Chords AB and CD of a circle are parallel to each other and lie on opposite sides of the centre of the circle. If AB = 36cm, CD= 48cm and the distance between the chords is 42cm: find the radius of the circle.
Answers
According to the given statement, e
figure will be like this :
Since; OP _|_ AB , OQ _|_ CD and AB//CD ;
it can be shown that POQ is a straight line.
As , AB = 36cm , CD = 48 cm and PQ = 42 cm
→ AP = PB = 1/2 of AB = 18 cm , CQ = DQ = 1/2 of CD = 24 cm
and if OQ = x cm , OP = ( 42 - x ) cm
Join OA and OC
OA = OC = r ( radius of the circle )
In right-angled triangle OAP,
OA² = OP² + AP² → r² = ( 42 - x )² + 18² ➖ 1
In right-angled triangle OCQ,
OC² = OQ² + CQ → r² = x² + 24² ➖ 2
After adding equations 1 and 2 we get :
( 42 - x )² + 18² = x² + 24²
1764 - 84x + x² + 324 = x² + 576
84x = 1512 and x = 18
r²= x² + 24²
r²= 18² + 24²
r = √900 = 30
Radius of circle = 30 cm