Question

In the figure, $ABCD$ is a unit square. A circle is drawn with centre $O$ on the extended 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

The correct option is D $\frac{6-\pi }{4}$


$ABCD$ is square with side lengths $=1$, so $AC=\sqrt{2}$
As $AC$ is tangent, so $\angle OAC={90}^{\circ }$
Let $OD=x$
In $△ADO$
$A{D}^2+O{D}^2=A{O}^2\Rightarrow 1+x^2=O{A}^2\Rightarrow OA=\sqrt{1+x^2}$

$In△OACO{A}^2+A{C}^2=O{C}^2\Rightarrow 1+x^2+2=(1+x{)}^2\Rightarrow x^2+3=x^2+2x+1\Rightarrow x=1$
Then $\angle AOD={45}^{\circ }$
Hence, Radius of the circle $r=\sqrt{1+1^2}=\sqrt{2}$
Required area $=area\left(△AOD\right)+area\left(ABCD\right)-area\left(AOX\right)=\frac{1}{2}\times 1\times 1+1\times 1-(\frac{\pi r^2}{2\pi }\times \frac{\pi }{4})=\frac{3}{2}-\frac{\pi }{8}(\sqrt{2}{)}^2=\frac{6-\pi }{4}$