SHW 101 Total: 30 marks


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1 SHW 0 Total: 30 marks 5. 5 PQR 80 (adj. s on st. line) PQR 55 x x In XYZ, a a 50 In PXY, b M+ 7. AB = AD and BC CD AC BD (prop. of isos. ) y 90 BD = ( + ) = AB BD DA x 60 (prop. of equil. ) M+ 8. DB = AB ADB 0 DBC ADB DAB DB = DC (given) x 40 M+ 9. BCD 65 alt. s, AB // CD ABC CBD BDC 80 int. s, AB // CD BDC 80 BDC 65 BCD BDC 65 BC = BD sides opp. equal s BCD is an isosceles triangle. 3A 0. (a) In ABE, (b) AB BE AE ( 6 8 ) 0 00 In ACD, (6 3) 00 AB AE BE ABE is a rightangled triangle, where CAD 90. (converse of Pyth. theorem) (8 x) x x x AC AD 5 5 6x 80 0 ( x 4)( x 0) 0 CD x 4 or x 0 (rejected) +A 3. (a) In ABC, (b) AC BC BC AB AB AC Consider BDC and BCA. BDC BCA = 90 (given) CBD ABC (common angle) In BDC, DCB CBD ( sum of ) 90 ABC In BCA, CAB ABC ( sum of ) 90 ABC DCB CAB BDC ~ BCA (AAA) BD BC BC BA (corr. sides, ~ s) BD BD 6 A+ 4. (a) Consider ABD and AED. DE is an angle bisector of ADC. ADE CDE AD = DC and ADE CDE DE AC prop. of isos. AED 90 ABD BD = DE given AD = AD common side ABD AED RHS (b) ABD AED ADB ADE (corr. s, s) CDE ADE ADB ADE CDE 80 (adj. s on st. line) 3ADE 80 ADE 60 3A
2 SHW A Total: 30 marks. AB (given) AM MB (line from centre chord bisects 3 chord) OA x AM. AM = MB (given) M+ AMO = 90 (line joining centre to midpt. of chord chord) AM 6 8 OA x AM M+ 3. BM = MD = 5 and AC BD The centre lies on the perpendicular bisector of the chord BD. AC is a diameter of the circle. AC 6 4. (a) AD (given) AM MD (line from centre chord bisects chord) 8 4 OA AM The radius of the circle is 5. M+ (b) OB = OA = 5 (radii) BN = NC (given) ON BC (line joining centre to midpt. of chord chord) In OBN, BN OB ON 5 BC BN +A 5. (a) BM = MC (given) BM AMB = 90 (line joining centre to midpt. of chord chord) In ABM, AB AM (b) (i) (ii) AM AO ( 5 r) In OBM, OB = + BM r (5 r) 8 r 5 30r r 30r 89 r 9.63 (cor.to d.p.) (a) BM = MC (given) (b) M+ +A CMO = 90 (line joining centre to midpt. of chord chord) In OCM, OC CM The length of is 4. AM (5 ) 4 33 AB AM BM AO ( 33 3).74 (cor.to d.p.) The length of AB is.74. M+M=A
3 7. (a) AM = MD (given) AMO = 90 (line joining centre to = BNO midpt. of chord chord) AD // BC (corr. s equal) A+A (b) (i) AM 4 OA 4 60 AM.65 (cor. to d.p.) Hence, the radius of the circle is.65. A (ii) ON BC (given) BN NC (line from centre chord bisects chord) 8 9 In OBN, ON OB BN ( 60) 9 79 MN = ON ( 79 4) 4.89 (cor. to d.p.) The length of MN is
