2 Basic Properties of Circles

Size: px
Start display at page:

Download "2 Basic Properties of Circles"

Transcription

1 Basic Properties of Circles Basic Properties of Circles Review Exercise (p..7 AC CE DC CB ACE DCB ACE ~ DCB common angle ratio of sides, inc.. (a. (a x ( s at a pt. x 75 (b CBP 60 (corr. s, PQ // RS (b y CBP 60 PQR 55 (vert. opp. s PQR + 5 (adj. s on st. line x + PQR PRB (ext. of x x 85 a ( sum of a 50 b a + 4 (ext. of. Yes, EF is parallel to GH. RSQ 0 (corr. s, AB // CD PQS + RSQ EF // GH (int. s supp. 4. BCD 65 alt. s, AB // CD BCD + BDC + 50 sum of BDC BCD BDC 65 BD BC sides opp. equal s BCD is an isosceles triangle. 5. PQR is an equilateral triangle. PRQ 60 PR is the median of QS in PQS. RS QR PR RPS RSP (base s, isos. RPS + RSP PRQ (ext. of RPS 60 RPS 0 6. (a In ACE and DCB, AC + DC CE CB (b ACE ~ DCB AE AC DB DC AE 5.5 AE (a In ABC, C 90 AC + BC (b In CAD, AC CD + AD 6 AC AB ( proved in (a corr. sides, ~ s (Pyth. theorem 676 AC + AD CD CAD is a right angle. (converse of Pyth. theorem (c Area of quadrilateral ABCD area of ABC + area of CAD ( (a In ABC and ACD, BCA CDA 90 CAB DAC common angle 80 BCA CAB sum of 80 CDA DAC proved ACD sum of ABC ~ ACD AAA (b (i In ABC, ACB 90 AB AC + BC (Pyth. theorem AB

2 NSS Mathematics in Action 5A Full Solutions (ii ABC ~ ACD AB AC (corr. sides, ~ s AC AD AD 64 AD 7 BD AB AD (cor. to sig. fig. 9. (a In ADE and CDE, DEA DEC 90 AD CD DE DE ADE CDE common side RHS (b Yes, ABD and AED are congruent. ACB ( sum of ACB ACB 0 ADE CDE (proved in (a EAD ECD (corr. s, s Activity 0 BAD In ABD and AED, ABD AED 90 BAD EAD 0 AD AD ABD AED proved common side AAS Activity. (p.. AOB. (b APB (c No matter where points B and P are, AOB APB. Reflex AOB. (b AQB (c No matter where points B and Q are, reflex AOB AQB. Activity. (p..8. (a AOB (i.e. c is the angle at the centre subtended by arc AB. (b APB and AQB are the angles at the circumference in the same segment. They are both subtended by arc AB.. (a APB (b AQB. APB AQB Activity. (p..4. (a Yes (b Yes (c Yes. Yes c c ( at centre twice at ce ( at centre twice at ce Activity.4 (p..46. (b A + C, B + D. P + R, Q + S. The sum of the opposite angles of a cyclic quadrilateral is 80. Activity.5 (p.. 5. (a Yes (b Yes.. Yes Classwork Classwork (p... Element Term AB minor arc region BEC major arc AFB chord region OBFC major sector region AFCEB major segment AB minor segment region OBEC minor sector. (a The two circles with the same centre are concentric circles. (b The circle is the circumcircle of ABC. (c The circle is the inscribed circle of PQR. 8

3 Basic Properties of Circles Classwork (p..4 (a x APB ( at centre twice at ce (b 0 60 x AOB ( at centre twice at ce (c Reflex AOB ( s at a pt. 0 x reflex AOB ( at centre twice at ce 0 0 Classwork (p..5. (a x 90 ( in semi-circle (b APB 90 ( in semi-circle APB x ( sum of x x 50. OA OP (radii OAP OPA (base s, isos. OPA 55 APB The line segment joining A and B is a diameter of the circle. (converse of in semi-circle Classwork (p..6 (a DOC AOB 4 ( DC AB (equal s, equal arcs x 4 (b CD AB ( CD AB (equal arcs, equal chords x 5 (c AB DC ( x 65 (equal chords, equal s Classwork (p..9 (a x BC AOB AB x (80 5 x (arcs prop. to s at centre (b (c (d y DOC (arcs prop. to s at centre AB AOB 48 y 5 80 x DOC (arcs prop. to s at centre AB AOB 50 x BOC AOB DOC (adj. s on st. line BOC y (arcs prop. to s at centre AOB AB 00 y x CED (arcs prop. to s at ce BC BEC 0 x y AB (arcs prop. to s at ce BEC BC 6 y (40 0 y 4 ACB AB (arcs prop. to s at ce DBC CD ACB ( x DBC ACB ( sum of 48 6 x 96 Classwork (p..47 (a DAB + BCD (opp. s, cyclic quad. x + 60 x 0 + CDA (opp. s, cyclic quad. y + 80 y 00 9

4 NSS Mathematics in Action 5A Full Solutions (b (c DAB + BCD (opp. s, cyclic quad. 9 + x y x 87 CDE (ext., cyclic quad. y 46 + (ext. of 78 x y (ext., cyclic quad. 78 Classwork (p..55. (a BAD + BCD (b (c A, B, C and D are not concyclic. + FBC (adj. s on st. line ABC ADE A, B, C and D are not concyclic. ABD ( sum of ABD 0 + ADC ( ( A, B, C and D are concyclic. (opp. s supp.. (a PSQ PRQ 50 PQRS is a cyclic quadrilateral. converse of s in the same segment PQR + RSP (opp. s, cyclic quad. (b x + ( x 60 QPS QRK QPS 90 PQRS is a cyclic quadrilateral. ext. int. opp. PRQ + PRS + QRK (adj. s on st. line PRQ PSQ PRQ x 50 Quick Practice Quick Practice. (p.. Join OQ. OQ (radius PRQ 50 ( s in the same segment In OQN, ON + NQ OQ NQ OQ ON ON PQ PN NQ PQ PN + NQ NQ ( 4 5 Quick Practice. (p..4 PN QN ( ON PQ (Pyth. theorem ( (line from centre chord bisects chord (line joining centre to mid-pt. of chord chord ONP 90 In OPN, ON PN OPN PON (base s, isos. OPN + PON + ONP ( sum of OPN + 90 OPN 45 Quick Practice. (p..4 PN NR ( ON PR (line joining centre to mid-pt. of chord chord OPN is a right-angled triangle. Let r be the radius of the circle. OP r and ON (r In OPN, OP ON + PN (Pyth. theorem r r ( r + 5 r r 6 r r The radius of the circle is. ON ( Quick Practice.4 (p..7 M and N are the mid-points of PQ and RQ respectively. OM PQ and ON RQ (line joining centre to mid-pt. of chord chord PQ PM + MQ ( + 6 ON OM RQ PQ 6 ( (chords equidistant from centre are equal 0

5 Basic Properties of Circles RN RQ DBC AC is a diameter of the circle. converse of in semi-circle Quick Practice.5 (p..7 (a AB CD, OM AB and ON CD OM ON equal chords, equidistant from centre MPN APC 90 vert. opp. s MON 90 sum of polygon All four interior angles are equal to 90 and two adjacent sides are equal. ONPM is a square. (b ONPM is a square. (proved in (a PM OM 4 OM AB ( AM MB (lines from centre chord bisects chord AP + PM PB PM AP + 4 ( 4 AP Quick Practice.6 (p..6 reflex AOC 60 8 ( s at a pt. 4 x reflex AOC ( at centre twice at ce x 4 x y AOC ( at centre twice at ce y 8 y 59 Quick Practice.7 (p..7 CAD 90 ( in semi-circle AB AD ( ABD ADB y (base s, isos. In ABD, ABD + ADB + BAD ( sum of y + y + ( y 50 y 5 In ACD, CAD + ACD + ADC ( sum of 90 + x + 5 x 65 Quick Practice.9 (p..9 BDC ( s in the same segment BCD 90 ( in semi-circle In ABC, BCA 0 ( sum of 8 ACD Quick Practice.0 (p..0 DEB DAB ( s in the same segment x AEB 90 ( in semi-circle In ACE, ECA + CAE + AEC ( sum of 4 + ( x (90 + x Quick Practice. (p..7 x + 64 x 6 x 8 Join OB, OC, OE, OF and OG. Let AOB x. AB BC CD DE EF FG GH ( AOB BOC COD DOE EOF FOG GOH x (equal chords, equal s AOB + BOC + COD 0 x 0 x 40 AOB + BOC + COD + DOE + EOF + FOG + GOH + AOH 60 ( s at a pt. 7x + AOH 60 7(40 + AOH 60 AOH 80 Quick Practice.8 (p..7 DOC DBC ODB DBC In ODP, ODP + DOP 90 DBC + DBC 90 DBC 90 DBC 0 at centre twice at ce alt. s, OD // BC ext. of

6 NSS Mathematics in Action 5A Full Solutions Quick Practice. (p..7 OP OQ OR, OP AB, OQ BC and OR AC AB BC AC chords equidistant from centre are equal AB BC AC equal chords, equal arcs Quick Practice. (p..40 Let COD x. BC CD ( BOC COD x (equal chords, equal s AOD AD (arcs prop. to s at centre BOC BC 0 AOD x 7 AOB + BOC + COD + DOA 60 ( s at a pt. Quick Practice.4 (p..4 ACB AB DBC CD ACB x 0 + x + x + x x 8 7 x 66.5 (arcs prop. to s at ce In BCK, BCK + CBK AKB (ext. of x + x 75 5 x 75 x 0 Quick Practice.5 (p..48 Join DC. AD AB ( ADB ABD (base s, isos. x BDC 90 + ADC ( x + + ( x + 90 x + x 68 x 4 ( in semi-circle (opp. s, cyclic quad. Quick Practice.6 (p..48 FAD x (ext., cyclic quad. In FAD, ADF + DFA + FAD 80 ( sum of ADF x ADF 7 x EDC ADF (vert. opp. s 7 x In DCE, CED + EDC BCD (ext. of 5 + (7 x x x 7 x 86 Quick Practice.7 (p..56 (a BAF + FEB (b 0 80 A, B, E and F are not concyclic. CDF BEF 00 CDF BEF F, E, C and D are not concyclic. (c ADB 0 and ACB 0 ADB ACB A, B, C and D are concyclic. (converse of s in the same segment Quick Practice.8 (p..57 FAB + BCD and BEF BCD FAB + BEF A, B, E and F are concyclic. int. s, AF // CD ext., cyclic quad. opp. s supp. Quick Practice.9 (p..58 (a ABC is an equilateral triangle. 60 prop. of equil. CDF is an equilateral triangle. DFC 60 prop. of equil. DFC B, C, E and F are concyclic. converse of s in the same segment (b B, C, E and F are concyclic. (proved in (a BEC BFC ( s in the same segment 60 (prop. of equil. FCE BEC (ext. of 60 7

