Page 1 A C B D Triangles 1) Let ABC DEF and their areas be, respectively, 64cm 2 and 121 cm 2. If EF=15.4 cm, find BC. Solution: Given: ABC DEF and A( ABC)=64cm 2 & A( DEF)= 121cm 2 and EF = 15.4cm. To find: BC = ? We know that, the ratio of the areas of two similar triangles is equal to the square of the ratio of their sides. i.e. BC 2 = = = 125.44 BC = 11.2 cm 2) In an equilateral triangle ABC, AD BC. Prove that 3AB 2 = 4AD 2 . Solution: Given: ABC is an equilateral triangle and AD BC. To prove that: 3AB 2 = 4AD 2 Proof: Since, AD BC therefore, BD = DC. Now, in right triangle ABD, we have AB 2 = AD 2 + BD 2 --- [Pythagoras theorem] Since BD = DC therefore, BD = BC AB 2 = AD 2 + AB 2 = AD 2 + AB 2 = AD 2 + AB 2 = AD 2 + ------ [Since, AB = BC = AC – sides of an equilateral triangle] AB 2 - = AD 2 = AD 2 3AB 2 = 4AD 2 Page 2 A C B D Triangles 1) Let ABC DEF and their areas be, respectively, 64cm 2 and 121 cm 2. If EF=15.4 cm, find BC. Solution: Given: ABC DEF and A( ABC)=64cm 2 & A( DEF)= 121cm 2 and EF = 15.4cm. To find: BC = ? We know that, the ratio of the areas of two similar triangles is equal to the square of the ratio of their sides. i.e. BC 2 = = = 125.44 BC = 11.2 cm 2) In an equilateral triangle ABC, AD BC. Prove that 3AB 2 = 4AD 2 . Solution: Given: ABC is an equilateral triangle and AD BC. To prove that: 3AB 2 = 4AD 2 Proof: Since, AD BC therefore, BD = DC. Now, in right triangle ABD, we have AB 2 = AD 2 + BD 2 --- [Pythagoras theorem] Since BD = DC therefore, BD = BC AB 2 = AD 2 + AB 2 = AD 2 + AB 2 = AD 2 + AB 2 = AD 2 + ------ [Since, AB = BC = AC – sides of an equilateral triangle] AB 2 - = AD 2 = AD 2 3AB 2 = 4AD 2 Triangles 3) In the given figure, PQR is a right triangle right angled at Q & QS PR. If PQ = 6cm and PS = 4cm, find the QS, RS & QR. Solution: Given: In PQR, PQR= 90 . PQ = 6cm, PS = 4cm. To find: QS , RS & QR? Since,QS PR, PS = SR And PS =4cm ----- (given ) SR = 4cm. PR = PS + SR = 4 + 4 =8 PQR is right angled triangle. by Pythagoras theorem, PR 2 = PQ 2 + QR 2 QR 2 = PR 2 - PQ 2 QR 2 = 8 2 - 6 2 QR 2 = 64 -36 = 28 QR = 2v QS 2 = PS SR -------- [ Since QS is perpendicular to PR] QS 2 = 4 4 = 16 QS = 4cm. R Q P S 6 4 Page 3 A C B D Triangles 1) Let ABC DEF and their areas be, respectively, 64cm 2 and 121 cm 2. If EF=15.4 cm, find BC. Solution: Given: ABC DEF and A( ABC)=64cm 2 & A( DEF)= 121cm 2 and EF = 15.4cm. To find: BC = ? We know that, the ratio of the areas of two similar triangles is equal to the square of the ratio of their sides. i.e. BC 2 = = = 125.44 BC = 11.2 cm 2) In an equilateral triangle ABC, AD BC. Prove that 3AB 2 = 4AD 2 . Solution: Given: ABC is an