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

Options

A. `60cm^(2)`
B. `32.5cm^(2)`
C. `65cm^(2)`
D. `30cm^(2)`

Explanation

Clearly, `OQ=OR=5cm`, `/_OQP=/_ORP=90^(@)` and `OP=13cm`
`:.PQ^(2)=OP^(2)-OQ^(2)=(13)^(2)-(5)^(2)=169-25=144impliesPQ=sqrt(144)=12cm`.
`:.ar(DeltaOQP)=(1)/(2)xxPQxxOQ=((1)/(2)xx12xx5)cm^(2)=30cm^(2)`.
Similarly, `ar(DeltaORP)=30cm^(2)`.
`:.ar("quad" PQOR)=(30+30)cm^(2)=60cm^(2)`