Alright, there's a way to figure this out. I don't know the answer, but I have some ideas of how to get there. If all else fails, we can employ some brute force (there are only so many combinations, right?)... Given how all of the figures in the problem are expressed only to the tenths digit, I'm inclined to think that
there is some rounding involved and that the exact geometric shapes as specified will not actually fill a square of any dimension. But it might get close enough for government work, so let's take a look.
I think it will be helpful to determine the dimension of the square. When arranging triangles to fill that square, the edges of the triangles must line up with the edges of the square. The area of the square is the same as the sum of the areas of all the triangles, meaning its side length will be the square root of whatever that turns out to be.
Triangles have three legs and three angles (duh). If we know any three of those numbers (except the three angles), we can calculate the other three. We can also craft some tailor-suited formulas for determining the areas of the given triangles.
Side-Side-Side
When all three side lengths \(\displaystyle a, b, c\) of a triangle are known,
Heron's formula gives us the area:
\(\displaystyle s = \frac{a + b + c}{2}\)
\(\displaystyle T = \sqrt{s(s-a)(s-b)(s-c)}\)
The following triangles are given in the problem as SSS:
14.5, 13.6, 13.6 \(\displaystyle = \frac{29\sqrt{52959}}{80} \approx 83.4215\)
8.9, 15.2, 12.7 \(\displaystyle = \frac{76\sqrt{\frac{69}{5}}}{5} \approx 56.4655\)
Side-Angle-Side
Given two side lengths \(\displaystyle a, b\) and their included angle \(\displaystyle C\), the area of the triangle can be found by applying the
law of sines in the following way:
\(\displaystyle T = \frac{a * b * sin(C)}{2}\)
The following triangles are given in the problem as SAS:
Angle-Side-Angle
This describes two angles \(\displaystyle A, C\) connected by a side \(\displaystyle b\). The law of sines helps us again here: first we determine the missing angle, then one of the missing sides, then use the Side-Angle-Side formula to calculate the area:
\(\displaystyle B = 180^{\circ} - A - C\)
\(\displaystyle a = \frac{b*sin(A)}{sin(B)}\)
\(\displaystyle T = \frac{a * b * sin(C)}{2} = \frac{b^2sin(A)sin(C)}{2sin(180^{\circ} - A - C)}\)
The following angles are given in the problem as ASA:
Total
The sum of the areas of these triangles is the "definitely not a natural number" \(\displaystyle \approx 309.3218\), making the side length of the square \(\displaystyle \approx 17.5875\). The original photograph depicts a triangle with a length of 17.3 that appears to be about the same as the square printed on the worksheet, so we might be in the same voting district as the ball park.
From here, we can test combinations of triangles where the side lengths add up to more-or-less 17.6cm. We also know that the angles of the triangles that meet in the corners of the square need to add up to somewhere in the neighborhood of 90°.