1.65. 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, resulting in a trapezoid. Find the area of this trapezoid.
1.65. Let $L$ and $M$ be the points of tangency (Fig. 1.51); then $N L = N M$, from which $A B = C D$, since $A D \parallel B C \parallel L M$. Draw $O K \perp L M$ and $B H \perp A D$. Then 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, $S = A B \cdot B H$. Further, $\triangle A B H \sim \triangle O L K (\angle L O K = \angle B A H$ as angles with mutually perpendicular sides),

$$
\text{so } \frac{L K}{L O} = \frac{B H}{A B} \text{, or } \frac{0.5 a}{R} = \frac{2 R}{A B}
$$

and, therefore, $A B = \frac{4 R^{2}}{a}$. Thus, $S = \frac{4 R^{2}}{a} \cdot 2 R = \frac{8 R^{3}}{a}$.

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