7 Further Practice Further Practice (p..5. (a CN ND ( ON CD (line joining centre to mid-pt. of chord chord ONC 90 In OND, NOD ONC ODN (ext. of (b In OME, OM + EM OE (Pyth. theorem EM OE OM OM EF ( MF EM (line from centre chord bisects chord 8. CD is the perpendicular bisector of the chord AB. CD is a diameter of the circle. CD CM + MD ( Radius of the circle CD 0 0 Let O be the centre of the circle. OA 0 OM OD MD (0 5 5 In OAM, AM AM OA OM AM AB AM + MB 75 OA OM (or 0 (Pyth. theorem Further Practice (p..8. (a ON OM 4, OM AB, ON CD ( CD AB (chords equidistant from 7 centre are equal CN ND ( 7.5. (b PB AP ( 4 OP AB Basic Properties of Circles (line joining centre to mid-pt. of chord chord Also, OQ OP and OQ BC ( BC AB (chords equidistant from 8 centre are equal QC BQ (line from centre chord bisects chord QC BC AB 4 (c OM AB, ON CD ( MB AM (line from centre chord bisects chord AB 5 0 and CD 5 0 AB CD, OM AB and ON CD OM ON (equal chords, equidistant.5 from centre Construct OM and ON such that OM PQ and ON RS. Note that MON is a straight line, since PQ // RS. PM MQ (line from centre chord bisects chord PQ 6 8 PQ RS, OM PQ and ON RS OM ON (equal chords, equidistant from centre 6 In PMO, OP PM + OM (Pyth. theorem OP PM + OM Radius of the circle 0

8 NSS Mathematics in Action 5A Full Solutions Further Practice (p..0. (a Consider ABD. ABD + BAD + ADB ( sum of ABD + ( ABD 50 x ABD ( s in the same segment 50.. (b BDC ( s in the same segment ( in semi-circle In ABC, + + BCA ( sum of x x 4 BOD BCD (opp. s of // gram 40 BAD ( at centre twice at ce 40 0 ADO BAD (alt. s, OD // AC p 0 Join PR. PRS 90 ( in semi-circle PSR PR (arcs prop. to s at ce SPR RS + PSR SPR PSR + SPR + PRS ( sum of PSR 90 PSR 45 Further Practice (p..49. (a DEC (ext., cyclic quad. 8 CDE DEC ECD ( sum of x ADC 90 ( in semi-circle In APD, PAD + PDA + APD ( sum of ( APD APD 4 BPC 4 Further Practice (p..4. Join OB. AB BC DE ( AOB BOC DOE (equal chords, equal s 55 AE CD ( AOE COD (equal arcs, equal s AOB + BOC + COD + ( s at a pt. DOE + EOA COD COD 60 COD 95 COD 97.5 (b Join OD. OD OA (radii ODA OAD (base s, isos. 8 OD OC (radii ODC OCD (base s, isos. 50 ADC ADC (opp. s, cyclic quad. x + 68 x. BC CD ( BC CD (equal chords, equal arcs BC (arcs prop. to s at ce CAD CD 8 8 ACB 90 ( in semi-circle 4

9 Basic Properties of Circles ACB ( sum of ADC + Exercise ADC + 6 ADC (opp. s, cyclic quad. Exercise A (p..8 Level. ON AB ( BN AN (line from centre chord bisects chord. AM MB (line from centre chord bisects chord AB 6 ON OM, OM AB and ON CD ( CD AB (chords equidistant from centre are equal 4. CN ND ( ON CD (line joining centre to mid-pt. of chord chord ONK 90 AM MB ( OM AB (line joining centre to mid-pt. of chord chord OMK 90 MKN DKB (adj. s on st. line 4 7 Consider quadrilateral OMKN. MON + OMK + ONK + MKN (4 80 ( sum of polygon MON Join OB. Consider NOB. OB BN + ON (Pyth. Theorem 0 The radius of the circle is AM MB ( OM AB (line joining centre to mid-pt. of chord chord MB 6 Consider OMB. OM OB MB 5 4 ON MN OM (7 4 (Pyth. theorem Join OB. OB OC ON + NC (5 + 8 (radii AN NB ( ON AB (line joining centre to mid-pt. of chord chord Consider ONB. NB OB ON 5 AN NB AB 4 (Pyth. theorem Join OD. Consider OND. OD OB ND 5 OD ON 5 (radii (Pyth. theorem 4 ON CD ( CN ND (line from centre chord bisects chord CD 4 8 5

10 NSS Mathematics in Action 5A Full Solutions 6. CM MD ( OM CD (line joining centre to mid-pt. of chord chord OMC 90 AOC OCM + OMC (ext. of + 90 Consider OAC. OC OA ( OCA OAC (base s, isos. OCA + OAC + AOC OCA + OCA 9 7. OM AP ( AM MP (line from centre chord bisects chord MP AP 4 Join OD. Consider OMD. OD OB MD 0 OD OM 0 6 (radii (Pyth. theorem 8 OM CD ( CM MD (line from centre chord bisects chord CD MD Join OP. OP OM + MP 9 + (Pyth. theorem 5 ON BP ( PN NB (line from centre chord bisects chord PN BP 8 9 Consider ONP. ON OP PN 5 9 OB AB ( AM + MB ( OM OB MB (0 4 6 (Pyth. theorem 9. (a PQ is the perpendicular bisector of the chord RS. PQ is a diameter of the circle. PQ 0 ( radius of the circle PQ 0 5 (b Join OR. OR 5 OT OR TR 5 4 OP 5 (radius PT OP + OT ( TQ PQ PT (0 9 (radius (Pyth. theorem 6

11 Basic Properties of Circles 0. BM MC 6 ( OM BC (line joining centre to mid-pt. of chord chord Consider OMB. OM OB BM Consider OMD. MD OD OM CD MD MC (5 6 9 (Pyth. theorem (Pyth. theorem ON AB (constructed AN NB (line from centre chord bisects chord AB 6 8 In OAN, ON OA AN 7 8 (Pyth. theorem 5 Distance between AB and CD ON OM ( Construct a circle with centre O lying on BH, such that the circle cuts AB at two points P and Q, and cuts BC at two points R and S as shown.. Level. Draw OM and ON such that OM AB and ON BC. ABH CBH OMB ONB 90 by construction OB OB common side OBM OBN AAS OM ON corr. sides, s PQ RS chords equidistant from centre are equal Construct OMN such that OM CD and ON AB. OMN is a straight line. OA OC 7 (radii OM CD (by construction CM MD (line from centre chord bisects chord CD 0 5 In OCM, OM OC CM (Pyth. theorem Let M be a point on AB such that OM AB. OM AB (by construction AM MB (line from centre chord bisects chord 4 Consider OMA. OM OA AM 5 9 Consider OMC. MC MB + BC OC ( OM + MC CD OC OD (4 5 6 (Pyth. theorem (Pyth. theorem 4. (a Consider ABP and ACP. AP AP common side BP CP OP BC line joining centre to mid-pt.of chord chord APB APC 90 ABP ACP SAS AB AC corr. sides, s ABC is an isosceles triangle. 7

12 NSS Mathematics in Action 5A Full Solutions (b OM AB ( AM MB (line from centre chord AB AM bisects chord 6 AC AB (proved in (a BC BP ( 6 AC BC, ON AC and OP BC OP ON (equal chords, equidistant from centre 7. OM CD ( CM MD (line from centre chord bisects chord 6 Let r be the radius of the circle. OM AM OA ( 8 r 5. (a Consider OAB and OAC. OA OA common side OB OC radii AB AC OAB OAC SSS OAB OAC corr. s, s AO bisects. (b Consider ABN and CAN. AB AC OAB OAC proved in (a AN AN common side ABN CAN SAS BN CN corr. sides, s ON BC line joining centre to mid-pt. of chord chord (c ON AN OA 6. (a (8 5 Consider ONC. NC OC ON 5 4 Consider ANC. AC AN + NC ON OY NY ( r (or 4 5 (Pyth. theorem (Pyth. theorem (b ON AB ( AN NB (line from centre chord bisects chord 8 9 Consider OAN. OA r (radius ON + AN 6r + 90 r r 5 ( r + 9 r r OA (Pyth. theorem 8 8. Join OC. Consider OCM. OC r (radius OM + CM OC (Pyth. theorem (8 r + 6 r 60 6r + r r 0 MB OB OM r ( 0 8 [ r (8 r] (r 8 Let M be a point on AB such that OM AB. OM AB (by construction MB AM (line from centre chord bisects chord 8 9 Join OB. OB (radius Consider OMB. OM OB MB 9 (Pyth. theorem 88 Let N be a point on CD such that ON CD. ON CD (by construction NC DN (line from centre chord bisects chord 4 ONK OMK 90 ONKM is a rectangle.

13 Basic Properties of Circles NK OM (property of rectangle KC NC NK NC OM (.6 88 (cor. to d.p. In BCN, NB BC NC (Pyth. theorem 9. (a (b PQ PR + RQ ( OP PQ 6 OR OP PR ( In ORM, RM OR OM OM RS RM MS RS RM Join OD. OD OP In OND, ND OD ON (Pyth. theorem ( (line from centre chord bisects chord (radii (Pyth. theorem 8 ON CD ( CN ND (line from centre chord bisects chord CD ND 8 0. (a MN MC (or 4 MN NC MN NC and BD MC BD is the perpendicular bisector of chord MC. BD is a diameter of the circle. (b (i NC MC 6 8 (ii Join OC. Let r be the radius of the circle. OC OB (radii r ON OB NB ( r 6 In OCN, OC r r ( r r r r r ON 8. + NC r (cor. to d.p. (Pyth. theorem The radius of the circle is 8.. BD OB AB Exercise B (p..0 Level. ACB + 8 BD AD (Pyth. theorem (adj. s on st. line ACB 4 x ACB ( at centre twice at ce ACB 90 ( in semi-circle CA CB ( x CBA (base s, isos. ACB + CBA + x ( sum of 90 + x x 45 9

14 NSS Mathematics in Action 5A Full Solutions ACD ABD ( s in the same segment 55 x + ACD 5 (ext. of x x 70 DAC DBE ( s in the same segment 5 x DAC + ADC (ext. of AOB ACB ( at centre twice at ce 5 06 x + AOB 8 (ext. of x x ACD CDB CAD (ext. of ACB 90 x ACB ACD ( in semi-circle 7. ADC 90 ( in semi-circle DCA ADC DAC ( sum of x DCA (x ( s in the same segment AOB ACB ( at centre twice at ce 4x OB OA (radii OBA x (base s, isos. x + OBA + AOB ( sum of x + x + 4x 6x x 0 ACB AOB ( at centre twice at ce 0 65 OC OA (radii OCA OAC (base s, isos. 0 OB OC (radii x OCB (base s, isos. ACB OCA DAC 90 ( in semi-circle ADC DAC ACD ( sum of AB AD ( ABD ADB (base s, isos. 5 ABD + 55 (ext. of (a (b OAB + AOC (int. s, BA // CO AOC 48 Reflex AOC 60 AOC ( s at a pt reflex AOC ( at centre twice at ce 06. BOD 6 (opp. s of // gram BCD BOD ( at centre twice at ce 6 8 ODC BCD (alt. s, DO // AC 8 BKD ODK + BOD (ext. of Join OC.. (a Reflex AOC ADC ( at centre twice at ce AOC + reflex AOC 60 ( s at a pt. AOC

