Page 1 Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 1 Exercise 6.2 Question 1: In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). (i) (ii) Answer 1: (i) Let EC = x cm It is given that DE || BC. By using basic proportionality theorem, we obtain Page 2 Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 1 Exercise 6.2 Question 1: In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). (i) (ii) Answer 1: (i) Let EC = x cm It is given that DE || BC. By using basic proportionality theorem, we obtain Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 2 (ii) Let AD = x cm It is given that DE || BC. By using basic proportionality theorem, we obtain Page 3 Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 1 Exercise 6.2 Question 1: In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). (i) (ii) Answer 1: (i) Let EC = x cm It is given that DE || BC. By using basic proportionality theorem, we obtain Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 2 (ii) Let AD = x cm It is given that DE || BC. By using basic proportionality theorem, we obtain Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 3 Question 2: E and F are points on the sides PQ and PR respectively of a ?PQR. For each of the following cases, state whether EF || QR. (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii)PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm Answer 2: (i) Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm Page 4 Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 1 Exercise 6.2 Question 1: In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). (i) (ii) Answer 1: (i) Let EC = x cm It is given that DE || BC. By using basic proportionality theorem, we obtain Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 2 (ii) Let AD = x cm It is given that DE || BC. By using basic proportionality theorem, we obtain Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 3 Question 2: E and F are points on the sides PQ and PR respectively of a ?PQR. For each of the following cases, state whether EF || QR. (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii)PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm Answer 2: (i) Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 4 (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm Page 5 Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 1 Exercise 6.2 Question 1: In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). (i) (ii) Answer 1: (i) Let EC = x cm It is given that DE || BC. By using basic proportionality theorem, we obtain Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 2 (ii) Let AD = x cm It is given that DE || BC. By using basic proportionality theorem, we obtain Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 3 Question 2: E and F are points on the sides PQ and PR respectively of a ?PQR. For each of the following cases, state whether EF || QR. (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii)PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm Answer 2: (i) Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 4 (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm Mathematics (www.tiwariacademy.com) (Chapter – 6) (Triangles) (Class – X) www.tiwariacademy.com 5 Question 3: In the following figure, if LM || CB and LN || CD, prove that Answer 3: In the given figure, LM || CB By using basic proportionality theorem, we obtainRead More

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