Question

In the adjoining figure ABC is a right angled at A . Find the area of the shaded region if $AB=6$ cm and $BC=10$ cm and I is the centre of in circle of $△$ ABC

Answer 1

Applying pythagoras theorem in $\Delta$ ABC, we have
$B{C}^2=A{B}^2+A{C}^2$
$\Rightarrow A{C}^2=B{C}^2-A{B}^2$
$\Rightarrow A{C}^2$ = $100-36=64$
$\Rightarrow$ $AC=8cm$

$\therefore$ Area of $\Delta$ ABC = $\frac{1}{2}\times AB\times AC$

$\Rightarrow$ Area of $\Delta$ ABC = $\frac{1}{2}\times 6\times 8c{m}^2=24c{m}^2$

Let r cm be the incircle.
Clearly,
Area of $\Delta$ ABC = Area of $\Delta$ IBC + Area of $\Delta$ ICA + Area of $\Delta$ IAB

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

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

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

$\Rightarrow$ $24=12r$

$\Rightarrow$ $r=2$

$\therefore$ Area of the shaded region = Area of $\Delta$ ABC - Area of the incircle

$\Rightarrow$ Area of the shaded region = 24 - $\pi r^2=(24-\frac{22}{7}\times 4)c{m}^2=\frac{80}{7}c{m}^2$