From a point P which is at a distance of 13 cm from the point 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 ______

#### Options

60 cm

^{2}65 cm

^{2}30 cm

^{2}32.5 cm

^{2}

#### Solution 1

From a point P which is at a distance of 13 cm from the point 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 **60 cm ^{2}.**

**Explanation:**

OP^{2} = OQ^{2} + PQ^{2}

169 = 25 + PQ^{2}

PQ^{2} = 144

PQ = 12

Area PQOR = ar(AOPQ) + ar(AOPR)

= `1/2 × 12 × 5 + 1/2 × 12 × 5` = 60 cm^{2}

#### Solution 2

From a point P which is at a distance of 13 cm from the point 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 **60 cm ^{2}**.

**Explanation:**

Firstly, draw a circle of radius 5 cm with centre O. P is a point at a distance of 13 cm from O. A pair of tangents PQ and PR are drawn.

Thus, quadrilateral PQOR is formed.

∵ OQ ⊥ QP .....[Since, QP is a tangent line]

In right angled ∆PQO,

OP^{2} = OQ^{2} + QP^{2}

⇒ 13^{2} = 5^{2} + QP^{2}

⇒ QP^{2} = 169 – 25 = 144

⇒ QP = 12 cm

Now, area of ∆OQP = `/2 xx QP xx QO`

= `1/2 xx 12 xx 5` = 30 cm^{2}

∴ Area of quadrilateral PQOR = 2 × ar ∆OQP

= 2 × 30

= 60 cm^{2}