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  • From a point P 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(A) 60 cm2 (B) 65 cm2 (C) 30 cm2 (D) 32.5 cm2

From a point P 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

(A) 60 cm2                   (B) 65 cm2                   (C) 30 cm2                   (D) 32.5 cm2

Answers (1)

Answer (A) 60 cm2
Solution
According to question

Given OP = 13 cm
OQ = OR = radius = 5 cm
  In \bigtriangleupPOQ, \angle Q= 90^{\circ}          (\because  PQ is tangent)                                    
Using Pythagoras theorem in \bigtriangleupPOQ       

\left ( PO \right )^{2}= \left ( PQ \right )^{2}+\left ( QO \right )^{2}      ( because H2 = B2 + P2)
\left ( 13 \right )^{2}= \left ( PQ \right )^{2}+\left ( 5 \right )^{2}
\left ( PQ\right )^{2}= 169-25
PQ= \sqrt{144}= 12
PQ= 12
Area of \bigtriangleupPOQ = \frac{1}{2}  × perpendicular × base
= \frac{1}{2}\times PQ\times OQ
= \frac{1}{2}\times 12\times 5= 30cm^{2}
Area of quadrilateral = 2 × area of \bigtriangleupPOQ
= 2 × 30 = 60 cm2

      

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