In the picture, chords AB and CD of the circle are extended to meet at P . Prove that P A × P B = P C × P D .

Join AC and BD . Let ∠ B A C = x ° and ∠ A C D = y ° . As A B C D is a cyclic quadrilateral. We know that, opposite angles of a cyclic quadrilateral is supplementary. ⇒ ∠ B A C + ∠ B D C = 180 ° ⇒ ∠ B D C = 180 ° - ∠ B A C ⇒ ∠ B D C = 180 ° - x Thus, ⇒ ∠ B D C + ∠ B D P = 180 ° (Linear pair) ⇒ ∠ B D P = 180 ° - ∠ B D C = x ° Similarly, ∠ A B D = 180 ° - ∠ A C D = 180 ° - y ∠ D B P = 180 ° - ∠ A B D = y ° In a ∆ A P C , ∆ P B D ∠ P A C = ∠ B D P = x ° and ∠ P C A = ∠ P B D = y ° The ∠ A P C is the common angle. So, ∆ A P C , ∆ D P B are similar by A . A . A similarity rule. A P P D = P C P B (C.P.C.T) ∴ P A × P B = P C × P D Hence, proved.

In the picture, chords AB and CD of the circle are extended to meet at P . Prove that if P B = P D , then A B C D is an isosceles trapezium.

Join AC and BD . Given, P B = P D . So, in ∆ P B D ∠ P B D = ∠ P D B = x ° ⇒ ∠ C D B + ∠ B D P = 180 ° (Linear pair) ⇒ ∠ C D B = 180 ° - x Similarly, ∠ A B D = 180 ° - x We know that, opposite angles of a cyclic quadrilateral are supplementary. ⇒ ∠ C A B + ∠ C D B = 180 ° ⇒ ∠ C A B = x And, ⇒ ∠ A C B + ∠ A B D = 180 ° ⇒ ∠ A C B = x In ∆ A P C , ∠ A C B = ∠ C A B = x We know that, sides opposite to equal angles are equal. So, A P = C P . Then, A B = C D a n d A C ∥ B D Therefore, A B C D will be an isosceles trapezium.

In the picture, a line through the centre of a circle meets a chord of the circle: What are the lengths of the two pieces of the chord?

From the figure: OA = OD = OE = 7 cm (Radius) PE = OE - OP = 7 - 3 = 4 cm . The intersection may be with in the circle. So, internal bisector theorem of chords C P × P B = P E × P D ⇒ C P × P B = 4 × ( 7 + 3 ) = 4 × 10 = 40 CP × PB = 40 … … … … … (i) CP + PB = 13 ....... (ii) From (i) and (ii), ⇒ ( 13 – P B ) P B = 40 By solving the above equation: ⇒ 13 P B - P B 2 - 40 = 0 ⇒ P B 2 - 13 P B + 40 = 0 ⇒ P B 2 - 8 P B - 5 P B + 40 = 0 ⇒ P B - 8 P B - 5 = 0 ∴ P B = 5 , 8 In the figure, P B > P C . P B = 8 c m , P C = 5 c m . Length of two pieces of chord are 8 cm, 5 cm .

E M B I B E

Standard 10 Mathematics Part - 1>Chapter 2 - Circles>Exercise 4

Kerala Board Solutions for Chapter: Circles, Exercise 4: Exercise 4

Author:Kerala Board

Kerala Board Mathematics Solutions for Exercise - Kerala Board Solutions for Chapter: Circles, Exercise 4: Exercise 4

Attempt the practice questions on Chapter 2: Circles, Exercise 4: Exercise 4 with hints and solutions to strengthen your understanding. Standard 10 Mathematics Part - 1 solutions are prepared by Experienced Embibe Experts.