Question

In the following figure, $AD$ is a straight line. $OP$ $\perp$ $AD$ and $O$ is the centre of both the circles. If $OA=34cm,OB=20cm$ and $OP=16cm$ ; find the length of $AB$ .

Answer 1

For the inner circle, $BC$ is a chord and $OP$ is perpendicular to $BC$

We know that the perpendicular to a chord, from the centre of a circle, bisects the chord.

$\therefore BP=PC$

By Pythagoras theorem,

${OA}^2={OP}^2+{BP}^2$

$\Rightarrow {BP}^2={20}^2-{16}^2=400-256=144$

$\therefore BP=12$cm

For the outer circle,$AD$ is the chord and $OP$ is perpendicular to $AD$

We know that the perpendicular to a chord, from the centre of a circle, bisects the chord.

$\therefore AP=PD$

By Pythagoras theorem,

${OA}^2={OP}^2+{AP}^2$

$\Rightarrow {AP}^2={34}^2-{16}^2=1,156‬-256=900$

$\Rightarrow AP=30$cm

$AB=AP-BP=30-12=18$cm