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 its radius be r cm. Draw $OP\perp AB$ and $OQ\perp CD$. Since $OP\perp AB,OQ\perp CDandAB||CD$, 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 teh circle bisets the chord.
$\therefore AP=PB=5cmandCQ=QD=12cm$
In right triangle OAP and OCQ , we have
$O{A}^2=O{P}^2+A{P}^{2}andO{C}^2=O{Q}^2+C{Q}^2$
$\Rightarrow r^2=x^2+5^2....\left(i\right)$
and, $r^2=(17-x^2)+{12}^2....\left(ii\right)$
$\Rightarrow x^2+5^2=(17-x{)}^2+{12}^2$
[On equating the valyes of $r^2$]
$\Rightarrow x^2+25=289-34x+x^2+144$
$\Rightarrow 34x=408\Rightarrow x=12cm$
Putting x = 12 cm in equation (i), we get
$r^2={12}^2+5^2=169$
$\Rightarrow r=13cm$
Hence, the radius of the circle is 13 cm.