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 opposite sides of the centre and the distance between them is 17 cm, find the radius of the circle.

• A. 10 cm
• B. 11 cm
• C. 12 cm
• D. 13 cm

Answer 1

The correct option is D

13 cm

Let O be the centre of the given circle and let its radius be $r$ cm. Draw OP $\perp$ AB and OQ $\perp$ CD. Since OP $\perp$ AB, OQ $\perp$ CD and AB $\parallel$ 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.

∴ $AP=PB=5cm$ and $CQ=QD=12cm$.

In right triangles $\Delta$OAP and $\Delta$OCQ, we have

${OA}^2$ = ${OP}^2$ + ${AP}^2$ and
${OC}^2$ = ${OQ}^2$ + ${CQ}^2$

$r^2$ = $x^2$ + $5^2$ ….(i)

and, $r^2$ = ${(17-x)}^2$ + ${12}^2$ ....(ii)

$x^2$ + $5^2$ = ${(17-x)}^2$ + ${12}^2$

[On equating the values of $r^2$]

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

$34x=408or,x=12cm$

Putting $x=12cm$ in equation (i), we get

$r^2=144+25=169$ or, $r=13cm$.

Hence, the radius of the circle is $13cm$.