Question

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

• A. $5\sqrtπ\bigg(\frac{3+9}{π}\bigg)$
• B. $3\sqrtπ\bigg(\frac{5}{2}+\frac{6}{π}\bigg)$
• C. $3\sqrtπ\bigg(\frac{5+12}{π}\bigg)$
• D. $4\sqrtπ\bigg(\frac{3+9}{\sqrt π}\bigg)$

Answer 1

Answer: (B)

$3\sqrtπ\bigg(\frac{5}{2}+\frac{6}{π}\bigg)$

Answer 2

The correct option is (B): $3\sqrtπ\bigg(\frac{5}{2}+\frac{6}{π}\bigg)$

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

As the circle touches the two opposite sides, its diameter will be same as the breadth of the rectangle.

Given, lb = $135$ and lb = $π(b/2)^2 = \frac{2}{3}×π(b/2)^2$

$⇒ \frac{5}{3}π\bigg(\frac{b^2}{4}\bigg) = 135 ⇒ b = \frac{18}{\sqrtπ}$

From this l = $\frac{15\sqrtπ}{2}$

∴ Required perimeter:

$2(l+b)=2\bigg[\frac{15\sqrtπ}{2}+\frac{18}{\sqrtπ}\bigg]=3\sqrtπ\bigg[\frac{5}{2}+\frac{6}{π}\bigg]$