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# CAT 2020 Quant Question [Slot 1] with Solution 04

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

1. $5\sqrt \pi \left( {3 + \frac{9}{\pi }} \right)$
2. $3\sqrt \pi \left( {\frac{5}{2} + \frac{6}{\pi }} \right)$
3. $3\sqrt \pi \left( {5 + \frac{{12}}{\pi }} \right)$
4. $4\sqrt \pi \left( {3 + \frac{9}{\pi }} \right)$
Option: 2
Solution:

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 $lb = \pi {(b/2)^2} = \frac{2}{3} \times \pi {(b/2)^2}$

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

From this $l = \frac{{15\sqrt \pi }}{2}$

$\therefore$ Required perimeter:

$2(l + b) = 2\left[ {\frac{{15\sqrt \pi }}{2} + \frac{{18}}{{\sqrt \pi }}} \right] = 3\sqrt \pi \left[ {\frac{5}{2} + \frac{6}{\pi }} \right]$

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