Question

A triangle PQR is drawn to circumsribe a circle of radius 8 cm such that the segements QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of $\Delta PQR$ is $336c{m}^2$, find the sides PQ and PR.

Answer 1

Since length of tangents drawn from a point to a circle are equal.

$\therefore$ $QS=QT=14cm$, $RU=RT=16cm$ and, $PS=PU=x$.

Thus, $PQ=x+14,PR=x+16$ and $QR=14+16=30$

Now, Area of $\Delta PQR$ = Area of $\Delta IQR$+Area of $\Delta OQP$ + Area of $\Delta IPR$

$\Rightarrow 336=\frac{1}{2}(QR\times 8)+\frac{1}{2}(14+2)\times 8+\frac{1}{2}(16+x)\times 8$

$\Rightarrow 84=30+14+x+16+x$

$\Rightarrow 24=2x\Rightarrow x=12$

Hence, $PQ=12+14=26cm$ and $PR=12+16=28cm$