A triangle PQR is drawn to circumscribe a circle of radius 8cm such that the segments QT and TR, into which QR is divided by the point of contact T, are of lengths 14cm and 16cm respectively. If area of  $\mathrm{△}\mathrm{PQR}$ is  336 cm², then the sides PQ and PR is

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

$\mathrm{QS}=\mathrm{QT}=14\mathrm{cm},\mathrm{RU}=\mathrm{RT}=16\mathrm{cm}\text{ and }\mathrm{PS}=\mathrm{PU}=\mathrm{x}$

Thus, $\mathrm{PQ}=\mathrm{x}+14,\mathrm{PR}=\mathrm{x}+16\text{ and }\mathrm{QR}=30$

Now, Area of $\mathrm{\Delta PQR}=\text{ Area of }\mathrm{\Delta IQR}+\text{ Area of }\mathrm{\Delta IQP}+\text{ Area of }\mathrm{\Delta IPR}$

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

$\begin{array}{l}\Rightarrow 84=30+14+x+16+x\\ \Rightarrow 24=2x\Rightarrow x=12\end{array}$

Hence, $\mathrm{PQ}=26\mathrm{cm}\text{ and }\mathrm{PR}=28\mathrm{cm}$