Question : Point O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, point P is taken. From this point, two tangents, PQ and PR, are drawn to the circle. Then, the area of quadrilateral PQOR is:
Option 1: $60\ \text{cm}^2$
Option 2: $32.5\ \text{cm}^2$
Option 3: $65\ \text{cm}^2$
Option 4: $30\ \text{cm}^2$
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Correct Answer: $60\ \text{cm}^2$
Solution : Given: Point O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, point P is taken. We know that the area of triangle is $\frac{1}{2} \times (\text{base}) \times (\text{height})$. $\angle OQP=\angle ORP=90°$(Angle at a point of contact of the tangent) Applying Pythagoras' theorem in $\angle OQP$, $(QP)^2+(OQ)^2=(OP)^2$ ⇒ $QP=\sqrt{(OP)^2–(OQ)^2}$ ⇒ $QP=\sqrt{(13)^2–(5)^2}$ ⇒ $QP=\sqrt{169–25}$ ⇒ $QP=\sqrt{144}$ ⇒ $QP=12\ \text{cm}$ The area of $\triangle OQP$ = $\frac{1}{2} \times 12 \times 5=30\ \text{cm}^2$ The area of quadrilateral $PQOR = 2 × 30 = 60\ \text{cm}^2$ Hence, the correct answer is $60\ \text{cm}^2$.
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