AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and distance between them is 17 cm, then the radius of the circle is :

Given : AB = 10 , CD = 24 and EF = 17 cm

To find : OB = OD = $$r$$ = ?

Solution : A line perpendicular to the chord from the centre of the circle bisects the chord.

=> $$AF = FB = \frac{AB}{2} = \frac{10}{2} = 5$$

Similarly, $$CE = ED = 12$$

Let OF = $$x$$ => OE = $$(17-x)$$

In right $$\triangle$$OFB

=> $$(OB)^2 = (OF)^2 + (FB)^2$$

=> $$r^2 = x^2 + 25$$

Now, in right $$\triangle$$OED

=> $$(OD)^2 = (OE)^2 + (ED)^2$$

=> $$r^2 = (17-x)^2 + 144$$

=> $$x^2 + 25 = x^2 - 34x + 289 + 144$$

=> $$34x = 408$$

=> $$x = \frac{408}{34} = 12$$

=> $$r^2 = 12^2 + 25$$

=> $$r = \sqrt{169} = 13$$ cm