10.074. One of the two parallel lines is tangent to a circle of radius $R$ at point $A$, while the other intersects this circle at points $B$ and $C$. Express the area of triangle $A B C$ as a function of the distance $x$ between the lines.
## Solution.

Let the distance between the parallel lines be $x$ (Fig. 10.72); then the area $S$ of triangle $ABC$ is $0.5 BC \cdot x$. Since $AM \perp BC$, then $BM = MC$ and $BM \cdot MC = AM \cdot MD = x(2R - x)$. Therefore, $0.25 BC^2 = x(2R - x)$, from which $S = 0.5 x \cdot 2 \sqrt{2Rx - x^2} = x \sqrt{2Rx - x^2}$.

Answer: $x \sqrt{2Rx - x^2}$.