Question

In figure, $PA$ and $PB$ are two tangent touching the circle with centre $O$ at $A$ and $B$ respectively. If the distance of point $P$ from the centre $O$ is $13cm$ and the diameter of the circle is $10cm$ , find the perimeter of quadrilateal $OAPB$.

Answer 1

$PA$ and $PB$ are tangent $\Rightarrow \angle OAP={90}^0$ and $\angle OBP={90}^0$

$OA$ and $OB$ are the radius

$OA=OB=5cm$

$OP=13cm$

In $\Delta OAP$

$OA$ is perpendicular to $AP$, as in a circle the radius and the tangent are perpendicular at a point on the circle.

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

$\Rightarrow {AP}^2={13}^2-5^2$

$\Rightarrow AP=12cm$

Similarly in $\Delta OBP$

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

$\Rightarrow BP=12cm$

$\therefore$ Perimeter of $OAPB=OA+AP+PB+OB=5+12+12+5$

$=10+24=34cm$