A company's logo must be created by superimposing an equilateral triangle and a square, as shown in the figure. If the side length of the triangle is $12 \mathrm{~cm}$ and the side of the triangle intersects the side of the square at its midpoint, what is the difference between the overlapping area (dark) and the sum of the non-overlapping areas (light)?

![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-17.jpg?height=478&width=553&top_left_y=1014&top_left_x=823)

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Solution

Let's call the side length of the square $2n$, then the four right triangles that are not overlapped (all congruent) have legs measuring $n$ and $(6-n)$, and the equilateral triangle not overlapped has a side measuring $(4n-12)$.

![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-17.jpg?height=450&width=516&top_left_y=1993&top_left_x=847)

Since the angles of the equilateral triangle measure $60^{\circ}$, then:

$$
\begin{aligned}
\frac{n}{6-n} & =\sqrt{3} \\
n & =6 \sqrt{3}-\sqrt{3} n \\
n(\sqrt{3}+1) & =6 \sqrt{3} \\
n & =\frac{6 \sqrt{3}}{\sqrt{3}+1} \\
n & =9-3 \sqrt{3} \text{ cm}
\end{aligned}
$$

Now let's calculate the areas:

$$
\begin{aligned}
A_{\text {dark }} & =(2 n)^{2}-n(6-n) \\
& =4 n^{2}+n^{2}-6 n \\
& =5 n^{2}-6 n \\
& =5(9-3 \sqrt{3})^{2}-6(9-3 \sqrt{3}) \\
& =5(81-54 \sqrt{3}+27)-54+18 \sqrt{3} \\
& =(486-252 \sqrt{3}) \text{ cm}^{2} \\
A_{\text {light }}= & 2 n(6-n)+\frac{(4 n-12)^{2} \sqrt{3}}{4} \\
= & 12 n-2 n^{2}+4 \sqrt{3}(n-3)^{2} \\
= & 12(9-3 \sqrt{3})-2(9-3 \sqrt{3})^{2}+4 \sqrt{3}(6-3 \sqrt{3})^{2} \\
= & 108-36 \sqrt{3}-2(81-54 \sqrt{3}+27)+4 \sqrt{3}(36-36 \sqrt{3}+27) \\
= & 108-36 \sqrt{3}-216+108 \sqrt{3}+252 \sqrt{3}-432 \\
= & 324 \sqrt{3}-540 \text{ cm}^{2}
\end{aligned}
$$

Therefore, the difference between the dark and light areas is $(1026-576 \sqrt{3}) \text{ cm}^{2}$.