On a rectangular metal sheet of area $135$ sq in, a circle is painted such that the circle touches two opposite sides. If the area of the sheet left unpainted is two$-$thirds of the painted area then the perimeter of the rectangle in inches is

- $5\sqrt{\pi }\left ( 3+\frac{9}{\pi} \right )$
- $3\sqrt{\pi }\left ( \frac{5}{2}+\frac{6}{\pi} \right )$
- $3\sqrt{\pi }\left ( 5+\frac{12}{\pi} \right )$
- $4\sqrt{\pi }\left ( 3+\frac{9}{\pi} \right )$

## 1 Answer

Given that, area of rectangular metal sheet $ = 135 \; \text{sq inches}.$

Now, we can draw the diagram.

Let the length and breadth of the rectangle be $l$ and $b$ respectively.

Let the radius of a circle be $r \; \text{inches}.$

$\Rightarrow lb = 135$

Let the painted area be $3x.$ Then left unpainted area will be $2x.$

So, $3x + 2x = 135$

$\Rightarrow 5x = 135$

$ \Rightarrow \boxed{x = 27}$

The painted area (area of circle) $ = 3 \times 27 = 81 \; \text {sq inches}.$

The left unpainted area $ = 2 \times 27 = 54 \; \text{sq inches}.$

Now, $ \pi r^{2} = 81 $

$ \Rightarrow r = \frac{9}{\sqrt{\pi}}$

$ \Rightarrow b = 2r $

$ \Rightarrow \boxed {b = \frac{18}{\sqrt{\pi}}}$

And, $lb = 135$

$ \Rightarrow l \left( \frac{18}{\sqrt{\pi}} \right) = 135 $

$ \Rightarrow \boxed{l = \frac{15 \sqrt{\pi}}{2}}$

$\therefore$ The perimeter of the rectangle $ = 2 ( l+b)$

$ \qquad \qquad = 2 \left( \frac{15 \sqrt{\pi}}{2} + \frac{18}{ \sqrt{\pi}}\right) $

$\qquad \qquad = 15 \sqrt{\pi} + \frac{36}{\sqrt{\pi}}$

$\qquad \qquad = 15 \sqrt{\pi} + \frac{36}{\sqrt{\pi}} \times \frac{\sqrt{\pi}}{\sqrt{\pi}}$

$\qquad \qquad = 3 \sqrt{\pi} \left( 5 + \frac{12}{\pi} \right) \; \text{inches}.$

Correct Answer$: \text{C}$