Q. 295.0( 1 Vote )

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


A. 60 cm2

B. 32.5 cm2

C. 65 cm2

D. 30 cm2

Answer :

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 cm2


So,


Area(PQOR) = 2(30) = 60 cm2

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