Question

In the figure, $O$ is the centre of a circle such that diameter $AB=13$ cm and $AC=12$ cm. $BC$ is joined. Find the area of the shaded region. (Take $\pi =3.14$).

Answer 1

In $\Delta ABC$,

Angle in a semicircle, $\angle C={90}^o$

Therefore, by Pythagoras theorem

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

$B{C}^2+{12}^2={13}^2$

$B{C}^2=169-144=25$

$BC=5$

Area of shaded region = Area of semicircle – Area of triangle

= $\frac{1}{2}\pi r^2-\frac{1}{2}bh$

= $\frac{1}{2}\times 3.14\times {\left(\frac{13}{2}\right)}^2-\frac{1}{2}\times 5\times 12$

= $\frac{3.14\times 13\times 13}{2\times 4}-30$

Area of shaded region $=36.39$ cm${}^2.$