Question

From a point $P$ which is at a distance of $13cm$ from the center$O$ of a circle of radius $5cm$, the pair of tangents $PQ$ and $PR$ to the circle are drawn. Then the area of the quadrilateral $PQOR$ is

• A. $60c{m}^2$
• B. $65c{m}^2$
• C. $30c{m}^2$
• D. $32.5c{m}^2$

Answer 1

The correct option is A

$60c{m}^2$

Step 1: Solve for area of the triangles

Given,

1. $OP=13cm$.
2. $OQ=OR=5cm$.
3. $PQ$ and $PR$ are tangents to the circle.

Since, $PQ$ and $PR$ are tangents to the circle, the radii of the circle $OQ\perp PQ$ and $OR\perp PR$.

$\therefore △OQP$ and $△ORP$ are right angled triangles.

According to Pythagoras theorem,

$$\begin{gathered} {\left(OQ\right)}^2+{\left(PQ\right)}^2={\left(OP\right)}^2 \\ \Rightarrow {\left(5\right)}^2+{\left(PQ\right)}^2={\left(13\right)}^2 \\ \Rightarrow {\left(PQ\right)}^2=169-25 \\ \Rightarrow {\left(PQ\right)}^2=144 \\ \Rightarrow PQ=12cm \end{gathered}$$

We know that, length of tangents drawn from a point to the circle are always equal i.e. $PQ=PR=12cm$

$\therefore$Area of $△OQP=$Area of $△ORP=\frac{1}{2}\times 5\times 12=30c{m}^2$ $[\because$Area of triangle$=\frac{1}{2}\times$base$\times$height$]$

Step 2: Solve for area of quadrilateral.

Area of quadrilateral $OQPR$ $=$ Area of $△OQP$ $+$ Area of $△ORP$

$=30+30=60c{m}^2$

Hence, area of the quadrilateral is $60c{m}^2$.