question_answer

In Fig, ABC is a right - angle triangle, \[\angle B=90{}^\circ ,AB=28\text{ }cm\] and \[BC=21\text{ }cm\]. With AC as diameter, a semicircle is drawn and with BC as radius, a quarter-circle is drawn. Find the area of the shaded region correct to two decimal places.

A)  \[428.75\,c{{m}^{2}}\]    

B)  \[857.50\,c{{m}^{2}}\]

C)  \[214.37\,c{{m}^{2}}\]               

D)  \[371.56\,c{{m}^{2}}\]

Correct Answer: A

Solution :

(a): Area of shaded portion = Area of ADC ? Area of sector DC + Area f triangle ADB ?  sector BED \[\Rightarrow \] Area of ADC \[=\pi \times {{(17.5)}^{2}}\times \frac{1}{2}=481\,\,c{{m}^{2}}\] \[\frac{\angle DBC}{\angle ABC}=\frac{21}{28}\Rightarrow \angle DBC=67.5\] and\[\angle DBA=22.5\] \[\Rightarrow \] Area of sector a\[DC=\left( \pi \times {{21}^{2}}\times \frac{67.5}{360} \right)\] \[-\left( \frac{1}{2}\times {{21}^{2}}\times \sin \,\,67.5 \right)=56\,\,c{{m}^{2}}\] \[\Rightarrow \]Area of ADE \[=\left( \frac{1}{2}\times 28\times 21 \right)-\] \[\left( 204+\frac{1}{2}\times {{21}^{2}}\times \sin \,\,22.5 \right)=56\,\,c{{m}^{2}}\]