Problem 4. A circle with radius 4 is inscribed in trapezoid $ABCD$, touching the base $AB$ at point $M$. Find the area of the trapezoid if $BM=16$ and $CD=3$.
Answer: 108.

Solution. Let $K, L, N$ be the points of tangency of the inscribed circle with the sides $AD, BC, CD$ respectively; let $I$ be the center of the inscribed circle. Immediately note that $BL = BM = 16$.

![](https://cdn.mathpix.com/cropped/2024_05_06_08868d1266709e60d590g-08.jpg?height=369&width=970&top_left_y=1666&top_left_x=543)

Since $I$ is the intersection point of the angle bisectors of the internal angles of the trapezoid, then $\angle IAD + \angle IDA = (\angle DAB + \angle ADC) / 2 = 180^\circ / 2 = 90^\circ$, where the penultimate equality follows from the parallelism of lines $AB$ and $CD$. Therefore, triangle $AID$ is a right triangle and $\angle AID = 90^\circ$. Similarly, triangle $BIC$ is also a right triangle.

Next, 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. Using the known fact that in a right triangle, the square of the altitude dropped to the hypotenuse equals the product of the segments into which it divides the hypotenuse. Then

$$
4^2 = IL^2 = CL \cdot LB = CL \cdot 16
$$

i.e., $CL = 1$. By the equality of tangent segments, we have $CN = CL = 1$, hence $DK = DN = CD - CN = 3 - 1 = 2$. In the right triangle $AID$, we get $4^2 = IK^2 = AK \cdot KD = AK \cdot 2$, i.e., $AK = 8$.

Now we have everything to find the area. Note that $LM$ is the height of the trapezoid and $LM = 2r = 8, AB + CD = AM + MB + CD = 8 + 16 + 3 = 27$, hence the answer $\frac{8 \cdot 27}{2} = 108$.