15 Basic Properties of Circles (b Join OB. OB OA (radii OBA OAB (base s, isos. 8 AOC ( at centre twice at ce OBC OBA 40 8 OC OB (radii BCO OBC (base s, isos. 4. AEF, AFE, BDC, DBC, EAC and CAF (any four of the above angles 5. Suppose OBC OAC. ACB AOB ( at centre twice at ce AKB AOK + OAC (ext. of 90 + OAC BKA BCK + OBC (ext. of 45 + OAC 90 + OAC 45 + OAC OAC 45 If OAC 45, then B and C will coincide, which contradicts the assumption that B and C are distinct points. OBC OAC (or any other reasonable answers Level 6. DCB + CKB 50 (ext. of DCB 50 8 DAB DCB ( s in the same segment ACB 90 ( in semi-circle ACB + CBA + CAB ( sum of ( CAD + CAD 8 7. DC DA ( DCA x (base s, isos. BD BC ( BDC DCA (base s, isos. x ADB 90 ( in semi-circle DAC + DCA + ADC ( sum of 8. BDC 0 DAC DBC x + x + (90 + x x 0 ( s in the same segment DAC + ACD + ADC ( sum of ADC ADC 90 AC is a diameter of the circle. (converse of in semi-circle 9. BAD BCD ( s in the same segment BCD (alt. s, CD // AB BAD ACB 90 ( in semi-circle + + ACB ( sum of (44 + BAD ABE AOE ( at centre twice at ce 4 6 ACE + BEC ABE (ext. of BEC BAD BEC ( s in the same segment 6 AKE BAD + ABE (ext. of Join BD. ABD 90 ( in semi-circle EBD ABD ABE 90 x ECD EBD 90 x ( s in the same segment 4

16 NSS Mathematics in Action 5A Full Solutions. (a OABC is a parallelogram. OA OC radii OABC is a rhombus. (b Reflex AOC 60 x s at a pt. reflex AOC at centre twice at ce (60 x x (c x (opp. s of // gram x 80 x (proved in (b x x 0. (a DK BK BK EK line from centre chord bisects chord In BKD and EKD, BK EK proved BKD EKD DK DK common side BKD EKD SAS (b Let BED x. KBD KED (corr. s, s x ABE ACE + BED 4 + x ABD 90 ABE + KBD x + x 90 x 4 BAD BED 4 (ext. of ( in semi-circle ( s in the same segment 4. (a In AKB and DKC, AKB DKC vert. opp. s BAK CDK s in the same segment ABK DCK s in the same segment AKB ~ DKC AAA AK DK corr. sides, ~ s BK CK AK CK BK DK (b AK CK BK DK (proved in (a ( 6 ( ( DK DK 4 5. Join AP. APB 90 in semi-circle ACQ APQ s in the same segment 90 i.e. QC AB 6. Join OA. ABQ AOQ s in the same segment ABP at centre twice at ce BP bisects ABQ. 7. (a EA EB EAB EBA base s, isos. BAD BED s in the same segment BCE EBA BED ext. of EAB BAD EAD BCE EAD (b ABE ADE ( s, in the same segment CBE ABE (adj. s on st.line ADE 55 AC 55 x 0 50 CDA (adj. s on st.line AED + DAE (ext. of > AED It is not possible that BCE ~ EAD. Exercise C (p..4 Level. Reflex AOB 60 AOB ( s at a pt x AOB Reflex AOB Major AB (arcs prop. to s at centre 80 x ACB ( sum of x (arcs prop. to s at ce 4

17 Basic Properties of Circles. 90 ( in semi-circle 80 ACB ( sum of x x 4. CB ACB AB 90 x 4 x x 4x 9x 450 x 50 BAC (arcs prop. to s at ce CD BOC BC 4 COD ( BOD BOC + COD COD (arcs prop. to s at centre BAD BOD ( at centre twice at ce x x ADC BAD alt. s, CD // AB AC : BD equal s, equal arcs 6. (a (b AOB AB BOC BC 4 AOB (48 64 (arcs prop. to s at centre ACB AOB ( at centre twice at ce BDC DBC DCB ( sum of EDB DBC DEB (ext. of ce AB : BC EDB : BDC (arcs prop. to s at 7 : 54 : ( in semi-circle BAD + CAD AB AD ( ABD ADB (base s, isos. ABD + ADB + BAD ( sum of ABD + 0 ABD 0 ACB ( sum of ce AB : BC ACB : (arcs prop. to s at 60 : 90 : 9. AC BA ( AC BA (equal arcs, equal chords 6 90 ( in semi-circle In ABC, 0. BC AB + AC Perimeter of ABC ( (cor. to sig. fig. APC APB + BPC CPD APC CD CPD AC CD BPD BPC + CPD (Pyth. theorem AC (arcs prop. to s at ce 5 EPF BPD EF EPF BD EF APD APB + BPC + CPD BD (arcs prop. to s at ce 0 FPG APD FG EPG AD FG (equal s, equal arcs AC CD, BD EF, AD FG (any two of the above answers AD (arcs prop. to s at ce 4

18 NSS Mathematics in Action 5A Full Solutions. ADB AB DAC CD (arcs prop. to s at ce ADB DAC KA KD (sides opp. equal s Similarly, KB KC AKD and BKC are isosceles triangles. AB BC CD ( AB BC CD (equal arcs, equal chords ABC and BCD are isosceles triangles. (any three of the above answers Level. BAD 90 ( in semi-circle BAD CAD BC (arcs prop. to s at ce CAD CD 40 x 5 50 Join BC. BCA BDA ( s in the same segment BCA ( sum of ADC BC 70 x (arcs prop. to s at ce Let AC and BD intersect at E. BAD 90 ( in semi-circle AB AD ( ABD ADB (base s, isos. ABD + ADB + BAD ( sum of ADB + 90 ADB 45 ADE + AED + DAE ( sum of DAE DAE 60 DAC CD (arcs prop. to s at ce ADB AB x x 4 5. (a OD OB (radii ODB OBD (base s, isos. 6. (b 0 BOC OBD + ODB (ext. of (arcs prop. to s at centre AOB BA BOC CB AOB ADB AOB ( at centre twice at ce 40 0 PRS PQS 0 PRQ QRS PRS QSR PRQ QR QP QSR PRQ ( s in the same segment (arcs prop. to s at ce 45 RQS QRS QSR ( sum of (a 44

19 Join OB, OC and CD. AB BC CD ( AOB BOC COD (equal arcs, equal s AOB + BOC + COD 0 BOC 0 BOC 40 BDC BOC ( at centre twice at ce 40 0 (b Reflex AOD 60 AOD ( s at a pt. ce ACD reflex AOD ( at centre twice at 40 0 CED + CDE + ECD ( sum of CED CED (a (i Basic Properties of Circles (b AOB COB corr. s, s AOD AOB adj. s on st. line COB COD adj. s on st. line AD DC equal s, equal arcs CD CAB BC 0. DAC arcs prop. to s at ce CAB DAC OC OA radii ACO CAB base s, isos. DAC OC // AD alt. s equal. (a OE BD BE ED line from centre chord bisects chord AE AE common side AEB AED 90 ABE ADE SAS (b DAC corr. s, s BC arcs prop. to s at ce CD BC CD. (a With the notations in the figure, (ii Join OT, OP and OQ. RS ST TP PQ QR ( ROS SOT TOP POQ QOR (equal chords, equal s ROS 60 5 ( s at a pt. 7 RPS ROS ( at centre twice at ce 7 6 (b Angle subtended at centre by any one side of the polygon 60 n Angle subtended at one of the vertices by any one side of the polygon 60 n 80 n 9. (a OA OC radii AB CB OB OB common side ABO CBO SSS DFE BDF + DBF (ext. of AGE CAG + ACG (ext. of x FGE GFE ( sum of (b AB : BC : CD : DE : EA ADB : BEC : CAD : DBE : ACE (c (arcs prop. to s at ce 0 : 40 : 40 : 0 : 50 : 4 : 4 : : 5 AB Circumference of the circle Circumference of the circle π 9π 45

20 NSS Mathematics in Action 5A Full Solutions Radius of the circle 9π π 4.5. If ABC is a right-angled triangle, then either one of A, B or C must be equal to 90. A : B : C a : b : c (arcs prop. to s at ce and A + B + C ( sum of a A 80, a + b + c b B 80 a + b + c c and C 80 a + b + c Since one of A, B or C is equal to 90, a b c one of the ratios, or is equal a + b + c a + b + c a + b + c to. The student s claim is correct. (or any other reasonable answers Exercise D (p..49 Level. BCD 95 (ext., cyclic quad. BCD + x (adj. s on st. line. x BCD + BAD (opp. s, cyclic quad. BCD CD CB ( BDC x (base s, isos. BCD + x + BDC ( sum of x + 04 x 8. ACB 90 ( in semi-circle + ACB + ( sum of x + (opp. s, cyclic quad. 46 x 50 0 x ABD ( s in the same segment y + BCD y + (54 + x y ( (opp. s, cyclic quad (a EBC + CDE (opp. s, cyclic quad. EBC 0 70 ECB + BAE (opp. s, cyclic quad. ECB 0 60 BEC + EBC + ECB ( sum of (b 96 BEC ADE CAD + ACD (ext. of DFE BCD (ext., cyclic quad. 60 DFE + FDE + DEF ( sum of DEF BAP PQD ext., cyclic quad. DCP + PQD opp. s, cyclic quad. DCP + BAP AB // CD int. s supp (a FCD + DEF FCD 0 50 x FCD (ext., cyclic quad. 50 (opp. s, cyclic quad. ABD y (ext., cyclic quad. AD AB ( ADB ABD (base s, isos. y BAD + ABD + ADB (b x + y y 50 y 65 ( sum of BPA AOB ( at centre twice at ce 40 0 BAP + BCP BAP 50 (opp. s, cyclic quad. 0 ABP + BAP + APB ( sum of ABP

21 0.. DAC + ADC + ACD ( sum of DAC ADC (opp. s, cyclic quad BC CD ( BC CD (equal chords, equal arcs BAC DAC BC CD DAC 5 ACB + + ( sum of ACB (arcs prop. to s at ce Basic Properties of Circles. ABD is an equilateral triangle. BAD 60 BCD is a right-angled triangle with C 90. BCD 90 BAD + BCD Level 4. Reflex 80 The polygon cannot be a cyclic quadrilateral. (or any other reasonable answers COA ( at centre twice at ce 0 0 COA 60 reflex COA ( s at a pt APB + COA (opp. s, cyclic quad. APB Join AD. + CDA ADE 90 + CDE + ( CDA + ADE ( + CDA + ADE (a Consider ABC and ADC. AB AD BC DC AC AC ABC ADC (b ABC ADC ADC + ADC opp. s, cyclic quad. in semi-circle common side SSS (proved in (a (corr. s, s (opp. s, cyclic quad. 90 AC is a diameter of the circle. (converse of in semi-circle (c If ABC and ADC are right-angled isosceles triangles, then BAD + DAC and BCD BCA + DCA In this case, BD is a diameter of the circle. (converse of in semi-circle 5. CEB DAE (corr. s, EC // AD 40 EC EB ( ECB EBC (base s, isos. CEB + ECB + EBC ( sum of 40 + EBC EBC 70 ADC + EBC (opp. s, cyclic quad. ADC (a KAD KCB ext., cyclic quad. KDA KBC ext., cyclic quad. AKD CKB common angle KAD ~ KCB AAA (b KA KD KC KB (corr. sides, ~ s KA KD KD + DC KA + AB + DC ( DC DC 7. DAB DCE (ext., cyclic quad. 65 ABD 90 ( in semi-circle ADB + ABD + DAB ( sum of ADB CBD ADB (alt. s, BC // AD 5 BDC + CBD DCE (ext. of BDC

