Question

If in a $\Delta ABC$, $b=12$ units, $c=5$ units and $\Delta =30$ sq. units,

then the distance between vertex $A$ and incentre of the triangle is equal to?

• A. $2$ units
• B. $2\sqrt{2}$ units
• C. $\sqrt{2}$ units
• D. none of these

Answer 1

The correct option is B $2\sqrt{2}$ units

$\Delta =\frac{1}{2}bcsinA\Rightarrow 30=\frac{1}{2}12\times 5\times \sin A$

$\Rightarrow \sin A=1\therefore A={90}^{\circ }$

$\Rightarrow a=13$

From the figure, $\frac{ID}{IA}=\sin \frac{A}{2}=\frac{1}{\sqrt{2}}$

$IA=\sqrt{2}ID=\sqrt{2}r$

Now, $s=15$ and $\Delta =30$

$\Rightarrow r=\frac{30}{15}=2$

$\therefore IA=2\sqrt{2}$