Problem 4. A circle is inscribed in trapezoid $ABCD$, touching the lateral side $BC$ at point $L$. Find the area of the trapezoid if $BL=4, CL=\frac{1}{4}$, and $AB=6$.
Answer: $6.75$.

Solution. Let $M, N, K$ be the points of tangency of the inscribed circle with the sides $AB, CD, AD$ respectively; let $I$ be the center of the inscribed circle. Denote the radius of the circle by $r$. Immediately note that $BM = BL = 4$ (the first equality follows from the equality of tangent segments), from which $AK = AM = AB - BM = 2$ (the first equality follows from the equality of tangent segments, the second is obvious).

![](https://cdn.mathpix.com/cropped/2024_05_06_08868d1266709e60d590g-14.jpg?height=372&width=985&top_left_y=1750&top_left_x=541)

Since $I$ is the point of intersection 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

$$
IL^2 = CL \cdot LB = \frac{1}{4} \cdot 4 = 1 = 1^2
$$

i.e., $r = IL = 1$, and also $1 = IK^2 = AK \cdot KD = 2 \cdot KD$, i.e., $KD = \frac{1}{2}$.

Now we have everything to find the area. Note that $MN$ is the height of the trapezoid and $MN = 2r = 2, AB + CD = 6 + CN + ND = 6 + \frac{1}{4} + \frac{1}{2} = 6.75$, from which the answer $\frac{2 \cdot 6.75}{2} = 6.75$.