Problem 4. A circle is inscribed in trapezoid $ABCD$, touching the lateral side $AD$ at point $K$. Find the area of the trapezoid if $AK=16, DK=4$ and $CD=6$.
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), from which $CL = CN = CD - DN = 2$ (the first equality follows from the equality of the segments of tangents, the second is obvious).

![](https://cdn.mathpix.com/cropped/2024_05_06_08868d1266709e60d590g-02.jpg?height=368&width=985&top_left_y=1769&top_left_x=541)

Since $I$ is the point of intersection of the 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 the 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

$$
IK^2 = AK \cdot KD = 16 \cdot 4 = 64 = 8^2
$$

i.e., $r = IK = 8$, and also $8^2 = IL^2 = CL \cdot LB = 2 \cdot LB$, i.e., $LB = 32$.[^0]

Now we have everything to find the area. Note that $MN$ is the height of the trapezoid and $MN = 2r = 16, AB + CD = (AM + MB) + 6 = (AK + BL) + 6 = 16 + 32 + 6 = 54$, from which the answer $\frac{16 \cdot 54}{2} = 432$.