Problem 90. Find the area of trapezoid $ABCD$ (Fig. 91).

Solution 1. The area of a trapezoid is equal to the product of the height of the trapezoid and its midline. The height of the given trapezoid is equal to the diameter of the inscribed circle (see Problem 69), i.e., $\frac{4 R_{1} R_{2}}{R_{1}+R_{2}}$. The midline $MN$ of the given trapezoid, according to Archimedes' problem, is $2 \sqrt{R_{1} R_{2}}$. Therefore, the area of the given trapezoid $ABCD$ is

$$
\frac{8 R_{1} R_{2} \sqrt{R_{1} R_{2}}}{R_{1}+R_{2}}
$$
Solution 2. It is obvious that the area of rectangle $C N E K$ (see Fig. 90) is half the area of triangle $A E B$, and the area of triangle $A E B$ (see Fig. 91) is half the area of trapezoid $A B P K$ (show this on your own!). It is further clear that the area of trapezoid $A B P K$ is half the area of the desired trapezoid $A B C D$. Thus, $S_{A B C D}=8 S_{C N E K}$. Taking into account the result of problem 88, we find that

$$
S_{A B C D} \frac{8 R_{1} R_{2} \sqrt{R_{1} R_{2}}}{R_{1}+R_{2}}
$$