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

• A. \(5\sqrtπ\bigg(\frac{3+9}{π}\bigg)\)
• B. \(3\sqrtπ\bigg(\frac{5}{2}+\frac{6}{π}\bigg)\)
• C. \(3\sqrtπ\bigg(\frac{5+12}{π}\bigg)\)
• D. \(4\sqrtπ\bigg(\frac{3+9}{\sqrt π}\bigg)\)

Answer 1

The Correct Option is B

The correct option is (B): \(3\sqrtπ\bigg(\frac{5}{2}+\frac{6}{π}\bigg)\)

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 = \(π(b/2)^2 = \frac{2}{3}×π(b/2)^2\)

\(⇒ \frac{5}{3}π\bigg(\frac{b^2}{4}\bigg) = 135 ⇒ b = \frac{18}{\sqrtπ}\)

From this l = \(\frac{15\sqrtπ}{2}\)

∴ Required perimeter:

\(2(l+b)=2\bigg[\frac{15\sqrtπ}{2}+\frac{18}{\sqrtπ}\bigg]=3\sqrtπ\bigg[\frac{5}{2}+\frac{6}{π}\bigg]\)

Answer 2

Let ABCD be the rectangle whose length is 2l and its width is 2b. 
Area of the painted zone (the circle) = \(𝜋 𝑏^ 2 \)\(\pi b^2\)

We have \(4lb - \pi b^2 = \frac{2}{3}\pi b^2\)\(......(1)\)
Solving we get \(l=\frac{5\pi}{12}b\)

Given, \(4lb=135\)
Equating in (1) we get \(4 \times \frac{5\pi}{12}b^2 = 135\)
\(⇒l = 15\sqrt{\frac{\pi}{2}}\)

Perimeter of rectangle\( = 2(l + b)\)
= \(2\left(15\frac{\sqrt{\pi}}{2} + \frac{18}{\sqrt{\pi}}\right)\)
= \(3\sqrt{\pi} \left(5 + \frac{12}{\pi}\right)\)