Class 10 Exam > Class 10 Notes > Mathematics (Maths) Class 10 > RD Sharma Solutions: Triangles (Exercise 6A)
Triangles (Exercise 6A) RD Sharma Solutions | Mathematics (Maths) Class 10 PDF Download
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Page 1 1. (i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC. (ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD. (iii) If = and AC = 6.6cm, find AE. (iv) If = and EC = 3.5cm, find AE. Sol: (i) BC. AD = 3.6 cm , AB = 10 cm, AE = 4.5cm DB = 10 3.6 = 6.4cm Or, Or, EC = Or, EC= 8 cm Thus, AC = AE + EC = 4.5 + 8 = 12.5 cm (ii) t is given that DE || BC. Adding 1 to both sides, we get : +1 (iii) BC. Adding 1 to both the sides, we get : EC = = 4.2 cm Therefore, Page 2 1. (i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC. (ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD. (iii) If = and AC = 6.6cm, find AE. (iv) If = and EC = 3.5cm, find AE. Sol: (i) BC. AD = 3.6 cm , AB = 10 cm, AE = 4.5cm DB = 10 3.6 = 6.4cm Or, Or, EC = Or, EC= 8 cm Thus, AC = AE + EC = 4.5 + 8 = 12.5 cm (ii) t is given that DE || BC. Adding 1 to both sides, we get : +1 (iii) BC. Adding 1 to both the sides, we get : EC = = 4.2 cm Therefore, AE = AC EC = 6.6 4.2 = 2.4 cm (iv) 8AE + 28 = 15AE 7AE = 28 AE = 4cm 2. D and E are the value of x, when (i) AD = x cm, DB = (x 2) cm, AE = (x + 2) cm and EC = (x 1) cm. (ii) AD = 4cm, DB = (x 4) cm, AE = 8cm and EC = (3x 19) cm. (iii) AD = (7x 4) cm, AE = (5x 2) cm, DB = (3x + 4) cm and EC = 3x cm. Sol: (i) || BC. : (ii) : = 4 (3x-19) = 8 (x-4) 12x 76 = 8x 32 4x = 44 x = 11 cm (iii) || BC. 3x (7x-4) = (5x-2) (3x+4) Page 3 1. (i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC. (ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD. (iii) If = and AC = 6.6cm, find AE. (iv) If = and EC = 3.5cm, find AE. Sol: (i) BC. AD = 3.6 cm , AB = 10 cm, AE = 4.5cm DB = 10 3.6 = 6.4cm Or, Or, EC = Or, EC= 8 cm Thus, AC = AE + EC = 4.5 + 8 = 12.5 cm (ii) t is given that DE || BC. Adding 1 to both sides, we get : +1 (iii) BC. Adding 1 to both the sides, we get : EC = = 4.2 cm Therefore, AE = AC EC = 6.6 4.2 = 2.4 cm (iv) 8AE + 28 = 15AE 7AE = 28 AE = 4cm 2. D and E are the value of x, when (i) AD = x cm, DB = (x 2) cm, AE = (x + 2) cm and EC = (x 1) cm. (ii) AD = 4cm, DB = (x 4) cm, AE = 8cm and EC = (3x 19) cm. (iii) AD = (7x 4) cm, AE = (5x 2) cm, DB = (3x + 4) cm and EC = 3x cm. Sol: (i) || BC. : (ii) : = 4 (3x-19) = 8 (x-4) 12x 76 = 8x 32 4x = 44 x = 11 cm (iii) || BC. 3x (7x-4) = (5x-2) (3x+4) 21 12x = 15 + 14x-8 6 26x + 8 = 0 (x-4) (6x-2) =0 x = 4, (as if x = x = 4 cm 3. D and E are points on the sides AB and AC (i) AD = 5.7cm, DB = 9.5cm, AE = 4.8cm and EC = 8cm. (ii) AB = 11.7cm, AC = 11.2cm, BD = 6.5cm and AE = 4.2cm. (iii) AB = 10.8cm, AD = 6.3cm, AC = 9.6cm and EC = 4cm. (iv) AD = 7.2cm, AE = 6.4cm, AB = 12cm and AC = 10cm. Sol: (i) We have: Hence, We conclude that DE || BC. (ii) We have: AB = 11.7 cm, DB = 6.5 cm Therefore, AD = 11.7 -6.5 = 5.2 cm Similarly, AC = 11.2 cm, AE = 4.2 cm Therefore, EC = 11.2 4.2 = 7 cm Now, Thus, We conclude that DE is not parallel to BC. (iii) We have: AB = 10.8 cm, AD = 6.3 cm Therefore, Page 4 1. (i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC. (ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD. (iii) If = and AC = 6.6cm, find AE. (iv) If = and EC = 3.5cm, find AE. Sol: (i) BC. AD = 3.6 cm , AB = 10 cm, AE = 4.5cm DB = 10 3.6 = 6.4cm Or, Or, EC = Or, EC= 8 cm Thus, AC = AE + EC = 4.5 + 8 = 12.5 cm (ii) t is given that DE || BC. Adding 1 to both sides, we get : +1 (iii) BC. Adding 1 to both the sides, we get : EC = = 4.2 cm Therefore, AE = AC EC = 6.6 4.2 = 2.4 cm (iv) 8AE + 28 = 15AE 7AE = 28 AE = 4cm 2. D and E are the value of x, when (i) AD = x cm, DB = (x 2) cm, AE = (x + 2) cm and EC = (x 1) cm. (ii) AD = 4cm, DB = (x 4) cm, AE = 8cm and EC = (3x 19) cm. (iii) AD = (7x 4) cm, AE = (5x 2) cm, DB = (3x + 4) cm and EC = 3x cm. Sol: (i) || BC. : (ii) : = 4 (3x-19) = 8 (x-4) 12x 76 = 8x 32 4x = 44 x = 11 cm (iii) || BC. 3x (7x-4) = (5x-2) (3x+4) 21 12x = 15 + 14x-8 6 26x + 8 = 0 (x-4) (6x-2) =0 x = 4, (as if x = x = 4 cm 3. D and E are points on the sides AB and AC (i) AD = 5.7cm, DB = 9.5cm, AE = 4.8cm and EC = 8cm. (ii) AB = 11.7cm, AC = 11.2cm, BD = 6.5cm and AE = 4.2cm. (iii) AB = 10.8cm, AD = 6.3cm, AC = 9.6cm and EC = 4cm. (iv) AD = 7.2cm, AE = 6.4cm, AB = 12cm and AC = 10cm. Sol: (i) We have: Hence, We conclude that DE || BC. (ii) We have: AB = 11.7 cm, DB = 6.5 cm Therefore, AD = 11.7 -6.5 = 5.2 cm Similarly, AC = 11.2 cm, AE = 4.2 cm Therefore, EC = 11.2 4.2 = 7 cm Now, Thus, We conclude that DE is not parallel to BC. (iii) We have: AB = 10.8 cm, AD = 6.3 cm Therefore, DB = 10.8 6.3 = 4.5 cm Similarly, AC = 9.6 cm, EC = 4cm Therefore, AE = 9.6 4 = 5.6 cm Now, (iv) We have : AD = 7.2 cm, AB = 12 cm Therefore, DB = 12 7.2 = 4.8 cm Similarly, AE = 6.4 cm, AC = 10 cm Therefore, EC = 10 6.4 = 3.6 cm Now, This, heorem, We conclude that DE is not parallel to BC. 4. In a ABC, AD is the bisector of A. (i) If AB = 6.4cm, AC = 8cm and BD = 5.6cm, find DC. (ii) If AB = 10cm, AC = 14cm and BC = 6cm, find BD and DC. (iii) If AB = 5.6cm, BD = 3.2cm and BC = 6cm, find AC. (iv) If AB = 5.6cm, AC = 4cm and DC = 3cm, find BC. Sol: (i) It is give that AD