8,9 In trapezoid $A B C D$, points $K$ and $M$ are the midpoints of the bases $A B$ and $C D$, respectively. It is known that $A M$ is perpendicular to $D K$ and $C K$ is perpendicular to $B M$, and the angle $C K D$ is $60^{\circ}$. Find the area of the trapezoid if its height is 1.
Through vertices $C$ and $D$, draw lines parallel to lines $M B$ and $M A$ respectively.

## Solution

Let $A K=K B=a, D M=M C=b$. Draw a line through point $C$, parallel to $B M$, until it intersects line $A B$ at point $P$, and draw a line through point $D$, parallel to $A M$, until it intersects line $A B$ at point $Q$. Then

$$
K Q=A K+A Q=A K+M D=a+b, K P=K B+B P=K B+M C=a+b .
$$

Right triangles $Q D K$ and $P C K$ are congruent by the hypotenuse and the altitude to the hypotenuse.

Assume that $\angle D Q K=\angle C K P$. Then $D Q \| C K$. Therefore, $A Q D M$ is a parallelogram, which means $K Q=C D$, i.e., $b + a = 2 b$. From this, it follows that $a=b$, i.e., $A B C D$ is a parallelogram, which contradicts the problem's condition ( $A B C D$ is a trapezoid).

Thus,

$$
\angle D K Q=\angle C K P=60^{\circ} .
$$

The area of each of these triangles is half the area of the trapezoid (since $Q K=A K+D M$ and $P K=K B+M C$). Therefore,

$$
S_{\triangle K C P}=\frac{1}{2} K P \cdot 1=\frac{1}{2} \cdot \frac{4}{\sqrt{3}}=\frac{2}{\sqrt{3}}
$$

$$
S_{\mathrm{ABCD}}=2 S_{\triangle \mathrm{KCP}}=\frac{4}{\sqrt{3}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-17.jpg?height=314&width=918&top_left_y=1924&top_left_x=570)

## Answer

$\frac{4 \sqrt{3}}{3}$.