Question

$ABCD$ is a square with side $a$. With centres $A,B,C$ and $D$ four circles are drawn such that each circle touches externally two of the remaining three circles. Let $\delta$ be the area of the region in the interior of the square and exterior of the circles. Then the maximum value $\delta$ is:

• A. $a^2(1-\pi )$
• B. $a^2\left(\frac{4-\pi }{4}\right)$
• C. $a^2(\pi -1)$
• D. $\frac{\pi a^2}{4}$

Answer 1

The correct option is B $a^2\left(\frac{4-\pi }{4}\right)$

Area of Interior region $=$ Area of square $-$ Total area of circles inside the square
Area of square $=$ $sid{e}^2$ $=$ $a^2$
Area of portion of 4 circles $=$ $4\times \left(\frac{1}{4}\pi r^2\right)$ =$4\times \left(\frac{1}{4}\pi \right(\frac{a}{2}{)}^2)$ $=$ $\frac{\pi a^2}{4}$
Thus, area of Interior region $=$ $a^2-\frac{\pi a^2}{4}$
$=$ $a^2\left(\frac{4-\pi }{4}\right)$