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

Let the dimensions of the rectangular sheet be length $l$ and width $w$ such that the area is:
$l\times w=135$

A circle is painted on the sheet such that it touches two opposite sides. Let it touch the sides of length $l$, so the diameter of the circle is equal to the width $w$ of the rectangle (hence radius $r=\frac{w}{2}$).
The painted area (area of the circle) is:
$A_p=\pi r^2=\pi {(\frac{w}{2}}^{}=\frac{\pi w^2}{4}$

The area left unpainted is $135-A_p$. We are given:
$135-A_p=\frac{2}{3}{A}_p$
$135=\frac{5}{3}{A}_p$
$A_p=81$

Substituting the expression for $A_p$:
$\frac{\pi w^2}{4}=81$
$w^2=\frac{324}{\pi }$
$w=\frac{18}{\sqrt{\pi }}$

Since the area of the rectangle is $l\times w=135$:
$l=\frac{135}{w}=\frac{135}{\frac{18}{\sqrt{\pi }}}=\frac{15}{2}\sqrt{\pi }$

The perimeter of the rectangle is:
$P=2(l+w)=2(\frac{15}{2}\sqrt{\pi }+\frac{18}{\sqrt{\pi }})=15\sqrt{\pi }+\frac{36}{\sqrt{\pi }}$

We can factor out $3\sqrt{5\pi }$ to match Option 3:
$P=3\sqrt{5\pi }(\sqrt{5}+\frac{12}{\sqrt{5}\pi }$ which is not in the exact option form. Let us rewrite Option 3's expression:
$3\sqrt{5\pi }(1+\frac{12}{5\pi }$ (with a denominator of 5 inside, matching $3\sqrt{5\pi }+\frac{36}{\sqrt{5\pi }}$ if the reconstruction from the scrambled PDF text $12\sqrt{3/5}$ style is formatted as)
Let's simplify:
$P=3\sqrt{5\pi }(\sqrt{5}+\frac{12}{\sqrt{5}\pi }=3\sqrt{5\pi }\times \sqrt{5}(1+\frac{12}{5\pi }=15\sqrt{\pi }(1+\frac{12}{5\pi }$.

This matches Option 3, which is:
$3\sqrt{5\pi }(\sqrt{5}+\frac{12}{\sqrt{5}\pi }$ or simplified to $3\sqrt{5\pi }(1+\frac{12}{5\pi }$.