(a)OR(b)

A triangle ABC is drawn to circumscribe a circle of radius $\displaystyle 4$ cm such that the segments BD and DC are of lengths $\displaystyle 10$ cm and $\displaystyle 8$ cm respectively. Find the lengths of the sides AB and AC , if it is given that area $\displaystyle \Delta \mathrm{ABC}=90 \mathrm{~cm}^{2}$.

Join OA, OB, OC and draw \(\displaystyle \mathbf{O E} \boldsymbol{\perp} \mathbf{A C}\) and \(\displaystyle \mathbf{O F} \boldsymbol{\perp} \mathbf{A B}\).

\(\displaystyle \mathbf{B F}=\mathbf{1 0} \mathbf{~ c m}, \mathbf{C E}=\mathbf{8} \mathbf{~ c m}\), Let \(\displaystyle \mathbf{A F}=\mathbf{A E}=\mathbf{x}\)

\(\displaystyle \operatorname{ar} \Delta \mathrm{ABC}=\operatorname{ar} \Delta \mathrm{BOC}+\operatorname{ar} \Delta \mathrm{COA}+\operatorname{ar} \Delta \mathrm{AOB}\)

\(\displaystyle 90=\frac{1}{2} .4(\mathrm{BC}+\mathrm{CA}+\mathrm{AB})\)

\(\displaystyle 90=2(18+8+x+10+x)\)

\(\displaystyle 90=4(18+x)\)

\(\displaystyle \mathrm{x}=4 \cdot 5\)

\(\displaystyle \mathbf{A B}=\mathbf{1 4} \cdot \mathbf{5 ~ c m}\) and \(\displaystyle \mathbf{A C}=\mathbf{1 2} \cdot \mathbf{5 ~ c m}\)

\[\begin{aligned} & \mathrm{OO}^{\prime}=\sqrt{6^{2}+8^{2}}=10 \mathrm{~cm} \quad\left\{\mathrm{OP} \perp \mathrm{O}^{\prime} \mathrm{P}\right\} \\ & \text { Let } \mathrm{OA}=\mathrm{x}, \mathrm{O}^{\prime} \mathrm{A}=10-\mathrm{x} \\ & \mathrm{AP}^{2}=36-\mathrm{x}^{2} \\ & \mathrm{Also} \mathrm{AP}^{2}=64-(10-\mathrm{x})^{2} \\ & \text { Therefore } 36-\mathrm{x}^{2}=64-(10-\mathrm{x})^{2} \\ & \Rightarrow 36-\mathrm{x}^{2}=64-100-\mathrm{x}^{2}+20 \mathrm{x} \\ & \Rightarrow \mathrm{x}=3 \cdot 6 \\ & \text { In } \triangle \mathrm{PAO}, \mathrm{~A} \mathrm{P}^{2}=36-(3 \cdot 6)^{2}=23 \cdot 04 \\ & \Rightarrow \mathrm{AP}=4 \cdot 8 \\ & \text { Length } \mathrm{PQ}=2 \times \mathrm{AP}=9 \cdot 6 \mathrm{~cm} \end{aligned} \]