5. Parallelogram $A B C D$ can be divided into four equilateral triangles with a side length

(8) of $2 \mathrm{~cm}$. What is the length of the longer diagonal of the parallelogram?
Solution. Let the foot of the perpendicular from vertex $C$ to line $AB$ be denoted as $E$.

Triangle $BCE$ is half of an equilateral triangle, so we have $\angle CBE = 60^\circ$, $|CB| = 2$ cm, $|CE| = \sqrt{3}$ cm, and $|BE| = 1$ cm.

![](https://cdn.mathpix.com/cropped/2024_05_30_fac652c8a845a57c6fa7g-02.jpg?height=319&width=711&top_left_y=1688&top_left_x=684)

By the Pythagorean theorem on $\triangle ACE$, we get: $|AC|^2 = |AE|^2 + |EC|^2$, \quad(*) i.e., $|AC| = \sqrt{5^2 + (\sqrt{3})^2} = \sqrt{28}$ (or $2\sqrt{7}$).

(full solution 8 points)

(applying the Pythagorean theorem $(*)$ 4 points)