Question

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 $∆$ ABC .

Answer 1

A is a point 13 cm from centre O. AP and AQ are the tangents to the circle with centre O.
AP = AQ
In ∆APO,
$$\begin{gathered} {\mathrm{AP}}^2+{\mathrm{OP}}^2={\mathrm{AO}}^2 \\ \Rightarrow {\mathrm{AP}}^2={\mathrm{AO}}^2-{\mathrm{OP}}^2 \\ \Rightarrow {\mathrm{AP}}^2={13}^2-5^2 \\ \Rightarrow {\mathrm{AP}}^2=169-25=144 \\ \Rightarrow \mathrm{AP}=12\mathrm{cm} \end{gathered}$$
Now In ∆ABC,
Perimeter = AB + BC + AC
= AB + BR + RC + AC
= AB + BP + CQ + AC ( BR = BP, RC = CQ)
= AP + AQ
= 12 + 12
= 24 cm