Answer: $\frac{5 \sqrt{2-\sqrt{3}}}{2}\left(5 \sin 15^{\circ}\right)$.

## Solution:

Let $M L-$ be the larger base of the trapezoid $M N K L$.

Consider the triangle $M N L: \angle M N L=90^{\circ}, Q$ - the midpoint of $M L$ (by condition) $\Rightarrow$ $Q$ is the midpoint of the hypotenuse $M L \Rightarrow N Q=M Q=Q L \Rightarrow M L=10$, since $N Q=5$ (by condition).

Let point $O$ be the intersection of the diagonals, and point $H$ be the foot of the perpendicular dropped from $K$ to the base $M L$, then $K H$ is the desired height.

Consider the triangle $M O L: \quad \angle K O L=15^{\circ}$ - the external angle in the isosceles triangle $M O L \Rightarrow \angle O M L=\angle O L M=7.5^{\circ}$.

Consider the triangle $M K L: \angle M K L=90^{\circ}, M K=M L \cdot \cos 7.5^{\circ}$.

Consider the triangle $M K H: \angle M H K=90^{\circ}$, then the desired height

$K H=M K \cdot \sin 7.5^{\circ}=M L \cos 7.5^{\circ} \sin 7.5^{\circ}=5 \sin 15^{\circ}=5 \sqrt{\frac{1-\cos 30^{\circ}}{2}}=\frac{5 \sqrt{2-\sqrt{3}}}{2}$.

Grading criteria are provided in the table:

| Points | Grading Criteria |
| :---: | :--- |
| $\mathbf{7}$ | Complete and justified solution. |
| $\mathbf{6}$ | Justified solution with minor flaws. |
| $\mathbf{5 - 6}$ | The solution contains minor errors, gaps in justification, but is generally correct and can become fully correct after minor corrections or additions. |
| $\mathbf{4}$ | The problem is more solved than not, for example, one of the two (more complex) significant cases is correctly considered. |
| $\mathbf{2 - 3}$ | The problem is not solved, but formulas, diagrams, considerations, or auxiliary statements relevant to the solution of the problem are provided. |
| $\mathbf{1}$ | The problem is not solved, but an attempt at a solution has been made, for example, individual (partial) cases have been considered in the absence of a solution or in the case of an incorrect solution. |
| $\mathbf{0}$ | The solution is absent, or the solution does not meet any of the criteria listed above. |
5. In an isosceles trapezoid $M N K L$ with bases $M L, N K$, the diagonals are perpendicular to the sides $M N$, $K L$ and intersect at an angle of $15^{\circ}$. Find the height of the trapezoid if the length $N Q=5$, where $Q-$ is the midpoint of the larger base.