A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the `Delta ABC`

Options

A. 12 cm
B. 20 cm
C. 24 cm
D. 30 cm

Answer

C

Explanation

Given Two tangents are drawn from an external point A to the circle with centre O.

Tangent BC is drawn at a point R. radius of circle equals 5cm.
To find perimeter of `DeltaABC`.
Proof `angleOPA=90^(@)`
[tangent at any point of a circle is perpendicular to the radius through the point of contact]
`:.OA^(2)=OP^(2)+AP^(2)` [by Pythagoras theorm]
`(13)^(2)=5^(2)+PA^(2)`
`impliesPA^(2)=144=12^(2)`
`impliesPA=12cm`
Now, perimeter of `DeltaABC=AB+BC+CA`
`=(AB+BR)+(RC+CA)`
`=AB+BP+CQ+CA`
[`:'BR=BP,RC=CQ` tangents from internal point to a circle are equal]
`=AP+AQ`
=2AP
=2(12)
=24cm
[AP=AQ tangnet from internal point to a circle are equal]
Hence, the perimeter of `DeltaABC=24cm`.