPC = 54 ( s in the same segment) BCQ = BAD (ext., cyclic quad.) = BAC + DAC = 54 + 20 = 74 38. In the figure, ABC and AEOD are straight lines. BA = BD and BDC = 33. Find BAE. A. 19 B. 20 C. 33 D. 57 The answer is A. Join BE. Let BAE = x. BD = BA (given) BDE = BAE = x (base s, isos. ) ABE = CDE = CDB + BDE (ext., cyclic quad.) 28
= 33 + x EBD = 90 ( in semi-circle) In ABD, BAE + ABD + BDE = 180 ( sum of ) x + (33 + x) + 90 + x = 180 3x = 57 x = 19 BAE = 19 39. In the figure, CT touches the circle at C. OC // AB and OAB = 58. Find BCT. A. 29 B. 32 C. 58 D. 61 The answer is A. AOC + OAB = 180 (int. s, OC // AB) AOC + 58 = 180 AOC = 122 Reflex AOC + AOC = 360 ( s at a pt.) Reflex AOC + 122 = 360 Reflex AOC = 238 Reflex AOC = 2 ABC ( at centre twice at circumference) 238 = 2 ABC ABC = 119 OCB + ABC = 180 (int. s, OC // AB) OCB + 119 = 180 OCB = 61 OCT = 90 (tangent radius) BCT + OCB = 90 BCT + 61 = 90 BCT = 29 29
40. In the figure, TA is the tangent to the circle at A. OT cuts the circle at B. OB = 20 and BT = 9. Find the length of TA. A. 20 B. 21 C. 22 D. 23 The answer is B. OA TA (tangent radius) OA = OB = 20 (radii) In OAT, OA 2 + TA 2 = OT 2 (Pyth. theorem) TA = 2 OT OA 2 = = 21 2 ( 20 + 9) 20 2 30
Level 3 41. D C E N G B In the figure, BCDG is a square and DEFG is a rhombus. CGF is a straight line. CE and DG intersect at N. Find DG : DN. A. 2 : 1 B. (1 + 2 ) : 1 C. ( 2 1) : 1 D. 2 : 1 The answer is B. In DGC, DG cos DGC = GC DG cos 45 = GC (property of square) DG 1 GC = 2 DEN ~ GCN (AAA) DN DE GN = GC DG DN F DG = GC 1 = 2 (corr. sides, ~ s) (by definition) DN + GN = DN GN 1+ = DN = 1+ 2 i.e. DG : DN = ( 1+ 2) :1 31
42. Q P A B R S In the figure, AQBS is a parallelogram, PQ = SR and PQA = RSB. Which of the following may not be true? A. PA = RB B. QR // PS C. QS bisects PSR. D. APS = QRB The answer is C. For A: In PQA and RSB, PQ = RS PQA = RSB AQ = BS given given opp. sides of //gram PQA RSB SAS PA = RB corr. sides, s For B: AQS = BSQ alt. s, AQ // SB and PQA = RSB given PQA + AQS = RSB + BSQ PQS = RSQ alt. s equal PQ // SR given PQ = SR 2 sides equal and // PQRS is a parallelogram. by definition QR // PS For C: There is no sufficient information to show that QS bisects PSR For D: PQA RSB proved QPA = SRB corr. s, s PQRS is a parallelogram. proved QPS = QRS opp. s of //gram 32
QPS QPA = QRS SRB APS = QRB Only C may not be true. 43. In the figure, ABCD is a square, where AC and BD intersect at O. E and F are points on OB and OC respectively such that BE = CF. DF is produced to meet CE at G. Which of the following are true? I. CDF = BCE II. DG CE III. COE ~ CGF A. I and II only B. I and III only C. II and III only D. I, II and III The answer is D. For I: In CDF and BCE, CF = BE DCF = CBE = 45 CD = BC given property of square by definition CDF BCE SAS CDF = BCE corr. s, s For II: BCE + DCG = BCD BCE + DCG = 90 property of square BCE = CDF proved in I CDF + DCG = 90 In DCG, DGC = 180 CDF DCG sum of = 180 90 = 90 DG CE For III: 33
In COE and CGF, COE = 90 CGF = 90 COE = CGF OCE = GCF OEC = 180 COE OCE GFC = 180 CGF GCF = 180 COE OCE OEC = GFC COE ~ CGF I, II and III are true. property of square proved in II common angle sum of sum of AAA 44. 3 In the figure, ABEF and BCDE are squares. P and Q are points on BE and CD respectively such that APQ is a straight line. AF = 3 and CQ = 2. Find the ratio of the area of trapezium BCQP to that of trapezium APEF. A. 1 : 2 B. 2 : 3 C. 3 : 5 D. 4 : 9 The answer is C. ABEBCD = 90 (property of square) BE // CD (corr. s equal) AB = BE = BC (by definition) AP = PQ (intercept theorem) 1 BP = 2 CQ (mid-pt. theorem) 1 (2) = 2 = 1 PE = 3 1 = 2 ( 1+ 2) 3 Area of trapezium BCQP = 2 = 4.5 34
