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+1 vote

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?

- 1096
- 1111
- 1243
- 2486

Migrated from GO Mechanical 11 months ago by Arjun

+2 votes

Best answer

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

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

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

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