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 of $\delta$ 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 $a^2\left(\frac{4-\pi }{4}\right)$

Let $ABCD$ is a square with side $a$.

$\therefore A\left(\square ABCD\right)=a^2$

Let's draw four circles with centre $A,B,C$ and $D$ such that each circle touches two of the remaining circles externally.

As the circles drawn, their radius will be $r=a/2$ and one fourth part of a circle lie in the square.

Area of a each circle $=\pi r^2=\frac{a^2}{4}\pi$

So, for each circle the area covered by the rectangle $=\frac{\frac{a^2\pi }{4}}{4}=\frac{a^2\pi }{16}$.

$\therefore$ the area covered by four circles $=4\times \frac{a^2\pi }{16}=\frac{a^2\pi }{4}$.

$\therefore$ $\delta =A\left(\square ABCD\right)-\left(Areaof\square ABCDcoveredby4circle\right)$

$=a^2-\frac{a^2\pi }{4}=\frac{4a^2-a^2\pi }{4}$

$=a^2\left(\frac{4-\pi }{4}\right)$

$\therefore$ $\delta =a^2\left(\frac{4-\pi }{4}\right)$