( 2 + 3) 3 Area of trapezium APEF = 2 = 7.5 The required ratio = 4.5 : 7.5 = 3 : 5 45. P S T Q N R In the figure, PSQ, QNR and PTR are straight lines. PS = PT and SN = TN. Which of the following may not be true? I. PN is an angle bisector of PQR. II. PN is an altitude of PQR. III. PN is a perpendicular bisector of PQR. A. I only B. II only C. I and III only D. II and III only The answer is D. For I, PS = PT SN = TN PN = PN given given common side PSN PTN SSS SPN = TPN corr. s, s PN is an angle bisector of PQR. I must be true. For II and III, There is no sufficient information to prove that PN is an altitude or a perpendicular bisector of PQR. II and III may not be true. Only II and III may not be true. 46. 35
B C P A In the figure, APC is a straight line and BP is an altitude of ABC. It is known that BP passes through the centroid of ABC. Which of the following must be true? I. BP also passes through the orthocentre of ABC. II. BP also passes through the in-centre of ABC. III. The circumcentre of ABC lies on BP. A. II only B. III only C. I and II only D. I and III only The answer is A. For II, In APB and CPB, BP is an altitude of ABC. given APB = CPB = 90 BP passes through the centroid. given BP is a median and AP = CP. BP = BP common side APB CPB SAS ABP = CBP corr. s, s BP is an angle bisector of ABC and hence it passes through the in-centre of ABC. II must be true. For I and III, The orthocentre and circumcentre may lie outside ABC. They may not lie on BP. I and III may not be true. Only II must be true. 47. 36
In the figure, ABCD, GFC and EFB are straight lines. GB = EC, GBC = ECB and GAB = EDC = 90. How many pairs of congruent triangles are there in the figure? A. 2 B. 3 C. 4 D. 5 The answer is C. In GBC and ECB, GB = EC BC = CB GBC = ECB given common side given GBC ECB SAS In GFB and EFC, GBC ECB proved BGF = CEF corr. s, s GFB = EFC GB = EC vert. opp. s given GFB EFC AAS In GAB and EDC, GB = EC ABG = 180 GBC DCE = 180 ECB given adj. s on st. line adj. s on st. line GBC = ECB given ABG = DCE GAB = EDC = 90 given GAB EDC AAS In GCA and EBD, GBC ECB proved GC = EB corr. sides, s GAB EDC proved GA = ED corr. sides, s GAB = EDC = 90 given GCA EBD RHS There are 4 pairs of congruent triangles in the figure. 48. 37
In the figure, S, Q and U are points on RT, PR and PT respectively. PS, RU and TQ intersect at V. RPT = 60, RVT = 120, RUT = 83 and VTR = 37. Which of the following is/are angle bisector(s) of PRT? I. RU II. QT III. PS A. I only B. II only C. I and II only D. I, II and III The answer is D. I: In PUR, PRU + RPU PRU + 60 = 83 PRU = 23 In RVT, = RUT VRT + RVT + VTR = 180 VRT + 120 + 37 = 180 VRT = 23 PRU = VRT = 23 RU is an angle bisector of PRT. ext. of sum of II: In UVT, VTU + VUT = RVT VTU + 83 = 120 VTU = 37 QTR = VTU = 37 QT is an angle bisector of PRT. ext. of III: V is the point of intersection of the angle bisectors RU and QT. V is the in-centre of PRT. PS passes through V. PS is an angle bisector of PRT. All of RU, QT and PS are angle bisectors of PRT. 38
49. In the figure, O is the centre and BCD = 120. If AOD. AB = 2 BC = 3 CD, find A. 60 B. 96 C. 110 D. 120 The answer is B. Join OB and OC. Let COD = x. 2 BC = 3CD BC CD 3 = 2 BOC : COD = BOC x 3 = 2 3x BOC = 2 BOD = BOC + COD 3x = 2 + x 5x = 2 BC : CD (arcs prop. to s at centre) 39
