8,9 The base $M Q$ of the trapezoid $M N P Q (M Q \| N P, M Q > N P)$ is the diameter of a circle that touches the line $M N$ at point $M$ and intersects the side $P Q$ at point $K$, with $P Q = 4 \sqrt{3} K Q$. The radius of the circle is $R$, and $\angle N Q M = 60^{\circ}$. Find the area of the trapezoid.
$\angle K M Q=15^{\circ}$.

## Solution

Let $K Q=x, P Q=4 x \sqrt{3}$. Suppose $\angle K M Q=\alpha, F$ is the projection of point $P$ on $M Q$. Then $\angle F P Q=\angle K M Q=\alpha$.

In the right triangle $P F Q$, it is known that

$$
\cos \alpha=\frac{P F}{P Q}=\frac{M N}{P Q}=\frac{2 R \sqrt{3}}{4 x \sqrt{3}}=\frac{R}{2 x}
$$

Since $\sin ^{2} \alpha+\cos ^{2} \alpha=1$, then

$$
\frac{x^{2}}{4 R^{2}}+\frac{R^{2}}{4 x^{2}}=1
$$

From this, we find that

$$
\sin \alpha=\frac{x}{2 R}=\sqrt{2}-\frac{\sqrt{3}}{4}
$$

Therefore, $\alpha=15^{\circ}$. Then

$$
\begin{gathered}
N P=M F=M Q-F Q=2 R-\frac{R(3+\sqrt{3})}{\sqrt{2}}=4 R(2-\sqrt{3}) \\
S_{\mathrm{MNPQ}}=\frac{1}{2}(M Q+N P) \cdot M N=2 R^{2}(5 \sqrt{3}-6)
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-18.jpg?height=440&width=834&top_left_y=1455&top_left_x=612)

Answer

$2 R^{2}(5 \sqrt{3}-6)$