Question

Find the area of the shaded region in Fig., if $AC=24cm,BC=10cm$ and O is the centre of the circle. (Use $\pi$ $=3.14$).

Answer 1

Given $AC=24cm,BC=17cm$

$\Rightarrow$ $\angle ACB={90}^o$ [ Angle inscribe in semicircle ]

$\therefore$ $△ACB$ is right angled triangle.

$\Rightarrow$ $(AB{)}^2=(AC{)}^2+(BC{)}^2$ [ By Pythagoras theorem ]

$\Rightarrow$ $(AB{)}^2=(24{)}^2+(10{)}^2$

$\Rightarrow$ $(AB{)}^2=576+100$

$\Rightarrow$ $(AB{)}^2=676$

$\therefore$ $AB=26cm$

$\therefore$ Radius of circle $\left(r\right)=\frac{AB}{2}=\frac{26}{2}=13,cm$

$\Rightarrow$ Area of circle $=\pi r^2=3.14\times (13{)}^2$

$\therefore$ Area of circle $=3.14\times 169=530.66c{m}^2$

$\Rightarrow$ Area of semicircle $=\frac{530.66}{2}=265.33c{m}^2$

$\Rightarrow$ Area of $△ABC=\frac{1}{2}\times AC\times BC$

$\Rightarrow$ Area of $△ABC=\frac{1}{2}\times 24\times 10$

$\therefore$ Area of $△ABC=120c{m}^2$

$\Rightarrow$ Area of shaded region = Area of circle - ( Area of semicircle +Area of $△ABC$ )

$\Rightarrow$ Area of shaded region $=530.66-(265.33+120)$

$\therefore$ Area of shaded region $=530.66-385.33=145.33c{m}^2$