Question

In the given figure $\angle B={90}^0$, AC is diameter of a semicircle and with BC as radius, a quarter circle is drawn, find the area of the shaded portion. (Write answer in the nearest integer).

Answer 1

In $△ABC$,

$AC=\sqrt{{AB}^2+{BC}^2}$

$=\sqrt{{\left(28\right)}^2+{\left(21\right)}^2}$

$=\sqrt{784+441}$

$=\sqrt{1225}$

$=35cm$

Area of shaded region $=$ ar $△ABC$ + ar semi-circle drawn on $AC$ - ar quadrant $DBC$

$=(\frac{1}{2}\times BC\times AB)+[\frac{1}{2}\times \pi \times {\left(\frac{AC}{2}\right)}^2]-[\frac{\theta }{{360}^0}\times \pi \times {BC}^2]$

$=(\frac{1}{2}\times 21\times 28)+[\frac{1}{2}\times \frac{22}{7}\times \frac{35}{2}\times \frac{35}{2}]-[\frac{{90}^0}{{360}^0}\times \pi \times 21\times 21]$

$=294+481.25-346.5$

$=428.75{cm}^2$