Question

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 PQR is

• A. $60c{m}^2$
• B. $65c{m}^2$
• C. $30c{m}^2$
• D. $32.5c{m}^2$

Answer 1

The correct option is A $60c{m}^2$

from the figure the area of the quadrilateral PQOR is sum of the area of triangles PQO and POR

now in triangle PQO $P{Q}^2+Q{O}^2=P{O}^2$[since it is aright angles triangle]

therefore $P{Q}^2={13}^2-5^2=144$

$⟹PQ=12$ and $PQ=PR$[since lengths of tangents drawn from a same point are same]

so $PR=12$

now the area of two triangles are

$\frac{1}{2}\times PQ\times QO+\frac{1}{2}\times PR\times RO=6\times 5+6\times 5=60$

therefore area is $60c{m}^2$