question_answer

In the adjoining figure ACB is a quadrant with radius 'a'. A semicircle is drawn outside the quadrant taking AB as a diameter. What is the area of shaded region?

A)  \[\frac{1}{4}\left( \pi -2{{a}^{2}} \right)\]                     

B)  \[\left( \frac{1}{4} \right)\left( \pi {{a}^{2}}-{{a}^{2}} \right)\]

C)  \[\frac{{{a}^{2}}}{2}\]          

D)  \[{{a}^{2}}\left( \frac{\pi -2}{2} \right)\]

Correct Answer: C

Solution :

(c): Area of quadrant \[=\frac{1}{4}\pi {{a}^{2}}\]; Area of triangle \[ACB=\frac{{{a}^{2}}}{2}\] Area of segment AQB = Area of sector AQBC ? Area of  \[\Delta \,ACB=\frac{\pi {{a}^{2}}}{4}-\frac{{{a}^{2}}}{2}=\frac{{{a}^{2}}}{4}[\pi -2]\] Area of semi ? circle \[=\frac{1}{2}\pi {{\left( \frac{a\sqrt{2}}{2} \right)}^{2}}=\frac{\pi {{a}^{2}}}{4}\] \[\therefore \]  Area of required region \[=\frac{\pi {{a}^{2}}}{4}-\frac{{{a}^{2}}}{4}[\pi -2]\] \[=\frac{{{a}^{2}}}{2}\] sq. unit