10.293. A circle of radius $R$ is inscribed in some angle, and the length of the chord connecting the points of tangency is $a$. Two tangents parallel to this chord are drawn, as a result of which a trapezoid is formed. Find the area of this trapezoid.
## Solution.

Let $L$ and $M$ be the points of tangency (Fig. 10.86); then $S L = S M$, from which $A B = C D$, since $A D \parallel B C \parallel L M$. Draw $O K \perp L M$ and $B H \perp A D$. The required area $S = \frac{1}{2}(A D + B C) B H$. For the circumscribed trapezoid, we have $A D + B C = A B + C D = 2 A B$; therefore, $S = A B \cdot B H$. Furthermore, $\triangle A B H \sim \triangle O L K$ ( $\angle L O K = \angle B A H$ as angles with mutually perpendicular sides), from which $\frac{L K}{L O} = \frac{B H}{A B}$ or $\frac{a / 2}{R} = \frac{2 R}{A B}$ and $A B = \frac{4 R^{2}}{a}$.

We obtain $S = \frac{4 R^{2}}{a} \cdot 2 R = \frac{8 R^{3}}{a}$.

Answer: $\frac{8 R^{3}}{a}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0775.jpg?height=454&width=753&top_left_y=350&top_left_x=302)

Fig. 10.87