22 NSS Mathematics in Action 5A Full Solutions ACD + ADC + CAD ( sum of ACD 40 ADC 40 ADC + ADC 80 (opp. s, cyclic quad. ADC AED + ACD (opp. s, cyclic quad. AED ACD (40 ADC 40 + ADC + AED ( 80 ADC + (40 + ADC 0 Join BD. Let CBE x. CE CB ( CEB CBE (base s, isos. x BEA + BAE CBE (ext. of BEA x 7 DBE 90 ( in semi-circle DBC + DEC (opp. s, cyclic quad. (90 + x + ( x 7 + x CBE 9 x 7 x 9 0. Let CO D x. CO D x ext., cyclic quad. OC OB radii OCB OBC base s, isos. x OAD OCB x O CD BAO ext., cyclic quad. ext., cyclic quad. x O D O C radii O DC O CD base s, isos. x O DC + O CD + CO D x x 60 O DC O CD CO D 60 CDO is an equilateral triangle.. (a APC + PCB ( x + x + y x + y ( y + y + x x + y ext. of ARB ACB + RBC ext. of sum of (b APC + ARB opp. s, cyclic quad. (x + y + ( x + y x + y x + y ACB sum of + x + y + ( x + y Exercise E (p..58 Level. (a BAD + BCD ( ( (b. (a (b A, B, C and D are concyclic. opp. s supp. CDB CAB ( s in the same segment x 8 DBC BDC DCB sum of DBC DAC A, B, C and D are concyclic. BDC ( s in the same segment x 5. (a CDB 90 A, B, C and D are concyclic. (b ACB ADB x 5 converse of s in the same segment converse of s in the same segment ADB EDA CDB (adj. s on st. line 4. (a Refer to the figure. ( s in the same segment Consider ABG. ABG BGA GAB sum of ABG + GBC ADF A, B, C and D are concyclic. ext. int. opp. 48

23 (b CAD CBD ( s in the same segment 5 DCE DAB (ext., cyclic quad. + CAD x Basic Properties of Circles. D is any point on AC, where AB is a diameter of the circle and the mid-point of AB is the centre. 5. Consider PBC. PCB + PBC APB ext. of 6 + PCB 6 PCB 46 ACD BCD PCB ABD ACD A, B, C and D are concyclic. converse of s in the same segment 6. ADC opp. s of // gram PQB PDC ext., cyclic quad. DPQ PQB alt. s, AD // BC DPQ A, B, Q and P are concyclic. ext. int. opp. 7. AB EB BAE BEA base s, isos. BAE BCD opp. s of // gram BEA BCD B, C, D and E are concyclic. ext. int. opp. 8. BPT 90 in semi-circle SQC 90 in semi-circle AQS SQC adj. s on st. line AQR BPR 90 A, P, R and Q are concyclic. ext. int. opp. 9. BAD (x + y ( sum of BAD + BCD (opp. s, cyclic quad. 80 ( x + y + z x + y z (or any other equivalent relationship 0. AB AC ACB base s, isos. AM MB and AN NC MN // BC mid-pt. theorem AMN corr. s, MN // BC AMN ACB B, C, N and M are concyclic. ext. int. opp. Level. (a ABC is an equilateral triangle. AB BC CA Consider ABD and ACD. AB AC proved BD CD AD AD common side ABD ACD SSS Consider ABE and CBE. AB CB proved AE CE BE BE common side ABE CBE SSS (b ABD ACD proved in (a ADB ADC corr. s, s ADB + ADC adj. s on st. line ADB ADC 90 Similarly, BEA BEC 90 FDC + FEC D, C, E and F are concyclic. opp. s supp. ADB BEA 90 A, B, D and E are concyclic. converse of s of the same segment. (a APB 90 in semi-circle APM + APB adj. s on st. line APM + 90 APM 90 AOM 90 APM AOM O, P, M and A are concyclic. converse of s in the same segment (b OA OP radii OAP OPA base s, isos. OAP OMP s in the same segment OPA OMB 4. AQS BRS ext., cyclic quad. BRS CPS ext., cyclic quad. AQS CPS AQSP is a cyclic quadrilateral. ext. int. opp. 49

24 NSS Mathematics in Action 5A Full Solutions 5. (a (b Since N is the mid-point of BD, N is the centre of the circle. AP PC NP AC line joining centre to mid-pt. of chord chord (b Join PB and let ARP θ. APQ ARP θ PBA ARP θ APB 90 BPQ APB APQ s in the same segment in semi-circle 90 θ PQB PBQ BPQ sum of θ (90 θ 90 Join RB. TQB 90 TRB 90 TQB + TRB R, T, Q and B are concyclic. proved in (a in semi-circle opp. s supp. 9. (a Join MP and NP. Let ABD x. ACD ABD s in the same segment x ACB BCD ACD 90 x AM MB and AP PC MP // BC mid-pt. theorem APM ACB corr. s, MP // BC 90 x NP AC proved in (a APN 90 NPM APN APM 90 (90 x x ABD 6. Consider ACB and DBC. AC DB ACB DBC BC CB common side ACB DBC SAS CDB corr. s, s A, B, C and D are concyclic. converse of s in the same segment 7. (a ACB 90 in semi-circle ECP ACB adj. s on st. line 90 ADB 90 in semi-circle ECP ADB P, D, E and C are concyclic. ext. int. opp. (b COD CAD at centre twice at ce CAD 80 ACP APB sum of 90 APB COD (90 APB APB 8. (a BAD + BCD BCDA is a cyclic quadrilateral. opp. s supp. BAD 90 BD is a diameter of the circle. converse of s in the same segment 50 Join AE, BF, CG and DH. EAC + CGE 80 CGE EAC If A, E, F and B are concyclic, then EAB + BFE (b opp. s, cyclic quad. opp. s, cyclic quad. BFE EAC CGE BFE BFE BDH ext., cyclic quad. CDH CGE C, G, H and D are concyclic. ext. int. opp. Join PU, QT, RW and SV. RSV PUT ext., cyclic quad. VWR PQT ext., cyclic quad. If R, S, W and V are concyclic, then RSV VWR s in the same segment PUT PQT P, Q, U and T are concyclic. converse of s in the same segment

25 Basic Properties of Circles Revision Exercise (p..66 Level. 4. With the notations in the figure, Join OF. Draw ON such that ON FE. OF OB (radii BC 0 0 ON AB (property of rectangle 6 Consider ONF. FN OF ON (Pyth. theorem ON FE (by construction FN NE (line from centre chord bisects chord FE 8 6. AEC 90 ( in semi-circle EBC EAC ( s in the same segment 0 AEB EDC + EBC (ext. of BEC AEC AEB POR + OPQ (int. s, OR // PQ POR 4 8 Reflex POR 60 POR ( s at a pt PQR reflex POR ( at centre twice at ce ORQ + PQR (int. s, OR // PQ ORQ 69 BCA AOB ( at centre twice at ce 54 7 ONC OBC + BCA (ext. of OAC + AOB ONC (ext. of ACB AB 5. CAB OAC BC ACB CAB 5 (arcs prop. to s at ce 5 DCA 90 ( in semi-circle DCB DCA + ACB DAB + DCB (opp. s, cyclic quad. ( DAC DAC 0 6. EDC EAB ( s in the same segment DAC DAB EAB EAB ADE ADC EDC 80 EAB AEB DAC + ADE 00 (60 EAB + (80 EAB EAB EAB 0 CDB CBD BC CD (ext. of (arcs prop. to s at ce CBD CDB + ADC (6 + CBD + (58 + CDB 0 + CDB + CDB CDB 60 CDB 40 KDC 40 (opp. s, cyclic quad. 5

26 NSS Mathematics in Action 5A Full Solutions 8. + ADC (opp. s, cyclic quad. ACB 5 AB BC 65 (arcs prop. to s at ce ACB ACB ( sum of. ADC + BCD (int. s, AD // BC (8 + ODC + (6 + OCD 64 + OCD OCD 58 BCD BAD + BCD (opp. s, cyclic quad. BAD + 94 BAD (a OC OB (radii OCB OBC (base s, isos. OCB + OBC + BOC ( sum of 0.. (b OCB + 80 OCB 50 BAC BOC ( at centre twice at ce DAB + BCD (opp. s, cyclic quad. ( DAC + + ( OCB + OCD OCD OCD 54 Join BC. DBC ( at centre twice at ce ACB 90 ( in semi-circle In BCE, BCE + EBC + CEB ( sum of CEB CEB 45 Draw OM such that OM BC. OM BC (by construction BM MC (line from centre chord bisects chord MC 6 Consider OMC. OM OC MC (Pyth. theorem 5 4 Consider OAM. 7 4 AB AM BM AM OA OM (Pyth. theorem AM MC (.74 (cor. to d.p.. (a CF FD ( OF CD (line joining centre to mid-pt. of chord chord FD CD 4 In OFD, OF OD FD (Pyth. theorem ( 5 Join DC. OC OD (radii OCD ODC (base s, isos. 5

27 Basic Properties of Circles (b 4. (a 5. Join AB. OE OF (proved in (a and OE AC and OF CD (proved in (a AC CD 4 OE AC AE EC AC 4 5 OB OE (chords equidistant from centre are equal ( OB OD (radii BE ( 5 (line from centre chord bisects chord Area of ABE BE AE ( 5.4 (cor. to d.p. AE 69 AD + DE [( ] AE AD + DE ADE 90 (converse of Pyth. theorem AE is a diameter of the circle. (converse of in semi-circle The centre of the circumcircle lies on AE. (b CB AD and AB BD ( CB is the perpendicular bisector of AD. The centre lies on CB. The centre also lies on AE. The intersection of CB and AE, i.e. C, is the centre of the circumcircle. AD AD common side AB AC ABD ACD RHS BAD CAD corr. s, s AD bisects. 6. ADC p + q (ext. of BCD p ( s in the same segment r ADC + BCD (ext. of 7. p + q + p p + q Let O be the centre of the circle and r be the radius. Join OA. OA r OM OC MC 8. (a ( r OM AB ( AM MB 0 5 Consider OAM. 5 + ( r 5 + r r 6 r + (line from centre chord bisects chord OA AM + OM (Pyth. theorem r r r The radius of the circle is. (b BCA sum of ADE 6 A, B, C and D are concyclic. ext. int. opp. ABCD is a cyclic quadrilateral. CAD ADE ACD (ext. of CBD CAD ( s in the same segment 6 Join BD and DC. ABD 90 ACD 90 ABD ACD in semi-circle in semi-circle CD EBD DE 9. DAC arcs prop. to s at ce EBD DAC P, A, B and Q are concyclic. converse of s in the same segment 5

28 NSS Mathematics in Action 5A Full Solutions 0. Draw a line segment PQ in rectangle ABCD, then draw another line segment RS which is perpendicular to PQ as shown in the following figure. (or any other reasonable answers Level. AOB + DOB (adj. s on st. line AOB OBC AOB (alt. s, DA // CB (a ACB 90 ( in semi-circle DBA DCA ( s in the same segment CEB + CBE + ECB ( sum of 5 + ( DBA ( DCA + 90 (b 5. CFB CFB 5 DCA 8 DCA 4 CFB + CBD + BCA ( sum of. (a BCA AOB ( at centre twice at ce 40 0 OKC OBC + BCA (ext. of AKO + OKC (adj. s on st. line (b AKO AFC (opp. s, cyclic quad. AFC 0 78 CDE AFC (ext., cyclic quad. 78 COE CDE ( at centre twice at ce BAF + ABO + COE + FEO (4 80 ( sum of polygon BAF BAF 64. (a ADC 90 ( in semi-circle APD + ADC + BAD ( sum of (b ( BDC ( s in the same segment 6 ADB ADC BDC (ext. of AKD + ADB + CAD ( sum of AKD Join MN. ABM MNC (ext., cyclic quad. AEM MND (ext., cyclic quad. MNC + MND (adj. s on st. line ABM + AEM MNC + MND BAE + ABM + AEM + BME (4 80 ( sum of polygon BME 60 BME 5 Join BE. BEC BC CAD (arcs prop. to s at ce CD BEC (8 4 DE DBE (arcs prop. to s at ce CAD CD 4 DBE (8 56 BKE + KEB + KBE ( sum of BKE

