Question

In the given figure, ABC is a right angle triangle right-angled at A. Find the area of shaded region, if AB = $6$m, BC = $10$m and O is the centre of the incircle of triangle ABC (Use $\pi =3.14$)

Answer 1

$ABC$ is a right angled triangle where $\angle A={90}^{\circ }$

$BC=10$cm and $AB=6$cm

Let $O$ be the centre and $r$ be the radius of the in-circle.

$AB,BC$ and $CA$ are the tangents to the circle at $P,M$ and $N$

$\therefore IP=IM=IN=r$(radius of the circle)

In $△BAC,$

${BC}^2={AB}^2+{AC}^2$(by pythagoras theorem)

$\Rightarrow {10}^2=6^2+{AC}^2$

$\Rightarrow {AC}^2=100-36=64$

$\therefore AC=8$cm

Area of $△ABC=\frac{1}{2}bh=\frac{1}{2}\times AC\times AB=\frac{1}{2}\times 8\times 6=24$sq.cm

Area of $△ABC=$Area of $△IAB+$Area of $△IBC+$ Area of $△ICA$

$\Rightarrow 24=\frac{1}{2}r\left(AB\right)+\frac{1}{2}r\left(BC\right)+\frac{1}{2}r\left(CA\right)$

$\Rightarrow 24=\frac{1}{2}r(AB+BC+CA)$

$\Rightarrow 24=\frac{1}{2}r(6+8+10)$

$\Rightarrow 24=12r$

$\therefore r=\frac{24}{12}=2$cm

Area of the circle$=\pi r^2=\frac{22}{7}\times 2^2=12.56$sq.cm

Area of shaded region$=$Area of $△ABC-$Area of the circle.

$=24-12.56=11.44$sq.cm