Question

$ABC$ is an isosceles right angled triangle with $\angle ABC={90}^o$. A semi-circle is drawn with $AC$ as the diameter. If $AB=BC=7cm$, find the area (in $c{m}^2$) of shaded region. $(\text{Take}\pi =\frac{22}{7})$

Answer 1

$△ABC$ is a right angled triangle. So, by Pythagoras theorem

$A{C}^2=A{B}^2+B{C}^2=7^2+7^2=98$

$\therefore AC=\sqrt{98}=7\sqrt{2}cm$

Arc $A-B-C$ is a semicircle. $\therefore AC=7\sqrt{2}cm$ is the diameter of the semicircle.

Area of the shaded portion $=$ Area of semi-circle $\left(A-B-C\right)-$ Area of $△ABC$

$=\frac{1}{2}\times \frac{\pi d^2}{4}-\frac{1}{2}\times AB\times BC$

$=\frac{22\times 98}{2\times 7\times 4}-\frac{7\times 7}{2}$

$=14c{m}^2$