4 SHW B Total: 5 marks. BAC ( at centre twice at ce ) 56 8 In ABD, x BAC 80 x 8 80 x 5. BAC 90 In ABC, x x In CDE, BDC 0 38 (ext. of ) 7 x BDC 7 4. ABC 90 In ABC, BAC BAC 47 BD BA (given) BDA BAC x ADB ACB 8 ADC 90 8 BDC 90 BDC 7 6. BOD ( s at a pt.) BOD 5 DCB BOD ( at centre twice at ce ) 5 y 76 In PBC, ( x 4) x 7 +A 7. BAC ( at centre twice at ce ) 6 OB = OC (radii) OBC OCB In OBC, OBC OCB 80 OBC 58 OBC 9 AC = AB (given) ACB ABC x 9 In ABC, ( x 9) ( x 9) 6 80 x 30.5 M++A 8. BOC BAC ( at centre twice at ce ) 5 04 OB = OC (radii) OBC OCB In OBC, OBC OCB OBC 38 BCD 90 OCD OCB 90 OCD M++A 9. AOE ABE ( at centre twice at ce ) AOE 58 (alt. s, OA // DE) ABE 58 ABE 9 BAD BED 58 In ABP, 58 (9 EBP ) EBP 45 M++A 7
5 SHW C Total: 5 marks. AB BC (given) AB BC (equal chords, equal arcs) AB AC 30 x x x AOD BC BOC x x 5 AD AD DC BC BC DC (arcs prop. to s at centre) (given) AC BD AC = BD 3 (equal arcs, equal chords) 7. Join BD. ADB 45 (property of square) AE ADB AB ADE 45 ADE.5 In ADF, x.5 90 ADE (arcs prop. to s at ce ).5 4. ABC is an equilateral triangle. ACB = ABC =BAC (prop. of equil. ) AB AC BC 7 (arcs prop. to s at ce ) Major arc BC 7 54 BDC BC 5. BAD (arcs prop. to s at ce ) BCD BDC 88 BDC 44 CD BC CBD BDC 44 (arcs prop. to s at ce ) In BCD, BCD BCD 9 +A 6. AB BC CD ACB BAC CBD 4 (arcs prop. to s at ce ) BDC BAC 4 In BCD, ( x 4) x 54 0
6 SHW D Total: 5 marks. ABC ADC 80 x ( x 4) 80 x 76 x 88. BAD DCE 4x 3x 0 x 3 3. CDE ABC 48 In DCE, DCB CDE DEC AD DC ABD DBC (arcs prop. to s at ce ) 4 48 ADC ABC 80 ADC (48 4) 80 ADC Join AD. AB = BC = CD AB BC CD (equal chords, equal arcs) BCA BAC CAD 8 (arcs prop. to s at ce ) BCD BAD 80 ( ACD 8) (8 8) 80 ACD 96 AED ACD 80 AED AED 84 M+ 5. Join AC. Consider ABC and ADC. AB = AD (given) BC = DC (given) AC = AC (common side) ABC ADC (SSS) ABC ADC (corr. s, s) ABC ADC 80 ABC 80 ABC 90 M+ (for the SSS) +A 3
7 SHW E Total: 5 marks. ACB ADB 4 A, B, C and D are concyclic. (converse of s in the same segment). BAD DCE A, B, C and D are not concyclic. 3. In BCD, BCD BCD 86 BCD EAD 86 A, B, C and D are concyclic. (ext. = int. opp. ) 4. (a) EBD ECD 43 B, C, D and E are concyclic. (converse of s in the same segment) (b) EBC ADE 43 DBC 80 DBC Join OA and OB. AB CD (given) ADB CAD (arcs prop. to s at ce ) In ADE, AEB CAD ADB ADB AOB ADB ( at centre twice at ce ) AEB A, B, E and O are concyclic. (converse of s in the same segment) +A 5. In ADF, DFB 9 9 In CDE, DCE 33 9 DCE 59 DFB DCB B, C, D and F are concyclic. (opp. s supp.) +A 6. (a) CAB FDB 43 A, F, C and D are concyclic. (converse of s in the same segment) (b) ACB 90 ECD (adj. s on st. line) 90 AFE ECD 90 AFE ACB 90 E, F, B and C are concyclic. (ext. = int. opp. ) +A 7. (a) EDB ECB 0 B, C, D and E are concyclic. (converse of s in the same segment) (b) EBC ADE and AED ADE AED EBC ED // BC (corr. s equal) 6
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