29 7. (a AOB AB (arcs prop. to s at centre AOC ABC + AOB AOC 90 0 BCA AOB ( at centre twice at ce 0 5 (b OC OA (radii ACO CAO (base s, isos. CAO + ACO + AOC ( sum of CAO 90 CAO 45 CEO CAO + AOB (ext. of Basic Properties of Circles (b KM KN corr. sides, s Also, BM CN KM BM KN CN KB KC (c KB KC proved in (b Also, AB DC KA KD KAD KDA base s, isos. BCD + KAD opp. s, cyclic quad. BCD + KDA BC // AD int. s supp. 0. NBP MDP ext., cyclic quad. BNP NBP NPB sum of. (a MDP DPM DMP sum of NMC vert. opp. s QM QN sides opp. equal s 8. (a APD CPB common angle PAD PCB ext., cyclic quad. PDA PBC ext., cyclic quad. PAD ~ PCB AAA (b AKB DKC vert. opp. s BAK CDK s in the same segment ABK DCK s in the same segment AKB ~ DKC AAA (c PAD ~ PCB (proved in (a PA PD PC PB (corr. sides, ~ s PA PD PD + DC PA + AB DC ( DC DC 4 AKB ~ DKC (proved in (b AB BK DC CK (corr. sides, ~ s 0 BK 4 BK (a AM MB and CN ND OMK ONK 90 line joining centre to mid-pt of chord chord AB DC OM ON equal chords, equidistant from centre OK OK common side OMK ONK RHS (b Join OC. AOC ABC (arcs prop. to s at centre AOD ABCD AOC OA OC (radii OAC OCA (base s, isos. OAC + OCA + AOC ( sum of OAC + 45 OAC 67.5 Join OB and AD. AB BC AOB BOC (equal arcs, equal s AOB AOC

30 NSS Mathematics in Action 5A. Full Solutions ADB AOB ( at centre twice at ce.5.5 CD ADB AB CAD.5.5 AED + EAD + EDA CAD (arcs prop. to s at ce AED AED 46.5 ( sum of Join BO and OE. Let CAE x. BOE CAE at centre twice x at ce ACE + BOE opp. s, cyclic quad. ACE x ACE + CAE + CEA 80 sum of (80 x + x + CEA CEA x CAE CEA CA CE sides opp. equal s. (a CE CD CED CDE base s, isos. ext., cyclic quad. AE AB sides opp. equal s ABE is an isosceles triangle. (b Let ABD x. AD DBC DC (arcs prop. to s at ce ABD DBC ABD x ABE ABD + DBC x AEB ABE (base s, isos. x EDC AEB (base s, isos. x DCB AEB + EDC 4x KK( (ext. of BDC 90 ( in semi-circle DBC + DCB + BDC ( sum of DCB x x From ( and (, we have 4x 90 x x 8 08 KK( DCB 4(8 7 BAD + DCB (opp. s, cyclic quad. BAD 7 4. BDC ABD alt. s, CF // BA DEF ABD ext., cyclic quad. KAE DEF corr. s, CA // DE BDC KAE A, K, D and F are concyclic. ext. int. opp. 5. (a ADC 90 in semi-circle FBE adj. s on st. line 90 EDF ADC adj. s on st. line 90 FBE EDF BEFD is a cyclic quadrilateral. converse of s in the same segment (b BDC 5 ( s in the same segment BFE BDC 5 ( s in the same segment DCF CEF + CFE (ext. of DCF + CFD ADC (ext. of 6 + CFD 90 CFD 8 6. (a AD DC AD DC equal arcs, equal chords DE DA DE DC DEC DCE base s, isos. (b AD ED DAE DEA base s, isos. DEC DCE proved in (a DEA + DEC + CEB adj. s on st. line CEB DEA DEC 7. (a DAE DCE BCD + DAE opp. s, cyclic quad. ( BCE + DCE + DAE BCE DAE DCE CEB BCE BC BE sides opp. equal s ADC opp. s of // gram 50 CAE + ACE + AEC sum of AEC

31 Basic Properties of Circles (b ADC + AEC A, E, C and D are concyclic. opp. s supp. (b Join DE. CDE CAE ( s in the same segment 7 DC AB and AB EC DC EC (opp. sides of // gram ( DEC CDE (base s, isos. 7 ECD + CDE + CED ECD ( sum of ECD 06 BCD + (int. s, CD // BA ( ECB + ECD + 50 ECB Join LM and NM. ECB 4 8. (a (i Consider ABM and CDM. ABM CDM s in the same segment BAM DCM s in the same segment AMB CMD vert. opp. s ABM ~ CDM AAA AB AM CD CM corr. sides, ~ s (ii OL AB AL LB line from centre chord bisects chord AL AB ON CD CN ND line from centre chord bisects chord CN CD Consider ALM and CNM. AB AL CN CD AB CD AM from (a (i CM LAM NCM s in the same segment ALM ~ CNM ratio of sides, inc. Join OM, OR and OW. PM MQ OM PQ RLO + RMO RLOM is a cyclic quadrilateral. ROM RLM WNO + WMO line joining centre to mid-pt. of chord chord opp. s supp. s in the same segment WNOM is a cyclic quadrilateral. opp. s supp. WOM WNM s in the same segment ALM ~ CNM proved in (a(ii ALM CNM corr. s, ~ s ROM WOM (c Consider ROM and WOM. ROM WOM proved in (b OM OM common side OMR OMW 90 proved in (b ROM WOM ASA RM WM corr. sides, s M is also the mid-point of RW. Multiple Choice Questions (p..7. Answer: B OC AB ( AC CB (line from centre chord bisects chord OC OA AC 4 Area of AOB AB OC ( + 4 (Pyth. theorem. Answer: B For I, PQ CD ( I is true. For II, AB CD, OP AB and OQ CD OP OQ (equal chords, equidistant from centre OQ PQ II is true. 57

32 NSS Mathematics in Action 5A Full Solutions For III, OP AB ( AP PB (line from centre chord bisects chord AP AB 6 Consider AOP. OP OA AP 8 PQ OP III is not true. Only I and II are true. (Pyth. theorem. Answer: A BCD 90 ( in semi-circle DCA BCD ACB ABD DCA ( s in the same segment 9 x ABD + (ext. of Answer : D For I, DC : CB DC : CB in general I may not be true. For II, AC AD+ DC CB+ CB CB AC CB (equal arcs, equal chords AC : CB : II must be true. For III, ADB ACB 90 ( in semi-circle AC CB (proved in II CAB (proved in II + CAB + ACB ( sum of CAB + 90 CAB 45 ADB : CAB 90 : 45 : III must be true. Only II and III must be true. 5. Answer: B BED + EBD BDC (ext. of EBD Join AD. ADB 90 ( in semi-circle ADC + (opp. s, cyclic quad. ( (9 + x x 8 6. Answer: C For option C, ACB ADB ( s in the same segment x BDC ( s in the same segment θ In CDP, CPD + PDC + DCP ( sum of y + ( x + θ + ( z + x θ x y z C must be true. 7. Answer: C ADC + (opp. s, cyclic quad. ADC ADC (arcs prop. to s at ADC ce ABC ( Answer: A Join BD. ADB 90 ( in semi-circle CBD COD ( at centre twice at ce 48 4 x ADB + CBD (ext. of

33 Basic Properties of Circles 9. Answer: B DCB + BAD 80 (opp. s, cyclic quad. DCB y OD OC ODC OCD y DOC ABO (radii (base s, isos. (corr. s, OD // BA x DOC + ODC + OCD ( sum of x + (80 y + (80 y B must be true. 0. Answer: B x y 80 ADC + (opp. s, cyclic quad. ADC x ACD + AED ACD y ACD + ADC + CAD (80 y + (80 x Answer: B x + y 5 (opp. s, cyclic quad. CDB CAB 5 ABCD is a cyclic quadrilateral. (converse of s in the same segment ADC + CBA (5 + ADB + 7. Answer: D ADB 56 (opp. s, cyclic quad. For I, ADE + DEB (int. s, AD // BE ABE DEB ( ADE + ABE ABED is a cyclic quadrilateral. (opp. s supp. For II, ABE DEB ( DEB EFC (corr. s, BE // CF ABE EFC BCFE is a cyclic quadrilateral. (ext. int. opp. For III, DFC ABE (ext., cyclic quad. ABE + BAD (int. s, AD // BE DFC + BAD ACFD is a cyclic quadrilateral. (opp. s supp. HKMO (p..76. AC AB + BC 4 + (Pyth. theorem 5 Consider AMN and ABC. AMN NBC (ext., cyclic quad. ANM ACB (ext., cyclic quad. MAN (common angle AMN ~ ABC (AAA AM MN AN (corr. sides, ~ s AB BC AC Area of AMN AM MN Area of ABC AB BC Area of Area of AN AC AN 5 5 AN NB AB AN AM MN AMN ABC AB BC AM MN 4 AB BC AN AC 5 NBC 90 ( NC is a diameter of the circle. (converse of in semi-circle Radius of circle BNMC NC NC NB + BC (Pyth. theorem

34 NSS Mathematics in Action 5A Full Solutions. Let O be the centre of the circle.. Let O be the centre of the circle. Radius of circle 0 5 Join OM, OA and OF. Let MN intersect OA at P. Let BC a. Then AB AC BC a, and B C 60 (prop. of equil. AM MB a and AN NC a ( MN // BC and MN BC a (mid-pt. theorem AMP B (corr. s, MN // BC 60 Since ABC is an equilateral triangle, O is not only the circumcentre of ABC, but also its incentre, i.e. AO bisects. 60 MAP 0 and AP MN In AMP, a MP AM cos60 and AP AM sin 60 In AMO, a AM a a AO cos0 PO AO AP a a 6 In POF, a PF OF OP (Pyth. theorem Join OA, OB and OC. Construct OD BC and OE AB. OC OB (radii CD BD (line from centre chord bisects chord OD OD (common side OCD OBD (SSS COD BOD (corr. s, s OA OB (radii AE BE (line from centre chord bisects chord OE OE (common side OAE OBE (SSS AOE BOE (corr. s, s Let COD α and AOE β. Then, CD 5 sin α and AE 5 sin β AB + BC AE + CD 0 sin α + 0 sin β 0(sin α + sin β AB + BC attains its maximum when α β 45. x 0(sin 45 + sin The greatest possible value of x is 0. d a a 5 MF MP + PF MF MN a a a a a + 5 a 60

SHW 1-01 Total: 30 marks

SHW 1-01 Total: 30 marks SHW -0 Total: 30 marks 5. 5 PQR 80 (adj. s on st. line) PQR 55 x 55 40 x 85 6. In XYZ, a 90 40 80 a 50 In PXY, b 50 34 84 M+ 7. AB = AD and BC CD AC BD (prop. of isos. ) y 90 BD = ( + ) = AB BD DA x 60

More information

Fill in the blanks Chapter 10 Circles Exercise 10.1 Question 1: (i) The centre of a circle lies in of the circle. (exterior/ interior) (ii) A point, whose distance from the centre of a circle is greater

More information

Class IX Chapter 7 Triangles Maths. Exercise 7.1 Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure).

