From a point P which is at a distance of 13 cm from the center 0 of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle is drawn. Then, the area of the quadrilateral PQOR is

Let us draw a circle of radius 5 cm having center O. P is a point at a distance of 13 cm from O. A pair of tangents PQ and PR are drawn.

Also, OQ = OR = radius = 5cm [1]

And OP = 13 cm

As OQ ⏊ PQ and OR ⏊PR [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]

△POQ and △POR are right-angled triangles.

In △PQO By Pythagoras Theorem [ i.e. (base)^{2}+ (perpendicular)^{2}= (hypotenuse)^{2} ]

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

(PQ)^{2} + (5)^{2} = (13)^{2}

(PQ)^{2} + 25 = 169

(PQ)^{2} = 144

PQ = 12 cm

And

PQ = PR = 12 cm [ tangents through an external point to a circle are equal] [2]

Area of quadrilateral PQRS, A = area of △POQ + area of △POR

[Using 1 and 2]

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