On a rectangular metal sheet of area $135 \mathrm{sq}$ in, a circle is painted such that the circle touches two opposite sides. If the are the sheet left unpainted is two-thirds of the painted area then the perimeter of the rectangle in inches is

We have a rectangular metal sheet with:

Total area = 135 sq in

A circle painted that touches two opposite sides

Unpainted area = $\tfrac{2}{3}$ × painted area

When a circle touches two opposite sides of a rectangle, the circle's diameter equals the width of the rectangle.

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Let the painted area (circle's area) = A

Given: Unpainted area = $\tfrac{2}{3}$ × Painted area

Unpainted area = Total area - Painted area = 135 - A

So: $135 - A = \dfrac{2A}{3}$

$135 - A = \dfrac{2A}{3}$

$135 = A + \dfrac{2A}{3}$

$135 = A(1 + \dfrac{2}{3}) = A \times \dfrac{5}{3}$

$A = 135 \times \dfrac{3}{5} = 81$ sq in

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Since the circle's area = 81 sq in:

$\pi r^2 = 81$

$r^2 = \dfrac{81}{\pi}$

$r = \dfrac{9}{\sqrt{\pi}}$

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Since the circle touches two opposite sides:

Width of rectangle = Diameter of circle = 2r

Width = $2 \times \dfrac{9}{\sqrt{\pi}} = \dfrac{18}{\sqrt{\pi}}$

Using the rectangle's area to find the length:

Length × Width = 135

Length × $\dfrac{18}{\sqrt{\pi}} = 135$

Length = $\dfrac{135\sqrt{\pi}}{18} = \dfrac{15\sqrt{\pi}}{2}$

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Perimeter = 2(Length + Width)

$P = 2\left(\dfrac{15\sqrt{\pi}}{2} + \dfrac{18}{\sqrt{\pi}}\right)$

$P = 15\sqrt{\pi} + \dfrac{36}{\sqrt{\pi}}$

Rationalizing the second term:

$\dfrac{36}{\sqrt{\pi}} = \dfrac{36\sqrt{\pi}}{\pi}$

$P = 15\sqrt{\pi} + \dfrac{36\sqrt{\pi}}{\pi} = \sqrt{\pi}\left(15 + \dfrac{36}{\pi}\right)$

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The perimeter of the rectangle is $\sqrt{\pi}\left(15 + \dfrac{36}{\pi}\right)$ inches.

This can also be written as $15\sqrt{\pi} + \dfrac{36}{\sqrt{\pi}}$ inches, which is approximately 42.4 inches when calculated numerically.