GATE Mechanical 2018 Set 2 | GA | Question: 7

Question

A wire would enclose an area of $1936 \: m^2$, if it is bent to a square. The wire is cut into two pieces. The longer piece is thrice as long as the shorter piece. The long and the short pieces are bent into a square and a circle, respectively. Which of the following choices is closest to the sum of the areas enclosed by the two pieces in square meters?

• A. $1096$
• B. $1111$
• C. $1243$
• D. $2486$

Answer 1

$1936\; m^2 = a^2 \implies a = \sqrt {1936 } = 44\;m$

Total length of wire $=$ Perimeter $=4a$

$4 \times 44\; m$ is split to two parts of lengths $132\;m $ and $44\;m$ respectively.

$132\;m$ is bent into a square and $44\;m $ to a circle.

Side of the square $ = \frac{132}{4} = 33\;m.$

Radius of the circle $ = \frac{44}{2\pi} =\frac{22}{\pi} \approx 7$

Sum of the area enclosed by both $ = 33^2 + \pi 7^2 \approx1089 + 154 = 1243 .$

Correct Option: C

Answer 2

Hence C is Correct :)

Answer 3

Given, area of square=1936sq.metres.

so, a=44metres

Total length of wire=4a=4*44=176m.

Since, it is divided into two different lengths of y and 3y, so y+3y=176m

y=44m and 3y=132.

area of square with longer piece with side(132/4=33m) is 33*33=1089 sq.m.

area of circle with shorter piece with radius (4/3.14=7(approx)) is =154sq.m.

total area= 1089+154=1243 sq.m.