Question

In figure, square $JKLM$ is inscribed in circle $O$. If the radius is $6$, calculate the area of the shaded region, to the nearest tenth.

• A. $10.3$
• B. $18.2$
• C. $22.8$
• D. $28.3$
• E. $38.6$

Answer 1

Answer: (A)

$10.3$

Given the radius of circle is $6$ and square $JKLM$ is inscribed in circle $O$.

As per given figure, $△MOL$ is right angle triangle

Then $(ML{)}^2=(OL{)}^2+(OM{)}^2$

$\Rightarrow (ML{)}^2=(6{)}^2+(6{)}^2$

$\Rightarrow ML=6\sqrt{2}$ is the length of square

Then area of circle $=$ $\pi r^2=\frac{22}{7}\times (6{)}^2=3.1428\times 36=113.14$

Area of square $=$ $a^2=(6\sqrt{2}{)}^2=72$

Area of circle $-$ area of square $=$ $113.14-72=41.14$

Then area of shaded region $=$ $\frac{41.14}{4}=10.285\approx 10.30$