equilateral triangle and AD BC. To prove that: 3AB 2 = 4AD 2 Proof: Since, AD BC therefore, BD = DC. Now, in right triangle ABD, we have AB 2 = AD 2 + BD 2 --- [Pythagoras theorem] Since BD = DC therefore, BD = BC AB 2 = AD 2 + AB 2 = AD 2 + AB 2 = AD 2 + AB 2 = AD 2 + ------ [Since, AB = BC = AC – sides of an equilateral triangle] AB 2 - = AD 2 = AD 2 3AB 2 = 4AD 2 Triangles 3) In the given figure, PQR is a right triangle right angled at Q & QS PR. If PQ = 6cm and PS = 4cm, find the QS, RS & QR. Solution: Given: In PQR, PQR= 90 . PQ = 6cm, PS = 4cm. To find: QS , RS & QR? Since,QS PR, PS = SR And PS =4cm ----- (given ) SR = 4cm. PR = PS + SR = 4 + 4 =8 PQR is right angled triangle. by Pythagoras theorem, PR 2 = PQ 2 + QR 2 QR 2 = PR 2 - PQ 2 QR 2 = 8 2 - 6 2 QR 2 = 64 -36 = 28 QR = 2v QS 2 = PS SR -------- [ Since QS is perpendicular to PR] QS 2 = 4 4 = 16 QS = 4cm. R Q P S 6 4 Triangles 4) In the given figure, and If 1 , 4 BX BC ? find the ratio of areas of ?ABC and ?YCX. Solution: Since Similarly ACX CXY ? ? ? ? ?ABC ?YCX …….(AA – similarity criterion) 2 2 2 ) ( ) ( ? ? ? ? ? ? ? ? ? ? XC BC XC BC YCX ar ABC ar Since, 1 4 BX BC ? ….. (Given) ? ? 1 4 BC XC BX BC ? ? ? ? 4BC – 4XC = BC ? 3BC = 4XC ? 4 3 BC XC ? Putting values in eq.( i ), we get 9 : 16 9 16 3 4 ) ( ) ( 2 ? ? ? ? ? ? ? ? ? ? ? YCX ar ABC ar 5) PQR is an isosceles triangle right angled at B. Two equilateral triangles are constructed on side QR and PR as shown in the given figure. Prove that area of ?QRS = 1 2 area of ?PRT. Solution: In ?PQR, 90 Q ?? and consider PQ = QR = x ? PR 2 = PQ 2 + QR 2 --- (Pythagoras theorem) ? PR 2 = 2 x 2 ? PR = 2 x A B C X P Q R S T Page 4 A C B D Triangles 1) Let ABC DEF and their areas be, respectively, 64cm 2 and 121 cm 2. If EF=15.4 cm, find BC. Solution: Given: ABC DEF and A( ABC)=64cm 2 & A( DEF)= 121cm 2 and EF = 15.4cm. To find: BC = ? We know that, the ratio of the areas of two similar triangles is equal to the square of the ratio of their sides. i.e. BC 2 = = = 125.44 BC = 11.2 cm 2) In an equilateral triangle ABC, AD BC. Prove that 3AB 2 = 4AD 2 . Solution: Given: ABC is an equilateral triangle and AD BC. To prove that: 3AB 2 = 4AD 2 Proof: Since, AD BC therefore, BD = DC. Now, in right triangle ABD, we have AB 2 = AD 2 + BD 2 --- [Pythagoras theorem] Since BD = DC therefore, BD = BC AB 2 = AD 2 + AB 2 = AD 2 + AB 2 = AD 2 + AB 2 = AD 2 + ------ [Since, AB = BC = AC – sides of an equilateral triangle] AB 2 - = AD 2 = AD 2 3AB 2 = 4AD 2 Triangles 3) In the given figure, PQR is a right triangle right angled at Q & QS PR. If PQ = 6cm and PS = 4cm, find the QS, RS & QR. Solution: Given: In PQR, PQR= 