Problem 4. A circle with radius 2 is inscribed in trapezoid $ABCD$, touching the base $CD$ at point $N$. Find the area of the trapezoid if $DN=1$ and $AB=12$.
Answer: 27.

Solution. Let $K, L, M$ be the points of tangency of the inscribed circle with the sides $AD, BC, AB$ respectively; let $I$ be the center of the inscribed circle. Immediately note that $DK = DN = 1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_08868d1266709e60d590g-20.jpg?height=371&width=968&top_left_y=1665&top_left_x=541)

Since $I$ is the intersection of the angle bisectors of the internal angles of the trapezoid, $\angle IAD + \angle IDA = (\angle DAB + \angle ADC) / 2 = 180^\circ / 2 = 90^\circ$, 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, $\angle IKD = 90^\circ$ 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

$$
4 = IK^2 = AK \cdot KD = 1 \cdot AK
$$

i.e., $AK = 4$. By the equality of tangent segments, we have $AM = AK = 4$, hence $BL = BM = AB - AM = 12 - 4 = 8$. In the right triangle $BIC$, we get $4 = IL^2 = CL \cdot LB = 8 \cdot CL$, i.e., $CL = 0.5$.

Now we have everything to find the area. Note that $LM$ is the height of the trapezoid and $LM = 2r = 4, AB + CD = AB + DN + CN = 12 + 1 + 0.5 = 13.5$, hence the answer $\frac{4 \cdot 13.5}{2} = 27$.