CAT 2020 Quant Question [Slot 1] with Solution 04

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:

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