A wire would enclose an area of 1936 m² , if it is bent into 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?

The correct answer is 1243.

First, we find the side length of the square that the wire would enclose if it were bent as a whole. Let $a$ be the side length of the square. Given that the area of the square is 1936 m², we have:

$a^2=1936$

Solving for $a$ by taking the square root:

$a=\sqrt{1936}=44\text{m}$

The total length of the wire is the perimeter of this square, which is:

$\text{Total length}=4\times a=4\times 44=176\text{m}$

The wire is cut into two pieces where the longer piece is thrice as long as the shorter piece. If we let the shorter piece be $s$ and the longer piece be $l$, we have:

$l=3s$

Since the sum of the two pieces is the total length of the wire:

$s+l=176$

$s+3s=176$

$4s=176$

$s=44\text{m}$

Therefore, the longer piece is:

$l=3\times 44=132\text{m}$

Next, the longer piece of length 132 m is bent into a square. The side length $a_{\text{new}}$ of this new square is:

$a_{\text{new}}=\frac{132}{4}$

The area of this square is:

${\text{Area}}_{\text{square}}={33}^2=1089{\text{m}}^2$

The shorter piece of length 44 m is bent into a circle. The circumference of this circle is 44 m, so the radius $r$ is given by:

$2\times \pi \times r=44$

$r=\frac{44}{2\times \pi }$

The area of this circle is:

${\text{Area}}_{\text{circle}}=\pi \times r^2\approx \pi \times 7^2\approx \frac{22}{7}\times 49=154{\text{m}}^2$

Finally, we sum the two areas:

$\text{Total Area}={\text{Area}}_{\text{square}}+{\text{Area}}_{\text{circle}}\approx 1089+154=1243{\text{m}}^2$

This matches the correct option of 1243.