3. A circle is inscribed in trapezoid $A B C D$, touching the lateral side $A D$ at point $K$. Find the area of the trapezoid if $A K=16, D K=4$ and $C D=6$.
Answer: 432.

Solution. Let $L, M, N$ be the points of tangency of the inscribed circle with the sides $BC, AB, CD$ respectively; let $I$ be the center of the inscribed circle. Denote the radius of the circle by $r$. Immediately note that $DN = DK = 4$ (the first equality follows from the equality of the segments of tangents), hence $CL = CN = CD - DN = 2$ (the second is obvious). Since $I$ is the point of intersection of the bisectors of the internal angles of the trapezoid, then $\angle IAD + \angle IDA = \frac{\angle DAB + \angle DAC}{2} = \frac{180^{\circ}}{2} = 90^{\circ}$, where the penultimate equality follows from the parallelism of the lines $AB$ and $CD$. Therefore, triangle $AID$ is right-angled and $\angle AID = 90^{\circ}$. Similarly, triangle $BIC$ is also right-angled. Further, since $IK$ and $IL$ are radii drawn to the points of tangency, then $\angle IKD = 90^{\circ}$ and $\angle ILB = 90^{\circ}$. Therefore, $IK$ and $IL$ are altitudes in triangles $AID$ and $BIC$ respectively.

We use the known fact that in a right-angled triangle, the square of the altitude dropped to the hypotenuse equals the product of the segments into which it divides the hypotenuse. Then $IK^2 = AK \cdot KD = 16 \cdot 4 = 64 = 8^2$, i.e., $r = IK = 8$, and also $8^2 = IL^2 = CL \cdot LB = 2 \cdot LB$, i.e., $LB = 32$. Now we have everything to find the area. Note that $MN$ is the height of the trapezoid and $MN = 2r = 16$, $AB + CD = (AM + MB) + 6 = (AK + BL) + 6 = 16 + 32 + 6 = 54$, hence the answer $\frac{16 \cdot 54}{2} = 432$.

Comment. A complete and justified solution - 7 points. A generally correct reasoning containing minor gaps or inaccuracies - up to 5 points. If the solution is not completed, 1-2 points are given for proving useful auxiliary statements.