Page 1 Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 1 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 cm Answer 1: Let O be the centre of the circle. Given that, OQ = 25cm and PQ = 24 cm As the radius is perpendicular to the tangent at the point of contact, Therefore, OP ? PQ Applying Pythagoras theorem in ?OPQ, we obtain OP 2 + PQ 2 = OQ 2 OP 2 + 24 2 = 25 2 OP 2 = 625  576 OP 2 = 49 OP = 7 Therefore, the radius of the circle is 7 cm. Hence, alternative (A) is correct Page 2 Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 1 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 cm Answer 1: Let O be the centre of the circle. Given that, OQ = 25cm and PQ = 24 cm As the radius is perpendicular to the tangent at the point of contact, Therefore, OP ? PQ Applying Pythagoras theorem in ?OPQ, we obtain OP 2 + PQ 2 = OQ 2 OP 2 + 24 2 = 25 2 OP 2 = 625  576 OP 2 = 49 OP = 7 Therefore, the radius of the circle is 7 cm. Hence, alternative (A) is correct Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 2 Question 2: In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ?POQ = 110 , then ?PTQ is equal to (A) 60 (B) 70 (C) 80 (D) 90 Answer 2: It is given that TP and TQ are tangents. Therefore, radius drawn to these tangents will be perpendicular to the tangents. Thus, OP ? TP and OQ ? TQ ?OPT = 90º ?OQT = 90º In quadrilateral POQT, Sum of all interior angles = 360 ?OPT + ?POQ + ?OQT + ?PTQ = 360 ? 90 + 110º + 90 + PTQ = 360 ? PTQ = 70 Hence, alternative (B) is correct Page 3 Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 1 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 cm Answer 1: Let O be the centre of the circle. Given that, OQ = 25cm and PQ = 24 cm As the radius is perpendicular to the tangent at the point of contact, Therefore, OP ? PQ Applying Pythagoras theorem in ?OPQ, we obtain OP 2 + PQ 2 = OQ 2 OP 2 + 24 2 = 25 2 OP 2 = 625  576 OP 2 = 49 OP = 7 Therefore, the radius of the circle is 7 cm. Hence, alternative (A) is correct Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 2 Question 2: In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ?POQ = 110 , then ?PTQ is equal to (A) 60 (B) 70 (C) 80 (D) 90 Answer 2: It is given that TP and TQ are tangents. Therefore, radius drawn to these tangents will be perpendicular to the tangents. Thus, OP ? TP and OQ ? TQ ?OPT = 90º ?OQT = 90º In quadrilateral POQT, Sum of all interior angles = 360 ?OPT + ?POQ + ?OQT + ?PTQ = 360 ? 90 + 110º + 90 + PTQ = 360 ? PTQ = 70 Hence, alternative (B) is correct Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 3 Question 3: If tangents PA and PB from a point P to a circle with centre O are inclined to each other an angle of 80 , then ?POA is equal to (A) 50 (B) 60 (C) 70 (D) 80 Answer 3: It is given that PA and PB are tangents. Therefore, the radius drawn to these tangents will be perpendicular to the tangents. Thus, OA ? PA and OB ? PB ?OBP = 90º and ?OAP = 90º In AOBP, Sum of all interior angles = 360 ?OAP + ?APB + ?PBO + ?BOA = 360 90º + 80º +90º + ?BOA = 360 ?BOA = 100 In ?OPB and ?OPA, AP = BP (Tangents from a point) OA = OB (Radii of the circle) OP = OP (Common side) Therefore, ?OPB ? ?OPA (SSS congruence criterion) And thus, ?POB = ?POA Hence, alternative (A) is correct. Page 4 Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 1 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 cm Answer 1: Let O be the centre of the circle. Given that, OQ = 25cm and PQ = 24 cm As the radius is perpendicular to the tangent at the point of contact, Therefore, OP ? PQ Applying Pythagoras theorem in ?OPQ, we obtain OP 2 + PQ 2 = OQ 2 OP 2 + 24 2 = 25 2 OP 2 = 625  576 OP 2 = 49 OP = 7 Therefore, the radius of the circle is 7 cm. Hence, alternative (A) is correct Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 2 Question 2: In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ?POQ = 110 , then ?PTQ is equal to (A) 60 (B) 70 (C) 80 (D) 90 Answer 2: It is given that TP and TQ are tangents. Therefore, radius drawn to these tangents will be perpendicular to the tangents. Thus, OP ? TP and OQ ? TQ ?OPT = 90º ?OQT = 90º In quadrilateral POQT, Sum of all interior angles = 360 ?OPT + ?POQ + ?OQT + ?PTQ = 360 ? 