5.1. A paper rectangle $4 \times 8$ was folded along the diagonal as shown in the figure. What is the area of the triangle that is covered twice?

![](https://cdn.mathpix.com/cropped/2024_05_06_efe315bbc66e758f8537g-5.jpg?height=531&width=528&top_left_y=271&top_left_x=707)
Answer: 10.

Solution. By the Pythagorean theorem, $A C=\sqrt{A B^{2}+B C^{2}}=4 \sqrt{5}$. Triangle $A E C$ is isosceles, and if we drop the height $E H$ in it, then triangle $A E H$ will be similar to triangle $A C D$. Therefore, $A H / E H=A D / D C=2 . A H=2 \sqrt{5}$, since the height in an isosceles triangle is also a median. Thus, $E H=\sqrt{5}, S(A E C)=\sqrt{5} \cdot 4 \sqrt{5} / 2=10$.