Question

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

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

Answer 1

Answer: (D)

13 cm

Let $OP=xcm$

$\therefore OQ=(17-x)cm$
$\because AB=10cm⟹AP=5cm$ ........... [$\because OP$ is a straight line from centre]
In $\Delta AOP,$
$(OA{)}^2=(AP{)}^2+(OP{)}^2$
$r^2=(5{)}^2+x^2$
$r^2=x^2+25$ ...... $\left(i\right)$
$\because CD=24cm⟹CQ=12cm$
Again in $\Delta OCQ$,
$(OC{)}^2=(CQ{)}^2+(OQ{)}^2$
$r^2=(12{)}^2+(17-x{)}^2$
$r^2=144+289+x^2-34x$ ....... $\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$, we get
$x^2-34x+289+144=x^2+25$
$⟹34x=408$
$⟹x=12cm$
Substituting the value of $x$ in $\left(i\right),$ we get
$r^2=(12{)}^2+25=144+25=169c{m}^2$
$⟹r=13cm$

Hence, radius of circle is $13cm$