question_answer

If BC passes through the centre of the circle, then the area of the shaded region in the given figure is

A)  \[\frac{{{a}^{2}}}{2}(3-\pi )\]     

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

C)  \[2{{a}^{2}}(\pi -1)\]     

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

Correct Answer: D

Solution :

 Here O is the centre of the circle. From right angled \[\Delta \,\,BAC,\]                 \[B{{C}^{2}}=A{{C}^{2}}+A{{B}^{2}}=2{{a}^{2}}\] \[\therefore \]  \[BC=\sqrt{2a}\] \[\therefore \] Radius of circle, \[OB=\frac{1}{2}\sqrt{2a}=\frac{1}{\sqrt{2}}a\] \[\therefore \]  Area of semi-circle \[=\frac{1}{2}\pi {{(OB)}^{2}}\]                 \[=\frac{1}{2}\pi {{\left( \frac{1}{\sqrt{2}\alpha } \right)}^{2}}=\frac{1}{4}\pi {{a}^{2}}\] Area of \[\Delta \,BAC=\frac{1}{2}\times a\times a=\frac{{{a}^{2}}}{2}\] Hence, area of shaded region \[=\frac{1}{4}\pi \,{{a}^{2}}-\frac{{{a}^{2}}}{2}\]                                 \[=\frac{{{a}^{2}}}{2}\left( \frac{\pi }{2}-1 \right)\]