Class IX Chapter 7 Triangles Maths. Exercise 7.1 Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Exercise 7.1 Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD? In ABC and ABD, AC = AD (Given) CAB = DAB (AB bisects

More information

Class IX Chapter 7 Triangles Maths

Class IX Chapter 7 Triangles Maths Class IX Chapter 7 Triangles Maths 1: Exercise 7.1 Question In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD? In ABC and ABD,

More information

Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD?

Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD? Class IX - NCERT Maths Exercise (7.1) Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD? Solution 1: In ABC and ABD,

More information

Properties of the Circle

Properties of the Circle 9 Properties of the Circle TERMINOLOGY Arc: Part of a curve, most commonly a portion of the distance around the circumference of a circle Chord: A straight line joining two points on the circumference

More information

Class IX - NCERT Maths Exercise (10.1)

Class IX - NCERT Maths Exercise (10.1) Class IX - NCERT Maths Exercise (10.1) Question 1: Fill in the blanks (i) The centre of a circle lies in of the circle. (exterior/interior) (ii) A point, whose distance from the centre of a circle is greater

More information

21. Prove that If one side of the cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle.

21. Prove that If one side of the cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle. 21. Prove that If one side of the cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle. 22. Prove that If two sides of a cyclic quadrilateral are parallel, then

More information

EXERCISE 10.1 EXERCISE 10.2

EXERCISE 10.1 EXERCISE 10.2 NCERT Class 9 Solved Questions for Chapter: Circle 10 NCERT 10 Class CIRCLES 9 Solved Questions for Chapter: Circle EXERCISE 10.1 Q.1. Fill in the blanks : (i) The centre of a circle lies in of the circle.

More information

6 CHAPTER. Triangles. A plane figure bounded by three line segments is called a triangle.

6 CHAPTER. Triangles. A plane figure bounded by three line segments is called a triangle. 6 CHAPTER We are Starting from a Point but want to Make it a Circle of Infinite Radius A plane figure bounded by three line segments is called a triangle We denote a triangle by the symbol In fig ABC has

More information

Class IX Chapter 8 Quadrilaterals Maths

Class IX Chapter 8 Quadrilaterals Maths Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Answer: Let the common ratio between

More information

Class IX Chapter 8 Quadrilaterals Maths

Class IX Chapter 8 Quadrilaterals Maths 1 Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Let the common ratio between the angles

More information

TRIANGLES CHAPTER 7. (A) Main Concepts and Results. (B) Multiple Choice Questions

TRIANGLES CHAPTER 7. (A) Main Concepts and Results. (B) Multiple Choice Questions CHAPTER 7 TRIANGLES (A) Main Concepts and Results Triangles and their parts, Congruence of triangles, Congruence and correspondence of vertices, Criteria for Congruence of triangles: (i) SAS (ii) ASA (iii)

More information

Exercise 10.1 Question 1: Fill in the blanks (i) The centre of a circle lies in of the circle. (exterior/ interior)

Exercise 10.1 Question 1: Fill in the blanks (i) The centre of a circle lies in of the circle. (exterior/ interior) Exercise 10.1 Question 1: Fill in the blanks (i) The centre of a circle lies in of the circle. (exterior/ interior) (ii) A point, whose distance from the centre of a circle is greater than its radius lies

More information

SOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c)

SOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c) 1. (A) 1 1 1 11 1 + 6 6 5 30 5 5 5 5 6 = 6 6 SOLUTIONS SECTION A. (B) Let the angles be x and 3x respectively x+3x = 180 o (sum of angles on same side of transversal is 180 o ) x=36 0 So, larger angle=3x

More information

Udaan School Of Mathematics Class X Chapter 10 Circles Maths

Udaan School Of Mathematics Class X Chapter 10 Circles Maths Exercise 10.1 1. Fill in the blanks (i) The common point of tangent and the circle is called point of contact. (ii) A circle may have two parallel tangents. (iii) A tangent to a circle intersects it in

More information

CHAPTER 7 TRIANGLES. 7.1 Introduction. 7.2 Congruence of Triangles

CHAPTER 7 TRIANGLES. 7.1 Introduction. 7.2 Congruence of Triangles CHAPTER 7 TRIANGLES 7.1 Introduction You have studied about triangles and their various properties in your earlier classes. You know that a closed figure formed by three intersecting lines is called a

More information

Label carefully each of the following:

Label carefully each of the following: Label carefully each of the following: Circle Geometry labelling activity radius arc diameter centre chord sector major segment tangent circumference minor segment Board of Studies 1 These are the terms

More information

CHAPTER 10 CIRCLES Introduction

CHAPTER 10 CIRCLES Introduction 168 MATHEMATICS CIRCLES CHAPTER 10 10.1 Introduction You may have come across many objects in daily life, which are round in shape, such as wheels of a vehicle, bangles, dials of many clocks, coins of

More information

Class 7 Lines and Angles

Class 7 Lines and Angles ID : in-7-lines-and-angles [1] Class 7 Lines and Angles For more such worksheets visit www.edugain.com Answer the questions (1) ABCD is a quadrilateral whose diagonals intersect each other at point O such

More information

PLC Papers. Created For:

PLC Papers. Created For: PLC Papers Created For: ed by use of accompanying mark schemes towards the rear to attain 8 out of 10 marks over time by completing Circle Theorems 1 Grade 8 Objective: Apply and prove the standard circle

More information

9 th CBSE Mega Test - II

9 th CBSE Mega Test - II 9 th CBSE Mega Test - II Time: 3 hours Max. Marks: 90 General Instructions All questions are compulsory. The question paper consists of 34 questions divided into four sections A, B, C and D. Section A

More information

RD Sharma Solutions for Class 10 th

RD Sharma Solutions for Class 10 th RD Sharma Solutions for Class 10 th Contents (Click on the Chapter Name to download the solutions for the desired Chapter) Chapter 1 : Real Numbers Chapter 2 : Polynomials Chapter 3 : Pair of Linear Equations

More information

PRACTICE QUESTIONS CLASS IX: CHAPTER 4 LINEAR EQUATION IN TWO VARIABLES

PRACTICE QUESTIONS CLASS IX: CHAPTER 4 LINEAR EQUATION IN TWO VARIABLES PRACTICE QUESTIONS CLASS IX: CHAPTER 4 LINEAR EQUATION IN TWO VARIABLES 1. Find the value of k, if x =, y = 1 is a solution of the equation x + 3y = k.. Find the points where the graph of the equation

More information

3. AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is:

3. AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is: Solved Paper 2 Class 9 th, Mathematics, SA 2 Time: 3hours Max. Marks 90 General Instructions 1. All questions are compulsory. 2. Draw neat labeled diagram wherever necessary to explain your answer. 3.

More information

VAISHALI EDUCATION POINT (QUALITY EDUCATION PROVIDER)

VAISHALI EDUCATION POINT (QUALITY EDUCATION PROVIDER) BY:Prof. RAHUL MISHRA Class :- X QNo. VAISHALI EDUCATION POINT (QUALITY EDUCATION PROVIDER) CIRCLES Subject :- Maths General Instructions Questions M:9999907099,9818932244 1 In the adjoining figures, PQ

More information

LLT Education Services

LLT Education Services 8. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle. (a) 4 cm (b) 3 cm (c) 6 cm (d) 5 cm 9. From a point P, 10 cm away from the

More information

Geometry 3 SIMILARITY & CONGRUENCY Congruency: When two figures have same shape and size, then they are said to be congruent figure. The phenomena between these two figures is said to be congruency. CONDITIONS

More information

Chapter 3. - parts of a circle.

Chapter 3. - parts of a circle. Chapter 3 - parts of a circle. 3.1 properties of circles. - area of a sector of a circle. the area of the smaller sector can be found by the following formula: A = qº 360º pr2, given q in degrees, or!

More information

= ( +1) BP AC = AP + (1+ )BP Proved UNIT-9 CIRCLES 1. Prove that the parallelogram circumscribing a circle is rhombus. Ans Given : ABCD is a parallelogram circumscribing a circle. To prove : - ABCD is

More information

In the figure, ACD is a straight line. Find x. A. 40 B. 55 C. 85 D. 110 Solution: The answer is B. 140 = x 110 = 2x x = 55

In the figure, ACD is a straight line. Find x. A. 40 B. 55 C. 85 D. 110 Solution: The answer is B. 140 = x 110 = 2x x = 55 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

More information

Triangles. 3.In the following fig. AB = AC and BD = DC, then ADC = (A) 60 (B) 120 (C) 90 (D) none 4.In the Fig. given below, find Z.

Triangles. 3.In the following fig. AB = AC and BD = DC, then ADC = (A) 60 (B) 120 (C) 90 (D) none 4.In the Fig. given below, find Z. Triangles 1.Two sides of a triangle are 7 cm and 10 cm. Which of the following length can be the length of the third side? (A) 19 cm. (B) 17 cm. (C) 23 cm. of these. 2.Can 80, 75 and 20 form a triangle?

More information

SSC CGL Tier 1 and Tier 2 Program

SSC CGL Tier 1 and Tier 2 Program Gurudwara Road Model Town, Hisar 9729327755 www.ssccglpinnacle.com SSC CGL Tier 1 and Tier 2 Program -------------------------------------------------------------------------------------------------------------------

More information

Grade 9 Circles. Answer the questions. For more such worksheets visit

Grade 9 Circles. Answer the questions. For more such worksheets visit ID : ae-9-circles [1] Grade 9 Circles For more such worksheets visit www.edugain.com Answer the questions (1) Two circles with centres O and O intersect at two points A and B. A line PQ is drawn parallel

More information

Page 1 of 15. Website: Mobile:

Page 1 of 15. Website:    Mobile: Exercise 10.2 Question 1: From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5

More information

Grade 9 Circles. Answer t he quest ions. For more such worksheets visit

Grade 9 Circles. Answer t he quest ions. For more such worksheets visit ID : th-9-circles [1] Grade 9 Circles For more such worksheets visit www.edugain.com Answer t he quest ions (1) ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it

More information

Answer Key. 9.1 Parts of Circles. Chapter 9 Circles. CK-12 Geometry Concepts 1. Answers. 1. diameter. 2. secant. 3. chord. 4.

Answer Key. 9.1 Parts of Circles. Chapter 9 Circles. CK-12 Geometry Concepts 1. Answers. 1. diameter. 2. secant. 3. chord. 4. 9.1 Parts of Circles 1. diameter 2. secant 3. chord 4. point of tangency 5. common external tangent 6. common internal tangent 7. the center 8. radius 9. chord 10. The diameter is the longest chord in

More information

Triangles. Chapter Flowchart. The Chapter Flowcharts give you the gist of the chapter flow in a single glance.

