Question

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 the distance between them is 17 cm, find the radius of the circle.

Answer 1

Let O be the centre of the given circle and let it's radius be r cm. Draw OP ⊥ AB and OQ ⊥ CD.
Since AB || CD. Therefore, points P, O and Q are collinear. So, PQ = 17 cm.

Let Op = x cm. Then, OQ = (17 - x) cm
Join OA and OC. Then, OA = OC = r.
Since, the perpendicular from the centre to a chord of the circle bisects the chord.
∴ AP = PB = 5 cm and CQ = QD = 12 cm.

In right triangles OAP and OCQ, we have 
OA² = OP² + AP² and OC² = OQ² + CQ²
⇒ r² = x² + 5²                        ...(i)
and r² = (17 - x)² + 12²         ....(ii)
⇒ x² + 5² = (17 - x)² + 12²    ....(On equating the values of r²)

⇒ x² + 25 = 289 - 34x + x² + 144
⇒ 34x = 408 
⇒ x = 12 cm
Putting x = 12 cm in (i), we get
r² = 12² + 5² = 169 
⇒ r = 13 cm
Hence, the radius of the circle is 13 cm.