90 + 110º + 90 + PTQ = 360 ? PTQ = 70 Hence, alternative (B) is correct Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 3 Question 3: If tangents PA and PB from a point P to a circle with centre O are inclined to each other an angle of 80 , then ?POA is equal to (A) 50 (B) 60 (C) 70 (D) 80 Answer 3: It is given that PA and PB are tangents. Therefore, the radius drawn to these tangents will be perpendicular to the tangents. Thus, OA ? PA and OB ? PB ?OBP = 90º and ?OAP = 90º In AOBP, Sum of all interior angles = 360 ?OAP + ?APB + ?PBO + ?BOA = 360 90º + 80º +90º + ?BOA = 360 ?BOA = 100 In ?OPB and ?OPA, AP = BP (Tangents from a point) OA = OB (Radii of the circle) OP = OP (Common side) Therefore, ?OPB ? ?OPA (SSS congruence criterion) And thus, ?POB = ?POA Hence, alternative (A) is correct. Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 4 Question 4: Prove that the tangents drawn at the ends of a diameter of a circle are parallel. Answer 4: Page 5 Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 1 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 cm Answer 1: Let O be the centre of the circle. Given that, OQ = 25cm and PQ = 24 cm As the radius is perpendicular to the tangent at the point of contact, Therefore, OP ? PQ Applying Pythagoras theorem in ?OPQ, we obtain OP 2 + PQ 2 = OQ 2 OP 2 + 24 2 = 25 2 OP 2 = 625  576 OP 2 = 49 OP = 7 Therefore, the radius of the circle is 7 cm. Hence, alternative (A) is correct Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 2 Question 2: In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ?POQ = 110 , then ?PTQ is equal to (A) 60 (B) 70 (C) 80 (D) 90 Answer 2: It is given that TP and TQ are tangents. Therefore, radius drawn to these tangents will be perpendicular to the tangents. Thus, OP ? TP and OQ ? TQ ?OPT = 90º ?OQT = 90º In quadrilateral POQT, Sum of all interior angles = 360 ?OPT + ?POQ + ?OQT + ?PTQ = 360 ? 90 + 110º + 90 + PTQ = 360 ? PTQ = 70 Hence, alternative (B) is correct Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 3 Question 3: If tangents PA and PB from a point P to a circle with centre O are inclined to each other an angle of 80 , then ?POA is equal to (A) 50 (B) 60 (C) 70 (D) 80 Answer 3: It is given that PA and PB are tangents. Therefore, the radius drawn to these tangents will be perpendicular to the tangents. Thus, OA ? PA and OB ? PB ?OBP = 90º and ?OAP = 90º In AOBP, Sum of all interior angles = 360 ?OAP + ?APB + ?PBO + ?BOA = 360 90º + 80º +90º + ?BOA = 360 ?BOA = 100 In ?OPB and ?OPA, AP = BP (Tangents from a point) OA = OB (Radii of the circle) OP = OP (Common side) Therefore, ?OPB ? ?OPA (SSS congruence criterion) And thus, ?POB = ?POA Hence, alternative (A) is correct. Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 4 Question 4: Prove that the tangents drawn at the ends of a diameter of a circle are parallel. Answer 4: Mathematics (www.tiwariacademy.com) (Chapter – 10) (Circles) (Class – X) www.tiwariacademy.com 5 Question 5: Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre. Answer 5: Let us consider a circle with centre O. Let AB be a tangent which touches the circle at P. We have to prove that the line perpendicular to AB at P passes through centre O. We shall prove this by contradiction method. Let us assume that the perpendicular to AB at P does not pass through centre O. Let it pass through another point O’. Join OP and O’P.Read More
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