Triangles. Chapter Flowchart. The Chapter Flowcharts give you the gist of the chapter flow in a single glance. Triangles Chapter Flowchart The Chapter Flowcharts give you the gist of the chapter flow in a single glance. Triangle A plane figure bounded by three line segments is called a triangle. Types of Triangles

More information

AREAS OF PARALLELOGRAMS AND TRIANGLES

AREAS OF PARALLELOGRAMS AND TRIANGLES AREAS OF PARALLELOGRAMS AND TRIANGLES Main Concepts and Results: The area of a closed plane figure is the measure of the region inside the figure: Fig.1 The shaded parts (Fig.1) represent the regions whose

More information

Revision Question Bank

Revision Question Bank Revision Question Bank Triangles 1. In the given figure, find the values of x and y. Since, AB = AC C = B [angles opposite to the equal sides are equal] x = 50 Also, the sum of all angles of a triangle

More information

10. Circles. Q 5 O is the centre of a circle of radius 5 cm. OP AB and OQ CD, AB CD, AB = 6 cm and CD = 8 cm. Determine PQ. Marks (2) Marks (2)

10. Circles. Q 5 O is the centre of a circle of radius 5 cm. OP AB and OQ CD, AB CD, AB = 6 cm and CD = 8 cm. Determine PQ. Marks (2) Marks (2) 10. Circles Q 1 True or False: It is possible to draw two circles passing through three given non-collinear points. Mark (1) Q 2 State the following statement as true or false. Give reasons also.the perpendicular

More information

1. Observe and Explore

1. Observe and Explore 1 2 3 1. Observe and Explore 4 Circle Module - 13 13-.1 Introduction : Study of circle play very important role in the study of geometry as well as in real life. Path traced by satellite, preparing wheels

More information

A plane can be names using a capital cursive letter OR using three points, which are not collinear (not on a straight line)

A plane can be names using a capital cursive letter OR using three points, which are not collinear (not on a straight line) Geometry - Semester 1 Final Review Quadrilaterals (Including some corrections of typos in the original packet) 1. Consider the plane in the diagram. Which are proper names for the plane? Mark all that

More information

Chapter 7. Geometric Inequalities

Chapter 7. Geometric Inequalities 4. Let m S, then 3 2 m R. Since the angles are supplementary: 3 2580 4568 542 Therefore, m S 42 and m R 38. Part IV 5. Statements Reasons. ABC is not scalene.. Assumption. 2. ABC has at least 2. Definition

More information

Geometry: Introduction, Circle Geometry (Grade 12)

Geometry: Introduction, Circle Geometry (Grade 12) OpenStax-CNX module: m39327 1 Geometry: Introduction, Circle Geometry (Grade 12) Free High School Science Texts Project This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution

More information

Solution 1: (i) Similar (ii) Similar (iii) Equilateral (iv) (a) Equal (b) Proportional

Solution 1: (i) Similar (ii) Similar (iii) Equilateral (iv) (a) Equal (b) Proportional Class X - NCERT Maths EXERCISE NO: 6.1 Question 1: Fill in the blanks using correct word given in the brackets: (i) All circles are. (congruent, similar) (ii) All squares are. (similar, congruent) (iii)

More information

Grade 9 Quadrilaterals

Grade 9 Quadrilaterals ID : ww-9-quadrilaterals [1] Grade 9 Quadrilaterals For more such worksheets visit www.edugain.com Answer t he quest ions (1) ABCD is a rectangle and point P is such that PB = 3 2 cm, PC = 4 cm and PD

More information

Mathematics 2260H Geometry I: Euclidean geometry Trent University, Winter 2012 Quiz Solutions

Mathematics 2260H Geometry I: Euclidean geometry Trent University, Winter 2012 Quiz Solutions Mathematics 2260H Geometry I: Euclidean geometry Trent University, Winter 2012 Quiz Solutions Quiz #1. Tuesday, 17 January, 2012. [10 minutes] 1. Given a line segment AB, use (some of) Postulates I V,

More information

Circle and Cyclic Quadrilaterals. MARIUS GHERGU School of Mathematics and Statistics University College Dublin

Circle and Cyclic Quadrilaterals. MARIUS GHERGU School of Mathematics and Statistics University College Dublin Circle and Cyclic Quadrilaterals MARIUS GHERGU School of Mathematics and Statistics University College Dublin 3 Basic Facts About Circles A central angle is an angle whose vertex is at the center of the

More information

CIRCLES MODULE - 3 OBJECTIVES EXPECTED BACKGROUND KNOWLEDGE. Circles. Geometry. Notes

CIRCLES MODULE - 3 OBJECTIVES EXPECTED BACKGROUND KNOWLEDGE. Circles. Geometry. Notes Circles MODULE - 3 15 CIRCLES You are already familiar with geometrical figures such as a line segment, an angle, a triangle, a quadrilateral and a circle. Common examples of a circle are a wheel, a bangle,

More information

HKDSE2018 Mathematics (Compulsory Part) Paper 2 Solution 1. B 4 (2 ) = (2 ) 2. D. α + β. x x. α β 3. C. h h k k ( 4 ) 6( 2 )

HKDSE2018 Mathematics (Compulsory Part) Paper 2 Solution 1. B 4 (2 ) = (2 ) 2. D. α + β. x x. α β 3. C. h h k k ( 4 ) 6( 2 ) HKDSE08 Mthemtics (Compulsory Prt) Pper Solution. B n+ 8 n+ 4 ( ) ( ) n+ n+ 6n+ 6n+ (6n+ ) (6n+ ). D α β x x α x β ( x) α x β β x α x + β x β ( α + β ) x β β x α + β. C 6 4 h h k k ( 4 ) 6( ) h k h + k

More information

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE

MOCKTIME.COM ONLINE TEST SERIES CORRESPONDENCE COURSE GEOMETRY TRIANGLES AND THEIR PROPERTIES A triangle is a figure enclosed by three sides. In the figure given below, ABC is a triangle with sides AB, BC, and CA measuring c, a, and b units, respectively.

More information

Math 9 Chapter 8 Practice Test

Math 9 Chapter 8 Practice Test Name: Class: Date: ID: A Math 9 Chapter 8 Practice Test Short Answer 1. O is the centre of this circle and point Q is a point of tangency. Determine the value of t. If necessary, give your answer to the

More information

Day 6: Triangle Congruence, Correspondence and Styles of Proof

Day 6: Triangle Congruence, Correspondence and Styles of Proof Name: Day 6: Triangle Congruence, Correspondence and Styles of Proof Date: Geometry CC (M1D) Opening Exercise Given: CE bisects BD Statements 1. bisects 1.Given CE BD Reasons 2. 2. Define congruence in

More information

Similarity of Triangle

Similarity of Triangle Similarity of Triangle 95 17 Similarity of Triangle 17.1 INTRODUCTION Looking around you will see many objects which are of the same shape but of same or different sizes. For examples, leaves of a tree

More information

Euclidian Geometry Grade 10 to 12 (CAPS)

Euclidian Geometry Grade 10 to 12 (CAPS) Euclidian Geometry Grade 10 to 12 (CAPS) Compiled by Marlene Malan marlene.mcubed@gmail.com Prepared by Marlene Malan CAPS DOCUMENT (Paper 2) Grade 10 Grade 11 Grade 12 (a) Revise basic results established

More information

Question Bank Tangent Properties of a Circle

Question Bank Tangent Properties of a Circle Question Bank Tangent Properties of a Circle 1. In quadrilateral ABCD, D = 90, BC = 38 cm and DC = 5 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 7 cm. Find

More information

Class 9 Geometry-Overall

Class 9 Geometry-Overall ID : in-9-geometry-overall [1] Class 9 Geometry-Overall For more such worksheets visit www.edugain.com Answer t he quest ions (1) AB is diameter of the circle. If A and B are connected to E, circle is

More information

(A) 50 (B) 40 (C) 90 (D) 75. Circles. Circles <1M> 1.It is possible to draw a circle which passes through three collinear points (T/F)

(A) 50 (B) 40 (C) 90 (D) 75. Circles. Circles <1M> 1.It is possible to draw a circle which passes through three collinear points (T/F) Circles 1.It is possible to draw a circle which passes through three collinear points (T/F) 2.The perpendicular bisector of two chords intersect at centre of circle (T/F) 3.If two arcs of a circle

More information

Triangle Congruence and Similarity Review. Show all work for full credit. 5. In the drawing, what is the measure of angle y?

Triangle Congruence and Similarity Review. Show all work for full credit. 5. In the drawing, what is the measure of angle y? Triangle Congruence and Similarity Review Score Name: Date: Show all work for full credit. 1. In a plane, lines that never meet are called. 5. In the drawing, what is the measure of angle y? A. parallel

More information

RMT 2013 Geometry Test Solutions February 2, = 51.

RMT 2013 Geometry Test Solutions February 2, = 51. RMT 0 Geometry Test Solutions February, 0. Answer: 5 Solution: Let m A = x and m B = y. Note that we have two pairs of isosceles triangles, so m A = m ACD and m B = m BCD. Since m ACD + m BCD = m ACB,

More information

Postulates and Theorems in Proofs

Postulates and Theorems in Proofs Postulates and Theorems in Proofs A Postulate is a statement whose truth is accepted without proof A Theorem is a statement that is proved by deductive reasoning. The Reflexive Property of Equality: a

More information

Topic 2 [312 marks] The rectangle ABCD is inscribed in a circle. Sides [AD] and [AB] have lengths

Topic 2 [312 marks] The rectangle ABCD is inscribed in a circle. Sides [AD] and [AB] have lengths Topic 2 [312 marks] 1 The rectangle ABCD is inscribed in a circle Sides [AD] and [AB] have lengths [12 marks] 3 cm and (\9\) cm respectively E is a point on side [AB] such that AE is 3 cm Side [DE] is

More information

Chapter 2 Preliminaries

Chapter 2 Preliminaries Chapter 2 Preliminaries Where there is matter, there is geometry. Johannes Kepler (1571 1630) 2.1 Logic 2.1.1 Basic Concepts of Logic Let us consider A to be a non-empty set of mathematical objects. One

More information

1. Prove that the parallelogram circumscribing a circle is rhombus.

1. Prove that the parallelogram circumscribing a circle is rhombus. UNIT-9 1. Prve that the parallelgram circumscribing a circle is rhmbus. Ans Given : ABCD is a parallelgram circumscribing a circle. T prve : - ABCD is a rhmbus r ABBCCDDA Prf: Since the length f tangents

More information

Grade 9 Geometry-Overall

Grade 9 Geometry-Overall ID : au-9-geometry-overall [1] Grade 9 Geometry-Overall For more such worksheets visit www.edugain.com Answer t he quest ions (1) A chord of a circle is equal to its radius. Find the angle subtended by

More information

Visit: ImperialStudy.com For More Study Materials Class IX Chapter 12 Heron s Formula Maths

Visit: ImperialStudy.com For More Study Materials Class IX Chapter 12 Heron s Formula Maths Exercise 1.1 1. Find the area of a triangle whose sides are respectively 150 cm, 10 cm and 00 cm. The triangle whose sides are a = 150 cm b = 10 cm c = 00 cm The area of a triangle = s(s a)(s b)(s c) Here

More information

Maharashtra Board Class X Mathematics - Geometry Board Paper 2014 Solution. Time: 2 hours Total Marks: 40

Maharashtra Board Class X Mathematics - Geometry Board Paper 2014 Solution. Time: 2 hours Total Marks: 40 Maharashtra Board Class X Mathematics - Geometry Board Paper 04 Solution Time: hours Total Marks: 40 Note: - () All questions are compulsory. () Use of calculator is not allowed.. i. Ratio of the areas

More information

1) Exercise 1 In the diagram, ABC = AED, AD = 3, DB = 2 and AE = 2. Determine the length of EC. Solution:

1) Exercise 1 In the diagram, ABC = AED, AD = 3, DB = 2 and AE = 2. Determine the length of EC. Solution: 1) Exercise 1 In the diagram, ABC = AED, AD = 3, DB = 2 and AE = 2. Determine the length of EC. Solution: First, we show that AED and ABC are similar. Since DAE = BAC and ABC = AED, we have that AED is

More information

Set 6 Paper 1. Set 6 Paper 1. 1 Pearson Education Asia Limited Section A(1) (Pyth. Theorem) (b) 24units Area of OPQ. a b (4)

Set 6 Paper 1. Set 6 Paper 1. 1 Pearson Education Asia Limited Section A(1) (Pyth. Theorem) (b) 24units Area of OPQ. a b (4) Set Paper Set Paper Section A().. a b a b 8 ( a b) a b ( 8) a b a b a b k k k h k. The weight of Sam 5kg( %) 5kg The weight of Benny 5kg( %). 5. (a).85kg 5kg Benny is the heaviest one among them, his claim

More information

CONGRUENCE OF TRIANGLES

CONGRUENCE OF TRIANGLES Congruence of Triangles 11 CONGRUENCE OF TRIANGLES You might have observed that leaves of different trees have different shapes, but leaves of the same tree have almost the same shape. Although they may

More information

Maharashtra State Board Class X Mathematics Geometry Board Paper 2015 Solution. Time: 2 hours Total Marks: 40

Maharashtra State Board Class X Mathematics Geometry Board Paper 2015 Solution. Time: 2 hours Total Marks: 40 Maharashtra State Board Class X Mathematics Geometry Board Paper 05 Solution Time: hours Total Marks: 40 Note:- () Solve all questions. Draw diagrams wherever necessary. ()Use of calculator is not allowed.

