Question

Four identical coins are placed in a square. For each coin, the ratio of area of circumference is same as the ratio of circumference to area. Then, find the area of the square that is not covered by the coins

• A. $16(4-x)$
• B. $12(4-x)$
• C. $16(8-x)$
• D. $16(12-x)$

Answer 1

The correct option is A $16(4-x)$

Let $R$ be the radius of each circle.

Area of circle $=\pi R^2$

Circumference of circle $2\pi R$

According to the question,

$\Rightarrow$ $\frac{\pi R^2}{2\pi R}=\frac{2\pi R}{\pi R^2}$

$\Rightarrow$ $R^2=4$

$\Rightarrow$ $R=2$

In figure we can see that,

Side of the square $=2\times$ Diameter of one circle$=2\times 2R=4\left(2\right)=8$

$\Rightarrow$ Area of the square $=(Side{)}^2=(8{)}^2=64$

$\Rightarrow$ Area covered by $4$ coins $=4\pi R^2=4\pi (2{)}^2=16\pi$

Area which is not covered by the coins $=$ Area of square $-$ Area of $4$ coins

$=64-16\pi$

$=16(4-\pi )$