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MATHEMATICS IN EVERYDAY LIFE–7, ANSWER KEYS, , Chapter 14 : Congruence of Triangles, 4., , EXERCISE 14.1, P, , 1., , (i) No, the triangles equal in area may not be, congruent., Consider two triangles as shown in the figures, given below:, , 5 cm, , Six elements of PQR are its three sides and three, angles., Three sides: PQ, QR and PR., Three angles: P, Q and R., 2. XYZ RPQ under the correspondence, XYZ RPQ., Therefore, all corresponding congruent parts of the, triangles are:, X R, Y P, Z Q and side XY side, RP, side YZ side PQ, side ZX side QR., 3. DEF BCA, under the correspondence DEF BCA., This means D B; E C; F A., Therefore, the corresponding parts of BCA are :, D, , E, , B, , F, , C, , (i) F A, (ii) DE BC, (iii) D B, , 4 cm, , 3 cm, , R, , Q, , A, , 3 cm, , 4 cm, , These triangles are equal in area but they are not, congruent., (ii) Yes, congruent rectangles have equal area., Two rectangles are congruent, if their lengths, and breadths are same i.e., same area., (iii) Yes, the squares having equal area are, congruent., Two squares are congruent if they have, same side length i.e., same area., (iv) No, all squares are not congruent as they do not, have same side length., (v) No, circles with same centre are not congruent, as they have different radii., Circles with same radii are congruent., 5. Given that :, BOD AOC, To prove :, BOC AOD, Proof :, , BOD AOC, If two angles are congruent, their measures are same., , BOD = AOC, , (iv) EF CA, , A, , (v) DF BA, , C, D, O, , Mathematics In Everyday Life-7, , B, 1
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Adding COD on both sides, we get, BOD + COD = AOC + COD, , BOC = AOD, If two angles have the same measure, they are congruent., , BOC AOD., (Hence proved), , Median AD = Median XE, BD = YE, ABD XYE, , Thus,, , S, , 1. In ABC and ZXY,, AB = ZX = 2.6 cm, BC = XY = 3.5 cm, CA = YZ = 3 cm, , Q, , Now, in PQR and RSP,, , cm, , cm, , cm, , QR = SP (opposite sides of a parallelogram), , 3, , 2.6, cm, , Y, , Diagonal PR is common., , C, , 3.5 cm, , P, , PQ = RS (opposite sides of a parallelogram), , 2.6, , 3, , B, , R, , (given), (given), (given), 3.5 cm, , X, , (by SSS congruence rule), , 4. In the parallelogram PQRS, we have PQ = RS and, QR = SP. PR is diagonal., , EXERCISE 14.2, , A, , [from (i)], , Z, , Hence, PQR RSP, , Therefore, ABC and ZXY are congruent., ABC ZXY (By SSS congruence rule), 2. (i) In BAC and BDC,, BA = BD, (given), AC = DC, (given), BC is common., , 5., , (by SSS congruence rule), , (i) In ABC and PQR,, AB = PQ = 7 cm,, BC = QR = 5 cm,, CA = RP = 6 cm, 7 cm, , A, , Q, , B, , cm, , m, , C, , B, , D, , 3., , P, , C, , Hence, BAC BDC, (ii), ABC = DBC, Hence, ABC = CBD, , (By SSS congruence rule), (by C.P.C.T.), ( CBD = DBC), , A, , , , , 7, , cm, , 5c, , 6, , 5 cm, , A, , R, , 6 cm, , A P, B Q, C R, ABC PQR, , (By SSS congruence rule), , (ii) We have,, AB = QR = 4 cm,, , X, , BC = RP = 5.5 cm, CA = PQ = 5 cm, , Z, , 2, , cm, , AB = XY, , 5., 5, , 5, , D, E, Given that : In ABC and XYZ, AB = XY,, BC = YZ and median AD = median XE., To Prove : ABD XYE, Proof : BC = YZ, (given), , 2BD = 2YE ( AD and XE are median), BD = YE, ... (i), In ABD and XYE,, , B, , 5.5 cm, , , , C, , Q, 4 cm, , Y, , m, , C, , 4c, , B, , 5 cm, , P, , A, , cm, , R, , A Q, B R, C P, , Hence,ABC QRP (By SSS congruence rule), , (given), Answer Keys
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(i) In ABC and RPQ, AB = RP = 14 cm, BC = PQ = 17 cm, CA = QR = 16 cm, , c, 6.5, , 16, , 14, , C, , Y, P, , Q, , X, , R, , 5, , Z, , 4, , D, , B, , 7, , 1., , (i) In ABC and FDE,, AB = FD = 7 cm, ABC = FDE = 90°, BC = DE = 5.2 cm, , 8, , E, , D, 7 cm, B, , C, 5.2 cm, , 7c, m, , E, , F, , ABC FDE (By SAS congruence rule), (ii) In PRQ and ABC,, PR = AB = 8.5 cm, RQ = BC = 12.6 cm, But included R included B, , P, , 8, , 6, , EXERCISE 14.3, , 3, , PQR ZXY (By SSS congruence rule), , P Z, Q X, R Y., (iii) In ABC and DFE,, AB = DF = 6 cm,, BC = FE = 7 cm,, CA = ED = 8 cm, A, , m, , A, , 5, , 3, , c, 7.2, , C, Thus, ABC DBC(By SSS congruence rule), , A = D, ABC = DBC, ACB = DCB, (Corresponding