3. An isosceles trapezoid $A B C D$ with bases $A B, C D$ can have a circle inscribed in it with center $O$. Determine the area $S$ of the trapezoid, given the lengths of segments $O B$ and $O C$.
3. Let's denote the points of tangency of the inscribed circle with the sides $AB, BC, CD, DA$ as $K, L, M, N$ respectively (Fig. 2). Since $ABCD$ is an isosceles trapezoid, the internal angles at vertices $A, B, C, D$ are $\alpha, \alpha, 180^{\circ}-\alpha$, and $180^{\circ}-\alpha$ respectively. The segments $OA, OB, OC, OD$ lying on the axes of these angles, together with the four equal segments $OK, OL, OM, ON$, divide the entire trapezoid into eight right triangles, which are congruent in one leg and have acute internal angles $\frac{1}{2} \alpha$ and $90^{\circ}-\frac{1}{2} \alpha$. These eight triangles can thus be divided into two sets of four congruent triangles: one set consists of triangles $OAK, OAN, OBK, OBL$ and the other set consists of triangles $OCL, OCM, ODM, ODN$. From this, it follows that the area $S$ of trapezoid $ABCD$ is equal to four times the sum of the areas of triangles $OBL$ and $OCL$, i.e., four times the area of triangle $OBC$. According to the internal angles at vertices $B$ and $C$, we see that triangle $OBC$ is a right triangle with legs $OB$ and $OC$, so its area is $\frac{1}{2}|OB| \cdot|OC|$ and the desired total area $S$ is thus $S=2|OB| \cdot|OC|$.

Note. A small variation of part of the previous procedure: if $O$ is the center of the circle inscribed in the tangential quadrilateral $ABCD$, it is easy to show that its area is equal to twice the sum of the areas of triangles $OAB$ and $OCD$ as well as triangles $OBC$ and $ODA$. The latter two triangles are congruent in our isosceles trapezoid $ABCD$.

Another solution. For the height $v$ and sides $a, b, c, d$ of trapezoid $ABCD$ with the inscribed circle $k(O, r)$, the equations $v=2r$ and $a+c=b+d$ hold. From the first equation, it follows that the center $O$ lies on the midline of the trapezoid, whose length $\frac{1}{2}(a+c)$ is equal to $\frac{1}{2}(b+d)$ according to the second equation. In our case, however, $b=d$, so the midline is congruent to both legs and point $O$ is its midpoint, as the isosceles trapezoid is axially symmetric. Together, we get that point $O$ lies on the circle constructed over the diameter $BC$, and therefore $OBC$ is a right triangle with area $\frac{1}{2}|OB| \cdot|OC|$. The height from the hypotenuse $BC$ is the radius $r$ of the inscribed circle $k$, so the area of triangle $OBC$ is also equal to $\frac{1}{2} b \cdot r$. By comparing both expressions, we get the equality $|OB| \cdot|OC|=b \cdot r$. For the desired area $S$ of our trapezoid, we thus have

$$
S=\frac{a+c}{2} \cdot v=b \cdot 2r=2 \cdot|OB| \cdot|OC|.
$$

For a complete solution, award 6 points, of which 3 points for justifying that the given trapezoid is composed of two sets of four congruent triangles, or for an equality of the type $S_{OAB}+S_{OCD}=S_{OBC}+S_{ODA}$. Award 4 points for the deeper insight $S=4 \cdot S_{OBC}$. Award 1 point for recognizing that $OBC$ is a right triangle, as well as for the approach where the solver only divides the given trapezoid into four pairs of congruent triangles and does not make further progress in their area considerations.