More information

Triangles. Example: In the given figure, S and T are points on PQ and PR respectively of PQR such that ST QR. Determine the length of PR.

Triangles. Example: In the given figure, S and T are points on PQ and PR respectively of PQR such that ST QR. Determine the length of PR. Triangles Two geometric figures having the same shape and size are said to be congruent figures. Two geometric figures having the same shape, but not necessarily the same size, are called similar figures.

More information

It is known that the length of the tangents drawn from an external point to a circle is equal.

It is known that the length of the tangents drawn from an external point to a circle is equal. CBSE -MATHS-SET 1-2014 Q1. The first three terms of an AP are 3y-1, 3y+5 and 5y+1, respectively. We need to find the value of y. We know that if a, b and c are in AP, then: b a = c b 2b = a + c 2 (3y+5)

More information

THEOREMS WE KNOW PROJECT

THEOREMS WE KNOW PROJECT 1 This is a list of all of the theorems that you know and that will be helpful when working on proofs for the rest of the unit. In the Notes section I would like you to write anything that will help you

More information

1. B (27 9 ) = [3 3 ] = (3 ) = 3 2. D. = c d dy d = cy + c dy cy = d + c. y( d c) 3. D 4. C

1. B (27 9 ) = [3 3 ] = (3 ) = 3 2. D. = c d dy d = cy + c dy cy = d + c. y( d c) 3. D 4. C HKDSE03 Mathematics (Compulsory Part) Paper Full Solution. B (7 9 ) [3 3 ] (3 ) 3 n + 3 3 ( n + ) 3 n + 5 3 6 n + 5. D y y + c d dy d cy + c dy cy d + c y( d c) c + d c + d y d c 3. D hl kl + hm km hn

More information

CBSE MATHEMATICS (SET-2)_2019

CBSE MATHEMATICS (SET-2)_2019 CBSE 09 MATHEMATICS (SET-) (Solutions). OC AB (AB is tangent to the smaller circle) In OBC a b CB CB a b CB a b AB CB (Perpendicular from the centre bisects the chord) AB a b. In PQS PQ 4 (By Pythagoras

More information

Class 9 Quadrilaterals

Class 9 Quadrilaterals ID : in-9-quadrilaterals [1] Class 9 Quadrilaterals For more such worksheets visit www.edugain.com Answer t he quest ions (1) The diameter of circumcircle of a rectangle is 13 cm and rectangle's width

More information

GEOMETRY OF THE CIRCLE TANGENTS & SECANTS

GEOMETRY OF THE CIRCLE TANGENTS & SECANTS Geometry Of The Circle Tangents & Secants GEOMETRY OF THE CIRCLE TANGENTS & SECANTS SOLUTIONS www.mathletics.com.au Basics Page 3 questions 1. What is the difference between a secant and a tangent? A

More information

Abhilasha Classses. Class X (IX to X Moving) Date: MM 150 Mob no (Set-AAA) Sol: Sol: Sol: Sol:

Abhilasha Classses. Class X (IX to X Moving) Date: MM 150 Mob no (Set-AAA) Sol: Sol: Sol: Sol: Class X (IX to X Moving) Date: 0-6 MM 0 Mob no.- 97967 Student Name... School.. Roll No... Contact No....... If = y = 8 z and + + =, then the y z value of is (a) 7 6 (c) 7 8 [A] (b) 7 3 (d) none of these

More information

Figure 1: Problem 1 diagram. Figure 2: Problem 2 diagram

Figure 1: Problem 1 diagram. Figure 2: Problem 2 diagram Geometry A Solutions 1. Note that the solid formed is a generalized cylinder. It is clear from the diagram that the area of the base of this cylinder (i.e., a vertical cross-section of the log) is composed

More information

Question 1 ( 1.0 marks) places of decimals? Solution: Now, on dividing by 2, we obtain =

Question 1 ( 1.0 marks) places of decimals? Solution: Now, on dividing by 2, we obtain = Question 1 ( 1.0 marks) The decimal expansion of the rational number places of decimals? will terminate after how many The given expression i.e., can be rewritten as Now, on dividing 0.043 by 2, we obtain

More information

Nozha Directorate of Education Form : 2 nd Prep

Nozha Directorate of Education Form : 2 nd Prep Cairo Governorate Department : Maths Nozha Directorate of Education Form : 2 nd Prep Nozha Language Schools Geometry Revision Sheet Ismailia Road Branch Sheet ( 1) 1-Complete 1. In the parallelogram, each

More information

Class-IX CBSE Latest Pattern Sample Paper {Mathematics}

Class-IX CBSE Latest Pattern Sample Paper {Mathematics} Class-IX CBSE Latest Pattern Sample Paper {Mathematics} Term-I Examination (SA I) Time: 3hours Max. Marks: 90 General Instructions (i) All questions are compulsory. (ii) The question paper consists of

More information

AREAS OF PARALLELOGRAMS AND TRIANGLES

AREAS OF PARALLELOGRAMS AND TRIANGLES 15 MATHEMATICS AREAS OF PARALLELOGRAMS AND TRIANGLES CHAPTER 9 9.1 Introduction In Chapter 5, you have seen that the study of Geometry, originated with the measurement of earth (lands) in the process of

More information

Class IX Chapter 9 Areas of Parallelograms and Triangles Maths

Class IX Chapter 9 Areas of Parallelograms and Triangles Maths 1 NCRTSOLUTIONS.blogspot.com Class IX Chapter 9 Areas of Parallelograms and Triangles Maths 1: Exercise 9.1 Question Which of the following figures lie on the same base and between the same parallels.

More information

Honors Geometry Review Exercises for the May Exam

Honors Geometry Review Exercises for the May Exam Honors Geometry, Spring Exam Review page 1 Honors Geometry Review Exercises for the May Exam C 1. Given: CA CB < 1 < < 3 < 4 3 4 congruent Prove: CAM CBM Proof: 1 A M B 1. < 1 < 1. given. < 1 is supp to

More information

Grade 7 Lines and Angles

Grade 7 Lines and Angles ID : ph-7-lines-and-angles [1] Grade 7 Lines and Angles For more such worksheets visit www.edugain.com Answer t he quest ions (1) If CD is perpendicular to AB, and CE bisect angle ACB, f ind the angle

More information

Honors Geometry Mid-Term Exam Review

Honors Geometry Mid-Term Exam Review Class: Date: Honors Geometry Mid-Term Exam Review Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. Classify the triangle by its sides. The

More information

Mathematics. Exercise 6.4. (Chapter 6) (Triangles) (Class X) Question 1: Let and their areas be, respectively, 64 cm 2 and 121 cm 2.

Mathematics. Exercise 6.4. (Chapter 6) (Triangles) (Class X) Question 1: Let and their areas be, respectively, 64 cm 2 and 121 cm 2. () Exercise 6.4 Question 1: Let and their areas be, respectively, 64 cm 2 and 121 cm 2. If EF = 15.4 cm, find BC. Answer 1: 1 () Question 2: Diagonals of a trapezium ABCD with AB DC intersect each other

More information

HANOI OPEN MATHEMATICS COMPETITON PROBLEMS

HANOI OPEN MATHEMATICS COMPETITON PROBLEMS HANOI MATHEMATICAL SOCIETY NGUYEN VAN MAU HANOI OPEN MATHEMATICS COMPETITON PROBLEMS 2006-2013 HANOI - 2013 Contents 1 Hanoi Open Mathematics Competition 3 1.1 Hanoi Open Mathematics Competition 2006...

More information

Grade 9 Lines and Angles

Grade 9 Lines and Angles ID : ww-9-lines-and-angles [1] Grade 9 Lines and Angles For more such worksheets visit www.edugain.com Answer t he quest ions (1) If CD is perpendicular to AB, and CE bisect angle ACB, f ind the angle

More information

Class X Chapter 15 Similarity Maths

Class X Chapter 15 Similarity Maths EXERCISE 15(A) Book Name: Selina Concise Question 1: State, true or false: (i) Two similar polygons are necessarily congruent. (ii) Two congruent polygons are necessarily similar. (iii) all equiangular

More information

Geometry Honors Review for Midterm Exam

Geometry Honors Review for Midterm Exam Geometry Honors Review for Midterm Exam Format of Midterm Exam: Scantron Sheet: Always/Sometimes/Never and Multiple Choice 40 Questions @ 1 point each = 40 pts. Free Response: Show all work and write answers

More information

Chapter (Circle) * Circle - circle is locus of such points which are at equidistant from a fixed point in

Chapter (Circle) * Circle - circle is locus of such points which are at equidistant from a fixed point in Chapter - 10 (Circle) Key Concept * Circle - circle is locus of such points which are at equidistant from a fixed point in a plane. * Concentric circle - Circle having same centre called concentric circle.

More information

Ch 10 Review. Multiple Choice Identify the choice that best completes the statement or answers the question.

Ch 10 Review. Multiple Choice Identify the choice that best completes the statement or answers the question. Ch 10 Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. In the diagram shown, the measure of ADC is a. 55 b. 70 c. 90 d. 180 2. What is the measure

More information

Basic Quadrilateral Proofs

Basic Quadrilateral Proofs Basic Quadrilateral Proofs For each of the following, draw a diagram with labels, create the givens and proof statement to go with your diagram, then write a two-column proof. Make sure your work is neat

More information

Grade 9 Lines and Angles

Grade 9 Lines and Angles ID : sg-9-lines-and-angles [1] Grade 9 Lines and Angles For more such worksheets visit www.edugain.com Answer t he quest ions (1) If AB and PQ are parallel, compute the angle Z. (2) Find the value of a+b.

More information

Class IX Chapter 9 Areas of Parallelograms and Triangles Maths

Class IX Chapter 9 Areas of Parallelograms and Triangles Maths Exercise 9.1 Question 1: Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels. (i) (ii) (iii) (iv) (v) (vi) (i)

More information

Set 5 Paper 2. Set 5 Paper 2. 1 Pearson Education Asia Limited 2017

Set 5 Paper 2. Set 5 Paper 2. 1 Pearson Education Asia Limited 2017 Set Paper Set Paper. B. C. B. C. C 6. D 7. A. D. A. A. C. C. B. B. C 6. C 7. C. A. B. D. B. D. A. A. B 6. B 7. D. D. C. A. C. D. D. A. D 6. D 7. A. A. C. C. B. D. B. D. A Section A. B ( 7) 7 ( ) 7 ( )

More information