Reflex BOD = 2 BCD = 2 120 = 240 Reflex BOD + BOD = 360 5x 240 + 2 = 360 5x 2 = 120 x = 48 AB = 3CD AB CD = 3 AOB : COD = AOB x = 3 AOB = 3x AOD = 3 48 = 144 ( at centre twice at circumference) AB : = reflex BOD AOB = 240 144 = 96 ( s at a pt.) CD (arcs prop. to s at centre) 50. In the figure, AB is a diameter of the circle. If BAC = x, which of the following must be true? I. BC AB x = 90 II. BC AB = sin x III. AC AB = cos x 40
A. I only B. II only C. I and II only D. II and III only The answer is C. I: ACB = 90 ( in semi-circle) BC : AB = BAC : ACB (arcs prop. to s at circumference) II: BC AB x = 90 I must be true. BC sin x = AB II must be true. AC III: cos x = AB AC AB may not be equal to III may not be true. Only I and II must be true. AC AB, 51. A D B P O In the figure, O is the centre of the circle and OD AP. If AB = 8 cm, BP = 10 cm and OP = 15 cm, find the radius of the circle. A. 5 cm B. 13 cm C. 17 cm D. 3 5 cm The answer is D. 1 DB = 2 AB chord) (perpendicular from centre to chord bisects 41
1 = 2 (8) cm = 4 cm DP = DB + BP = 4 + 10 = 14 In ODP, by Pythagoras theorem, OD 2 = OP 2 DP 2 OD = 2 2 15 14 cm = 29 cm In ODB, by Pythagoras theorem, OB 2 = OD 2 + DB 2 OB = 2 29 + 4 cm = 45 cm = 3 5 cm The radius of the circleis 3 5 cm. 52. A B C D E In the figure, DC is the diameter of the circle BDC. ABC and EDC are straight lines. EBA + DAE = A. 45. B. 60. C. 90. D. 120. The answer is C. DC is a diameter, CBD = 90 o ( in semi-circle) 42
DAE = DBE ( s in the same segment) EBA + DAE = EBA + DBE = 180 o CBD (adj. s on st. line) = 180 o 90 o = 90 o 53. In the figure, AT touches the circle at A. BAT = 27. If the circumference of the circle is 40 cm, find the length of AB. A. 6 cm B. 7 cm C. 8 cm D. 9 cm The answer is A. OAT = 90 (tangent radius) OAB + BAT = 90 OAB + 27 = 90 OAB = 63 OB = OA (radii) OBA = OAB = 63 (base s, isos. ) In OAB, AOB + OBA + OAB = 180 ( sum of ) AOB + 63 + 63 = 180 AOB = 54 AB : circumference = AOB : 360 (arcs prop. to s at centre) AB 54 40 cm = 360 AB = 6 cm 43
54. In the figure, AC is a diameter of the circle. AT touches the circle at A and CT cuts the circle at B. D is a point lying on AT such that DB // AC. DT = 4.5, DB = 6 and BC = 12.5. Find the radius of the circle. A. 6 B. 7.5 C. 8 D. 16 The answer is C. AT AC (tangent radius) TDB = TAC = 90 (corr. s, DB // AC) In BDT, TB 2 = DT 2 + DB 2 (Pyth. theorem) TB = = 2 DT + 2 4.5 + DB 6 2 2 = 7.5 In BDT and CAT, TDB = TAC (proved) DTB = ATC (common angle) BDT ~ CAT (AAA) AC DB AC 6 TC = TB (corr. sides, ~ s) 7.5 + 12.5 = 7. 5 AC = 16 1 Radius of the circle = 2 AC 1 = 2 16 44
= 8 55. In the figure, CD is a diameter of the circle. BAC = 48, ABD = 158 and BDC = 68. Which of the following is/are true? I. AB is the tangent to the circle at B. II. AC is the tangent to the circle at C. III. AB = AC A. I only B. II only C. I and III only D. I, II and III The answer is A. I: Join OB. OB = OD (radii) OBD = ODB = 68 (base s, isos. ) OBA = ABD OBD = 158 68 = 90 AB is the tangent to the circle at B. (converse of tangent radius) I is true. II: In quadrilateral ABDC, ACD + BAC + ABD + BDC = (4 2) 180 ( sum of polygon) ACD + 48 + 158 + 68 = 360 ACD = 86 ACD 90 AC is not the tangent to the circle at C. 45
II is not true. III: AC is not the tangent to the circle at C. AB AC III is not true. Only I is true. 56. In the figure, a circle is inscribed in ABC, where X, Y and Z are points of contact. AB = 16 cm, AC = 34 cm and ABC = 90. Find the length of AZ. A. 6 cm B. 10 cm C. 24 cm D. 30 cm The answer is B. In ABC, AB 2 + BC 2 = AC 2 (Pyth. theorem) BC = = 2 AC AB 2 2 2 34 16 cm = 30 cm Let AZ = x cm. CZ = AC AZ = (34 x) cm CY = CZ = (34 x) cm (tangent properties) AX = AZ = x cm (tangent properties) BX = AB AX = (16 x) cm BY = BX = (16 x) cm (tangent properties) BY + CY = BC (16 x) + (34 x) = 30 2x = 20 x = 10 46