90 . PQ = 6cm, PS = 4cm. To find: QS , RS & QR? Since,QS PR, PS = SR And PS =4cm ----- (given ) SR = 4cm. PR = PS + SR = 4 + 4 =8 PQR is right angled triangle. by Pythagoras theorem, PR 2 = PQ 2 + QR 2 QR 2 = PR 2 - PQ 2 QR 2 = 8 2 - 6 2 QR 2 = 64 -36 = 28 QR = 2v QS 2 = PS SR -------- [ Since QS is perpendicular to PR] QS 2 = 4 4 = 16 QS = 4cm. R Q P S 6 4 Triangles 4) In the given figure, and If 1 , 4 BX BC ? find the ratio of areas of ?ABC and ?YCX. Solution: Since Similarly ACX CXY ? ? ? ? ?ABC ?YCX …….(AA – similarity criterion) 2 2 2 ) ( ) ( ? ? ? ? ? ? ? ? ? ? XC BC XC BC YCX ar ABC ar Since, 1 4 BX BC ? ….. (Given) ? ? 1 4 BC XC BX BC ? ? ? ? 4BC – 4XC = BC ? 3BC = 4XC ? 4 3 BC XC ? Putting values in eq.( i ), we get 9 : 16 9 16 3 4 ) ( ) ( 2 ? ? ? ? ? ? ? ? ? ? ? YCX ar ABC ar 5) PQR is an isosceles triangle right angled at B. Two equilateral triangles are constructed on side QR and PR as shown in the given figure. Prove that area of ?QRS = 1 2 area of ?PRT. Solution: In ?PQR, 90 Q ?? and consider PQ = QR = x ? PR 2 = PQ 2 + QR 2 --- (Pythagoras theorem) ? PR 2 = 2 x 2 ? PR = 2 x A B C X P Q R S T Triangles Area of equilateral triangle ?QRS = 2 3 4 x Area of equilateral triangle ?PRT = ? ? ? ? 2 2 33 2 44 PR x ? Therefore area of ?PRT = 2 3 2 x = 2 2 2 3 2 3 2 2 4 x x ?? = 2 area of ?QRS 6) The areas of two similar triangles are144 cm 2 and 81 cm 2 respectively. If the median of the first triangle is 14.4 cm find the corresponding median of the other. Solution: Let two triangles be LMN and XYZ, such that ?LMN ?XYZ 2 2 )] ( [ )] ( [ ) ( ) ( XYZ Median LMN Median XYZ ar LMN ar ? ? ? ? ? ? 2 ) ( 4 . 14 81 144 ? ? ? ? ? ? ? ? ? ? XYZ Median ? (Median of ?XYZ) 2 = 14.4 14.4 81 144 ?? Median = 10.8 cm. L N R M K P T S Page 5 A C B D Triangles 1) Let ABC DEF and their areas be, respectively, 64cm 2 and 121 cm 2. If EF=15.4 cm, find BC. Solution: Given: ABC DEF and A( ABC)=64cm 2 & A( DEF)= 121cm 2 and EF = 15.4cm. To find: BC = ? We know that, the ratio of the areas of two similar triangles is equal to the square of the ratio of their sides. i.e. BC 2 = = = 125.44 BC = 11.2 cm 2) In an equilateral triangle ABC, AD BC. Prove that 3AB 2 = 4AD 2 . Solution: Given: ABC is an equilateral triangle and AD BC. To prove that: 3AB 2 = 4AD 2 Proof: Since, AD BC therefore, BD = DC. Now, in right triangle ABD, we have AB 2 = AD 2 + BD 2 --- [Pythagoras theorem] Since BD = DC therefore, BD = BC AB 2 = AD 2 + AB 2 = AD 2 + AB 2 = AD 2 + AB 2 = AD 2 + ------ [Since, AB = BC = AC – sides of an equilateral triangle] AB 2 - = AD 2 = AD 2 3AB 2 = 4AD 2 Triangles 3) In the given figure, PQR is a right triangle