13. In a trapezoid, the lengths of the diagonals are 6 and 8, and the length of the midline is 5. Find the height of the trapezoid.
Answer: $4.8$.

Solution: Parallel to the bases AD and $BC$, translate one of the diagonals BD by the length of BC. We obtain a triangle $ACE$ with sides $6, 8, 10$, which, by the converse of the Pythagorean theorem, is a right triangle. Writing the area in two ways: $S=\frac{1}{2} AC \times CE = \frac{1}{2} AE \times CH$, we find its height $\mathrm{CH}=4.8$.

![](https://cdn.mathpix.com/cropped/2024_05_06_63faea101299db5464bfg-02.jpg?height=354&width=742&top_left_y=2390&top_left_x=1094)

## Variant 3-b