O is the center of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to P the circle. Then, the area of quad. PQOR is

In Given Figure,

PQ = PR…[1]

[Tangents drawn from an external point are equal]

In △QOP and △ROP

PQ = PR [By 1]

OP = OP [Common]

OQ = OR [radii of same circle]

△QOP ≅△ROP

[By Side - Side - Side Criterion]

area(Δ QOP) = area(Δ ROP)

[Congruent triangles have equal areas]

area(PQOR) = area(Δ QOP) + area(Δ ROP)

area(PQOR) = area(Δ QOP) + area(Δ QOP) = 2[area(Δ QOP)]

Now,

OQ ⏊PQ

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, QOP is a right - angled triangle at Q with OQ as base and PQ as height.

In △QOP,

By Pythagoras Theorem in △OPB

[i.e. (hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2}]

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

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

25 + (PQ)^{2} = 169

(PQ)^{2} = 144

PQ = 12 cm

Area(ΔQOP) = 1/2 × Base × Height

= 1/2 × OQ × PQ

= 1/2 × 5 × 12

= 30 cm^{2}

So,

Area(PQOR) = 2(30) = 60 cm^{2}

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