5. In an isosceles trapezoid $M N K L$ with bases $M L$, $N K$ the diagonals are perpendicular to the sides $M N, \quad K L$ and intersect at an angle of $22.5^{\circ}$. Find the height of the trapezoid if the length $N Q=3$, where $Q-$ is the midpoint of the larger base.
Answer: $\frac{3 \sqrt{2-\sqrt{2}}}{2}\left(3 \sin 22.5^{\circ}\right)$.

## Solution:

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

Consider the triangle $M N L: \angle M N L=90^{\circ}, Q$ is the midpoint of $M L$ (by condition) $\Rightarrow$ $Q$ is the midpoint of the hypotenuse $M L \Rightarrow N Q=M Q=Q L \Rightarrow M L=6$, since $N Q=3$ (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 $M O L: \quad \angle K O L=22.5^{\circ}$ - the external angle in the isosceles triangle $M O L \Rightarrow \angle O M L=\angle O L M=11.25^{\circ}$.

Consider the triangle $M K L: \angle M K L=90^{\circ}, M K=M L \cdot \cos 11.25^{\circ}$.

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

$K H=M K \cdot \sin 11.25^{\circ}=M L \cos 11.25^{\circ} \sin 11.25^{\circ}=3 \sin 22.5^{\circ}=3 \sqrt{\frac{1-\cos 45^{\circ}}{2}}=\frac{3 \sqrt{2-\sqrt{2}}}{2}$.

## Variant 2