angles of congruent triangles), , R, , Thus, ABC RPQ (By SSS congruence rule), , A = R, B = P, C = Q., (ii) In PQR and ZXY,, PQ = ZX = 4 cm, QR = XY = 5 cm, RP = YZ = 3 cm, , 4, , 7.2, cm, , 16, , D, , 8 cm, , Q, , 17, , 14, , 17, , 6.5, cm, , m, , A, , P, , A, , B, , B, , 5.2, cm, , 6., , 6, C, , 8.5, cm, , 7, F, , ABC DFE (By SSS congruence rule), A D, B F, C E, (Corresponding angles of congruent triangles), 7. In ABC and DBC,, AB = DB = 6.5 cm, BC is common., CA = CD = 7.2 cm, , Q, , 35°, 12.6 cm, A, , , , Mathematics In Everyday Life-7, , B, , cm, 8.5, 35°, , 12.6 cm, , R, , C, , So we cannot say that the triangles are, congruent., , 3
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2. Given that : In ABC, altitude AD bisects BC., To prove : ADB ADC, Proof : Since, altitude AD bisects BC., , BD = DC, ... (i), Now, in ADB and ADC, AD is common., , A, , B, , C, D, ADB = ADC = 90° ( AD is an altitude), DB = DC, [From (i)], , ADB ADC (By SSS congruence rule), , AB = AC, (Corresponding sides of congruent triangles), Equal pairs of sides of these two triangles are, AB = AC, DB = DC and AD common., 3. Given that : PQ = SR and PQ || SR., To prove : PSR RQP, Construction : Draw a diagonal PR., Q, R, , S, P, Proof : PQ SR, , QPR = SRP, ... (i) (Alternate angles), Now, in PSR and RQP,, SR = QP, (given), SRP = QPR, [from (i)], RP is common., , PSR RQP (By SAS congruence rule), Hence,, PS = QR, (Corresponding sides of congruent triangles), 4. Given that : In ABC, AD is the bisector of A and, AB = AC., To prove : B = C, A, , Proof : Since, AD bisects A, , BAD = CAD, ... (i), In ABD and ACD,, AB = AC, (given), BAD = CAD, [from (i)], AD is common., , ABD ACD (By SAS congruence rule), Hence,, B = C, (Corresponding angles of congruent triangles), Angles opposite to equal sides are equal., 5. Given that : ABCD is a quadrilateral and AC is a diagonal., To prove : ABC CDA, Proof : Since, diagonal AC divides the quadrilateral, ABCD in two triangles., In ABC and CDA,, BA = DC = 6.8 cm, (given), BAC = DCA = 35°, (given), AC is common., , A, , 6.8 cm, 35°, , 35°, D, , C, , 6.8 cm, , , ABC CDA (By SAS congruence rule), , BAC = DCA = 35°, (Alternate angles), AC is transversal., , AB || CD., , EXERCISE 14.4, 1. Given that : Line l m, and M is the mid point of, line segment PQ., To prove : M is also mid point of RS., Proof : l m, MPR = MQS and MRP = MSQ, ...(i), (Alternate angles), R, , P, , 4, , D, , l, , M, S, , Q, , M is mid point of PQ., MP = MQ, Now, in MPR and MQS,, MPR = MQS, , B, , B, , m, , ... (ii), [from (i)], , C, Answer Keys
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(iii), , (ii) In ABC and FED,, ABC = FED = 90°, hypotenuse CA = hypotenuse DF, side AB = side FE, , P, , X, 60°, 6 cm, , Y, , D, , E, , C, , F, , A, , 80°, , 60° 80°, R, 6 cm, , Q, , Z, , In YXZ and PQR,, , B, , , , X = Q = 60°, side XZ = side QR, Z = R = 80°, YXZ PQR (By ASA congruence rule), 5. Given that : In ABC, B = C, BL and CM bisect, B and C respectively., To prove : BL = CM, Proof : B = C, A, , ABC FED, (By RHS congruence rule), 2. Given that : In ABC, P is a point on side BC such, that PL AB and PM AC., To prove : AP bisects BAC., Proof : In ALP and AMP,, ALP = AMP = 90°, ( PL AB and PM AC), A, , L, , M, , M, , L, B, , C, , 2LBC = 2MCB, (BL and CM bisect B and C), LBC = MCB, .. (ii), BMC and CLB,, CBM = BCL, (given), BC is common., MCB = LBC, [from (i)], , BMC CLB (By ASA congruence rule), , BL = CM, (Corresponding sides of congruent triangles), , C, , B, , P, Hypotenuse AP is common., side LP = side MP, , , , , ALP AMP (By RHS congruence rule), PAL = PAM, (Corresponding angles of congruent triangles), Hence, AP bisects BAC., Proved, 3., , A, , D, , EXERCISE 14.5, 1., , (i) In PRQ and CBD,, PRQ = CBD = 90°, hypotenuse QP = hypotenuse DC., PR = CB, , R, , P, , D, , C, , Q, , (given), (given), (given), , B, , Hence,PRQ CBD, (By RHS congruence rule), , 6, , B, , L, , C E, , F, , M, , Given that : Two triangles ABC and DEF are such, that, AL BC, DM EF, AB = DE, AC = DF, AL = DM, To prove : ABC DEF, Proof : In ALB and DME,, ALB = DME = 90°, ( AL and DM are altitudes), AL = DM, (given), Answer Keys
