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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?

  1. 1096
  2. 1111
  3. 1243
  4. 2486
in Numerical Ability by Veteran
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Migrated from GO Mechanical 11 months ago by Arjun

2 Answers

+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
by Veteran
+1 vote

Hence C is Correct :)

by Loyal
Answer:

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