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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

edited | 109 views
Migrated from GO Mechanical 5 months ago by Arjun

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

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 (422k points)
+1 vote  Hence C is Correct :)

by Active (4.9k points)