Question

a b and CD are two parallel chords of a circle which are on opposite sides of the centre such that a b is equal to 24 cm and CD is equal to 10 cm and the distance between a b and CD is 17 cm find the radius of the circle

Answer 1

Hi mate...

Here is your answer...

Step-by-step explanation :

Let's the centre of the circle be O.

Let's the line joining CD and AB me et CD at P and AB at Q.

So,
AQ=BQ=AB/2 (perpendicular from the centre bisects the chord)

AQ=BQ=12

Similarly,
CP=DP=5

Taking triangle AOQ
$ao {}^{2} = aq {}^{2} + oq {}^{2}$
$ao {}^{2} = x {}^{2} + 144$
Taking triangle COP
$oc {}^{2} = cp {}^{2} + op {}^{2}$
$oc {}^{2} = (17 - x) {}^{2} + 25$
$oc {}^{2} = x {}^{2} - 34x + 264$

But OA and Oc Are equal as they are radius
$x {}^{2} - 34x+ 264 = x {}^{2} + 144$



Hope this helps you...