bisects A. Applying angle Page 5 1. (i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC. (ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD. (iii) If = and AC = 6.6cm, find AE. (iv) If = and EC = 3.5cm, find AE. Sol: (i) BC. AD = 3.6 cm , AB = 10 cm, AE = 4.5cm DB = 10 3.6 = 6.4cm Or, Or, EC = Or, EC= 8 cm Thus, AC = AE + EC = 4.5 + 8 = 12.5 cm (ii) t is given that DE || BC. Adding 1 to both sides, we get : +1 (iii) BC. Adding 1 to both the sides, we get : EC = = 4.2 cm Therefore, AE = AC EC = 6.6 4.2 = 2.4 cm (iv) 8AE + 28 = 15AE 7AE = 28 AE = 4cm 2. D and E are the value of x, when (i) AD = x cm, DB = (x 2) cm, AE = (x + 2) cm and EC = (x 1) cm. (ii) AD = 4cm, DB = (x 4) cm, AE = 8cm and EC = (3x 19) cm. (iii) AD = (7x 4) cm, AE = (5x 2) cm, DB = (3x + 4) cm and EC = 3x cm. Sol: (i) || BC. : (ii) : = 4 (3x-19) = 8 (x-4) 12x 76 = 8x 32 4x = 44 x = 11 cm (iii) || BC. 3x (7x-4) = (5x-2) (3x+4) 21 12x = 15 + 14x-8 6 26x + 8 = 0 (x-4) (6x-2) =0 x = 4, (as if x = x = 4 cm 3. D and E are points on the sides AB and AC (i) AD = 5.7cm, DB = 9.5cm, AE = 4.8cm and EC = 8cm. (ii) AB = 11.7cm, AC = 11.2cm, BD = 6.5cm and AE = 4.2cm. (iii) AB = 10.8cm, AD = 6.3cm, AC = 9.6cm and EC = 4cm. (iv) AD = 7.2cm, AE = 6.4cm, AB = 12cm and AC = 10cm. Sol: (i) We have: Hence, We conclude that DE || BC. (ii) We have: AB = 11.7 cm, DB = 6.5 cm Therefore, AD = 11.7 -6.5 = 5.2 cm Similarly, AC = 11.2 cm, AE = 4.2 cm Therefore, EC = 11.2 4.2 = 7 cm Now, Thus, We conclude that DE is not parallel to BC. (iii) We have: AB = 10.8 cm, AD = 6.3 cm Therefore, DB = 10.8 6.3 = 4.5 cm Similarly, AC = 9.6 cm, EC = 4cm Therefore, AE = 9.6 4 = 5.6 cm Now, (iv) We have : AD = 7.2 cm, AB = 12 cm Therefore, DB = 12 7.2 = 4.8 cm Similarly, AE = 6.4 cm, AC = 10 cm Therefore, EC = 10 6.4 = 3.6 cm Now, This, heorem, We conclude that DE is not parallel to BC. 4. In a ABC, AD is the bisector of A. (i) If AB = 6.4cm, AC = 8cm and BD = 5.6cm, find DC. (ii) If AB = 10cm, AC = 14cm and BC = 6cm, find BD and DC. (iii) If AB = 5.6cm, BD = 3.2cm and BC = 6cm, find AC. (iv) If AB = 5.6cm, AC = 4cm and DC = 3cm, find BC. Sol: (i) It is give that AD bisects A. Applying angle (ii) It is given that AD bisects . Applying angle bisector Let BD be x cm. Therefore, DC = (6- x) cm (iii) It is given that AD bisector Applying angle get: BD = 3.2 cm, BC = 6 cm Therefore, DC = 6- 3.2 = 2.8 cm (iv) It is given that AD bisects Applying angle 5. M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that (i) D M D C M N B N (ii) D N A N D M D C Sol: (i) Given: ABCD is a parallelogramRead More
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