### Instruction: In the diagram, two pairs of identical isosceles triangles are cut off of square $ABCD$, leaving rectangle $PQRS$. The total area cut off is $200 \text{ m}^2$. What is the length of $PR$, in meters? [asy]
size(5cm);

pair a = (0, 1); pair b = (1, 1); pair c = (1, 0); pair d = (0, 0);
pair s = (0, 0.333); pair p = (0.667, 1); pair q = (1, 0.667); pair r = (0.333, 0);

// Thicken pen
defaultpen(linewidth(1));

// Fill triangles
path tri1 = a--p--s--cycle;
path tri2 = p--q--b--cycle;
path tri3 = q--c--r--cycle;
path tri4 = s--r--d--cycle;
fill(tri1, gray(0.75));fill(tri2, gray(0.75));
fill(tri3, gray(0.75));fill(tri4, gray(0.75));

// Draw rectangles
draw(a--b--c--d--cycle); draw(p--q--r--s--cycle);

// Labels
label("$A$", a, NW); label("$B$", b, NE); label("$C$", c, SE); label("$D$", d, SW);
label("$P$", p, N); label("$Q$", q, E); label("$R$", r, S); label("$S$", s, W);

[/asy] Let's write a Python program to solve it.### Response: import math

# define the total area cut off
total_area_cut_off = 200  # in m^2

# the total area cut off is composed of 4 isosceles triangles
# so the area of one triangle is total_area_cut_off divided by 4
area_of_one_triangle = total_area_cut_off / 4  

# for an isosceles right triangle, the area is half the square of its side
# so the side of one triangle is the square root of (2 * area_of_one_triangle)
side_of_one_triangle = math.sqrt(2 * area_of_one_triangle)

# the length of PR is 2 times the side of one triangle
PR = 2 * side_of_one_triangle

# print the result
print(PR)