Perimeter of rectangular metal sheet

CAT 2020 Slot 1 · QA · Medium · Geometry

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On a rectangular metal sheet of area 135 sq in, a circle is painted such that the circle touches opposite two 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. A.

    3\sqrt{\pi}\left(5 + \frac{12}{\pi}\right)

  2. B.

    4\sqrt{\pi}\left(3 + \frac{12}{\pi}\right)

  3. C.

    5\sqrt{\pi}\left(3 + \frac{12}{\pi}\right)

  4. D.

    3\sqrt{\pi}\left(\frac{5}{2} + \frac{6}{\pi}\right)

Answer

A

Explanation

Total area of sheet A=135A = 135 sq in. Given: Unpainted Area =23×Painted Area= \frac{2}{3} \times \text{Painted Area}. Total Area =Painted Area+Unpainted Area=53×Painted Area=135= \text{Painted Area} + \text{Unpainted Area} = \frac{5}{3} \times \text{Painted Area} = 135. Painted Area (Circle)=135×35=81 sq in\text{Painted Area (Circle)} = 135 \times \frac{3}{5} = 81\text{ sq in}

Since circle touches two opposite sides of the rectangle, width w=d=2rw = d = 2r. Area of circle=πr2=81  ⟹  r=9π  ⟹  w=18π\text{Area of circle} = \pi r^2 = 81 \implies r = \frac{9}{\sqrt{\pi}} \implies w = \frac{18}{\sqrt{\pi}}

Since total area l×w=135l \times w = 135: l=13518/π=15π2l = \frac{135}{18/\sqrt{\pi}} = \frac{15\sqrt{\pi}}{2}

Perimeter of rectangle =2(l+w)=2(15π2+18π)=15π+36π=3π(5+12π)= 2(l + w) = 2\left(\frac{15\sqrt{\pi}}{2} + \frac{18}{\sqrt{\pi}}\right) = 15\sqrt{\pi} + \frac{36}{\sqrt{\pi}} = 3\sqrt{\pi}\left(5 + \frac{12}{\pi}\right).

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