The incircle of radius $4$ cm of $△ A B C$ touches the side $B C$ at $D$ . If $B D = 6$ cm and $D C = 8$ cm, then the area of $△ A B C$ is

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Solution

We have, $∠ I B D = \frac{B}{2}, ∠ I C D = \frac{C}{2}$
$B D = 6 c m, C D = 8 c m, \Rightarrow B C = a = 14 c m .$
Also, $I D = r = 4 c m$
Now, $tan \frac{B}{2} = \frac{4}{6} = \frac{2}{3}$ and $tan \frac{C}{2} = \frac{4}{8} = \frac{1}{2}$
$tan \frac{B}{2} tan \frac{C}{2} = \sqrt{\frac{(s - a) (s - c)}{s (s - b)}} \times \sqrt{\frac{(s - a) (s - b)}{s (s - c)}}$
$∴ \frac{2}{3} \times \frac{1}{2} = \frac{s - 14}{s} \Rightarrow s = 21 c m$
Area, $△ = r s = 4 \times 21 = 84 {c m}^{2}$

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