right angled at Q & QS PR. If PQ = 6cm and PS = 4cm, find the QS, RS & QR. Solution: Given: In PQR, PQR= 90 . PQ = 6cm, PS = 4cm. To find: QS , RS & QR? Since,QS PR, PS = SR And PS =4cm ----- (given ) SR = 4cm. PR = PS + SR = 4 + 4 =8 PQR is right angled triangle. by Pythagoras theorem, PR 2 = PQ 2 + QR 2 QR 2 = PR 2 - PQ 2 QR 2 = 8 2 - 6 2 QR 2 = 64 -36 = 28 QR = 2v QS 2 = PS SR -------- [ Since QS is perpendicular to PR] QS 2 = 4 4 = 16 QS = 4cm. R Q P S 6 4 Triangles 4) In the given figure, and If 1 , 4 BX BC ? find the ratio of areas of ?ABC and ?YCX. Solution: Since Similarly ACX CXY ? ? ? ? ?ABC ?YCX …….(AA – similarity criterion) 2 2 2 ) ( ) ( ? ? ? ? ? ? ? ? ? ? XC BC XC BC YCX ar ABC ar Since, 1 4 BX BC ? ….. (Given) ? ? 1 4 BC XC BX BC ? ? ? ? 4BC – 4XC = BC ? 3BC = 4XC ? 4 3 BC XC ? Putting values in eq.( i ), we get 9 : 16 9 16 3 4 ) ( ) ( 2 ? ? ? ? ? ? ? ? ? ? ? YCX ar ABC ar 5) PQR is an isosceles triangle right angled at B. Two equilateral triangles are constructed on side QR and PR as shown in the given figure. Prove that area of ?QRS = 1 2 area of ?PRT. Solution: In ?PQR, 90 Q ?? and consider PQ = QR = x ? PR 2 = PQ 2 + QR 2 --- (Pythagoras theorem) ? PR 2 = 2 x 2 ? PR = 2 x A B C X P Q R S T Triangles Area of equilateral triangle ?QRS = 2 3 4 x Area of equilateral triangle ?PRT = ? ? ? ? 2 2 33 2 44 PR x ? Therefore area of ?PRT = 2 3 2 x = 2 2 2 3 2 3 2 2 4 x x ?? = 2 area of ?QRS 6) The areas of two similar triangles are144 cm 2 and 81 cm 2 respectively. If the median of the first triangle is 14.4 cm find the corresponding median of the other. Solution: Let two triangles be LMN and XYZ, such that ?LMN ?XYZ 2 2 )] ( [ )] ( [ ) ( ) ( XYZ Median LMN Median XYZ ar LMN ar ? ? ? ? ? ? 2 ) ( 4 . 14 81 144 ? ? ? ? ? ? ? ? ? ? XYZ Median ? (Median of ?XYZ) 2 = 14.4 14.4 81 144 ?? Median = 10.8 cm. L N R M K P T S Triangles 7) The areas of two similar triangles are 144 cm 2 and 81 cm 2 respectively. If the altitude of bigger triangle is 4 cm, find the corresponding altitude of other triangle. Solution: Let two triangles be LMN and RST, 2 2 ) ( ) ( RP LK RST ar LMN ar ? ? ? ? 2 2 81 144 4 LK ? ? 2 81 16 144 LK ? ? ? LK = 3 cm. 8) If the diagonals of a quadrilateral PQRS divide each other proportionally, then prove that it is a trapezium. Solution: In quad. PQRS, diagonals PR and QS intersect at M such that MP MQ MR MS ? Draw MT SR Proof: In ?PSR, MT SR PT PM TS MR ? ( i ) [ B.P. Theorem ] But PM QM MR MS ? ( ii ) Comparing eq. ( i ) and ( ii ), we get PT QM TS MS ? ? MT PQ Also MT SR PQ SR Hence PQRS is a trapezium. P Q S R T MRead More

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