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

1. 60 cm²

2. 65 cm²

3. 30 cm²

4. 32.5 cm²

Given, the point P is 13 cm from O, the centre of the circle as shown in the figure below:

Radius of the circle (OQ) = 5 cm

PQ and PR are tangents from P to the circle.

PQ ⊥ OQ (∵ radius of a circle and tangent through that point are perpendicular to each other.)

∴ OQP = 90°.

So, in △OQP,

$OP^2 = OQ^2 + PQ^2 \\[1em] 13^2 = 5^2 + PQ^2 \\[1em] PQ^2 = 13^2 - 5^2 \\[1em] PQ^2 = 169 - 25 \\[1em] PQ^2 = 144 \\[1em] PQ = 12 \text{ cm}$

Area of △OPQ = $\dfrac{1}{2} \times PQ \times OQ$ = $\dfrac{1}{2} \times 12 \times 5 = 30$ cm².

Similarly,

PR ⊥ OR (∵ radius of a circle and tangent through that point are perpendicular to each other.)

∴ ∠ORP = 90°.

So, in △ORP,

$OP^2 = OR^2 + PR^2 \\[1em] 13^2 = 5^2 + PR^2 \\[1em] PR^2 = 13^2 - 5^2 \\[1em] PR^2 = 169 - 25 \\[1em] PR^2 = 144 \\[1em] PR = 12 \text{ cm}$

Area of △POR = $\dfrac{1}{2} \times PR \times OR$ = $\dfrac{1}{2} \times 12 \times 5 = 30$ cm².

Area of quadrilateral PQOR = Area of △POR + Area of △OPQ = 30 + 30 = 60 cm².

Hence, Option 1 is the correct option.