Question

In the figure, $ABCD$ is a unit square. A circle is drawn with centre $O$ on the external line $CD$ and passing through $A$. If the diagonal $AC$ is tangent to the circle, then the area of the shaded region is

• A. $\frac{9-\pi }{6}$
• B. $\frac{8-\pi }{6}$
• C. $\frac{7-\pi }{4}$
• D. $\frac{6-\pi }{4}$

Answer 1

Answer: (D)

$\frac{6-\pi }{4}$

$AC$ is tangent to the circle, therefore $AC\perp OA\Rightarrow \angle OAC={90}^o$

In $△OAC,\angle ACD={45}^o\Rightarrow \angle AOD={45}^o$

$\therefore AC=OA=\sqrt{2}=$radius of circle.

$ar\left[AXD\right]=ar\left[circle\right]-ar\left[△AOD\right]=\frac{\pi }{8}\times 2-\frac{1}{2}$

Shaded area $=ar[sq.ABCD]-ar\left[AXD\right]=1-\frac{\pi }{4}+\frac{1}{2}=\frac{6-\pi }{4}$