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?

To find the sum of the areas enclosed by the two pieces of wire, we begin by determining the total length of the wire from the initial shape.

Let the side length of the initial square be s. The area of this square is given as 1936 m².
Using the area formula for a square:
$s^2=1936$
Taking the square root of both sides, we find the side length of the square:
$s=\sqrt{1936}=44\text{meters}$

The total length of the wire is equal to the perimeter of this square:
$L=4\times s=4\times 44=176\text{meters}$

Next, the wire is cut into two pieces, a longer piece and a shorter piece, where the longer piece is thrice as long as the shorter piece.
Let the length of the shorter piece be $x$. The length of the longer piece is $3x$.
The sum of their lengths equals the total length of the wire:
$x+3x=176$
$4x=176$
$x=44\text{meters}$

Thus, the lengths of the two pieces are:
- Shorter piece: $L_{\text{short}}=44\text{meters}$
- Longer piece: $L_{\text{long}}=3\times 44=132\text{meters}$

The longer piece is bent into a square.
The perimeter of this new square is 132 meters. Let the side length of this square be $s_2$:
$4s_2=132$
$s_2=33\text{meters}$
The area enclosed by the longer piece (square) is:
$A_{\text{square}}={33}^2=1089{\text{m}}^2$

The shorter piece is bent into a circle.
The circumference of this circle is 44 meters. Let the radius of the circle be $r$:
$2\pi r=44$
Using the approximation $\pi \approx \frac{22}{7}$:
$2\times \frac{22}{7}\times r=44$
$\frac{44}{7}\times r=44$
$r=7\text{meters}$

The area enclosed by the shorter piece (circle) is:
$A_{\text{circle}}=\pi r^2\approx \frac{22}{7}\times 7^2=22\times 7=154{\text{m}}^2$

Finally, we calculate the sum of the areas enclosed by the two pieces:
$A_{\text{total}}=A_{\text{square}}+A_{\text{circle}}=1089+154=1243{\text{m}}^2$

Therefore, the closest value to the sum of the areas enclosed by the two pieces is 1243.