Question

Given a chord AB of length 5 cm, of a circle with centre O. OL is perpendicular to chord AB and OL= 4 cm. OM is perpendicular to chord CD such that OM= 4 cm. Then CM is equal to??
​

Answer 1

Step-by-step explanation:

A perpendicular from centre of the circle to any chord divides the chord in half.

Draw a perpendicular OL from centre to the circle. Join the centre of the circle to end A of the chord. In the right angle triangle OAL, OL is 4 cms long and hypotenuse OA is 5 cms long. Using the formula for right angle triangle a^2 + b^2 = h^2. In the question a = OL = 4 cms; h = OA = 5 cms, we have to find b = LA

4^2 + b^2 = 5^2

16 + b^2= 25

b^2 = 9

b = LA = 3

Since LA is half the length of the chord, AB = 2 x 3 = 6 cms.