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

The Correct Option is B

The correct option is (B): \(3\sqrt{\pi} \left(\frac{5}{2} + \frac{6}{\pi}\right)\) 

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

As the circle touches the two opposite sides, its diameter is equal to the breadth of the rectangle.

Given, \(l \cdot b = 135\)
and area covered by circle is \(\frac{2}{3} \cdot \pi \left(\frac{b}{2}\right)^2\)

So, \(\frac{5}{3} \cdot \pi \cdot \left(\frac{b^2}{4}\right) = 135 \Rightarrow b = \frac{18}{\sqrt{\pi}}\)

Then, \(l = \frac{135}{b} = \frac{15\sqrt{\pi}}{2}\)

Required perimeter:
\(2(l + b) = 2\left(\frac{15\sqrt{\pi}}{2} + \frac{18}{\sqrt{\pi}}\right) = 3\sqrt{\pi} \left(\frac{5}{2} + \frac{6}{\pi}\right)\)