The length of AZ is 10 cm. 57. In the figure, a circle is inscribed in ABC, where X, Y and Z are points of contact. AX = 14 cm, CZ = 15 cm and ABC = 90. Find the radius of the circle. A. 6 cm B. 7 cm C. 8 cm D. 10 cm The answer is A. Let O be the centre of the circle and x cm be the radius of the circle. Join OX and OY. OX = OY (radii) ABC = 90, i.e. AB BC (given) OX AB and OY BC (tangent radius) OXBY is a square. BY = BX = OX = x cm AZ = AX = 14 cm (tangent properties) CY = CZ = 15 cm (tangent properties) In ABC, AC 2 = AB 2 + BC 2 (Pyth. theorem) (AZ + CZ) 2 = (AX + BX) 2 + (BY + CY) 2 (14 + 15) 2 = (14 + x) 2 + (x + 15) 2 841 = 196 + 28x + x 2 + x 2 + 30x + 225 2x 2 + 58x 420 = 0 x 2 + 29x 210 = 0 (x 6)(x + 35) = 0 x = 6 or 35 (rejected) 47
The radius of the circle is 6 cm. 58. In the figure, PQ and QR are the tangents to the circle at two points A and B respectively. If BC = CD, BAD = 76 and ADC = 106, find AQB. A. 30 B. 40 C. 44 D. 53 The answer is C. Join AC. CD = BC (given) DAC = CAB (equal chords, equal angles) 1 i.e. CAB = 2 CBR 1 = 2 76 = 38 BAD = CAB = 38 ( in alt. segment) ABC + ADC = 180 (opp. s, cyclic quad.) ABC + 106 = 180 ABC = 74 QBA + ABC + CBR = 180 (adj. s on st. line) QBA + 74 + 38 = 180 48
QBA = 68 QA = QB (tangent properties) QAB = QBA (base s, isos. ) In ABQ, = 68 AQB + QBA + QAB = 180 ( sum of ) AQB + 68 + 68 = 180 AQB = 44 59. In the figure, O is the centre of the circle. ADO is a straight line. AC is the tangent to the circle at point C. B is a point lying on AC such that AB = BC. CED = 5 DC. Which of the following must be true? I. OCD is an equilateral triangle. II. AD = OD III. AD = 2DB A. II only B. I and III only C. II and III only D. I, II and III The answer is D. I: DC : circumference = DC : ( CED + DC ) = DC : (5 DC + DC ) = DC : 6 DC = 1 : 6 arcs prop. to s at centre radii base s, isos. 49
DOC 360 DOC : 360 = 1 = 6 DOC = 60 OD = OC ODC = OCD In OCD, DC : circumference OCD + ODC + DOC = 180 2 OCD + 60 = 180 2 OCD = 120 OCD = 60 DOC = OCD = 60 OD = DC OD = OC = DC OCD is an equilateral triangle. I must be true. sum of sides opp. eq. s II: OCA = 90 In OAC, OC cos AOC = OA OC cos 60 = OA tangent radius OA = OC cos 60 = 2OC AD = OA OD = 2OC OD = 2OD OD = OD II must be true. III: In OAC, OC = OD = AD AD = OD and AB = BC OC = 2DB mid-pt. theorem 50
i.e. AD = 2DB III must be true. Alternative method: DC = OD = AD DC = AD and AB = BC DB AC In OAC, OAC + OCA + DOC = 180 OAC + 90 + 60 = 180 OAC = 30 In ABD, BD sin OAC = AD BD sin 30 = AD axis of symmetry of isos. base sum of AD = BD sin 30 = 2BD III must be true. I, II and III must be true. 60. In the figure, AEC and BED are straight lines. AB // DC and DE = CE. Which of the following must be true? I. A, B, C and D are concyclic. II. AE = DE III. ADC = BCD A. I only B. III only C. I and III only D. II and III only The answer is C. 51
I: DE = CE BDC = ACD base s, isos. ABD = BDC alt. s, AB // DC = ACD A, B, C and D are concyclic. converse of s in the same I must be true. segment II: There are not enough conditions to prove that AE = DE. II may not be true. III: ABC + BCD = 180 int. s, AB // DC ABC = 180 BCD ABC + ADC = 180 opp. s, cyclic quad. ADC = 180 ABC = 180 (180 BCD) = BCD III must be true. BDC CBD BCD BCD Alternative method: In ADC and BCD, ACD = DAC = CD = DC ADC ADC = III must be true. proved s in the same segment common side AAS Only I and III must be true. corr. s, s 52