Question

In the figure, $AC=24$ cm, $BC=10$ and $O$ is the center of the circle. Find the area of shaded region.

Answer 1

In triangle $ABC$, applying Pythagoras theorem,

$A{B}^2=A{C}^2+B{C}^2$

$A{B}^2={24}^2+{10}^2$

$A{B}^2=576+100$

$A{B}^2=676$

$AB=26$

Area of the triangle $ABC=\frac{1}{2}\times BC\times AC$

$=\frac{1}{2}\times 24\times 10$

$=120$

Area of the semi circle $=\frac{\pi r^2}{2}$

here $r=\frac{AB}{2}$

$=\frac{26}{2}$

$=13$

So,

Area of semi circle $=\begin{array}{c}\frac{\pi {\left(13\right)}^2}{2}=\frac{22}{7}\times 169\times \frac{1}{2}=265.57\end{array}$

Area of shaded part = Area of semicircle - Area of triangle ABC

$=265.57-120$

$=145.57$

Thus area of shaded part is $145.57{\text{units}}^2$