2. (15 points) The area of an isosceles trapezoid is 100, and its diagonals are perpendicular to each other. Find the height of this trapezoid.
Solution. Let $ABCD$ be an isosceles trapezoid, the diagonals $AC$ and $BD$ of which are perpendicular to each other. Denote $O = AC \cap BD$, point $M$ is the midpoint of $BC$, and point $H$ is the midpoint of $AD$. Draw the segment $MH: O \in MH, MH \perp AD$. Since $MH$ is the axis of symmetry of the given trapezoid, then $OB = OC, OA = OD$. Therefore, the right triangles $BOC$ and $AOD$ are isosceles, meaning $\angle OBM = \angle OAH = 45^\circ$. Then the right triangles $MOB$ and $AOH$ are also isosceles: $OM = BM, OH = AH$.

![](https://cdn.mathpix.com/cropped/2024_05_06_aadf511cc872419fdfecg-07.jpg?height=340&width=457&top_left_y=201&top_left_x=1416)

We have:

$OM = BM = 0.5 BC, OH = AH = 0.5 AD \Rightarrow MH = OM + OH = 0.5 BC + 0.5 AD = 0.5(BC + AD)$, which means the height $MH$ of the trapezoid is equal to its midline. This means that for the area of the given trapezoid we get: $S_{ABCD} = 0.5(BC + AD) MH = MH \cdot MH = MH^2$. Then $MH = \sqrt{S_{ABCD}} = \sqrt{100} = 10$.

Answer: 10.

| Criterion Content | Points |
| ---: | :---: |
| The correct answer is obtained with justification. | 15 |
| The solution contains a computational error, possibly leading to an incorrect answer, but the solution has a correct sequence of all steps. | 10 |
| In the solution of the problem, one or two formulas are correctly written, which are the beginning of a possible solution. | 5 |
| The solution does not meet any of the criteria described above. | 0 |
| Maximum score | 15 |