The parallel sides of an isosceles trapezoid are $a=10$ and $c=15$ units long, and the radius of the circumscribed circle is $r=10$. What could be the lengths of the legs? What is the area?
Solution. According to the given data, two trapezoids exist. One contains the center of the circle inside it. In the other case, the center of the circle is outside the trapezoid. Let's calculate the lengths of the legs and the area of the trapezoid in both cases.

![](https://cdn.mathpix.com/cropped/2024_05_02_9f17add37fa2ed6803ebg-1.jpg?height=403&width=415&top_left_y=281&top_left_x=844)

First, let's examine the first case. Denote the vertices of the trapezoid as $A, B, C, D$, and the center as $O$ as shown in the figure. Draw perpendiculars from the center to the bases of the trapezoid. Denote the distances from the bases as $m_{1}$ and $m_{2}$, with $O C=10$.

$$
\begin{aligned}
& m_{1}^{2}=10^{2}-\left(\frac{15}{2}\right)^{2}=100-\frac{225}{4}=\frac{175}{4}, m_{1}=\frac{5 \sqrt{7}}{2} \\
& m_{2}^{2}=10^{2}-5^{2}=75, \quad m_{2}=5 \sqrt{3}
\end{aligned}
$$

The height of the trapezoid:

$$
m_{1}+m_{2}=\frac{5 \sqrt{7}+10 \sqrt{3}}{2}
$$

The area of the trapezoid:

$$
\frac{10+15}{2} \cdot \frac{5 \sqrt{7}+10 \sqrt{3}}{2}=\frac{125 \sqrt{7}+250 \sqrt{3}}{4} \approx 190.93
$$

The length of the leg of the trapezoid is calculated using the Pythagorean theorem: $B C^{2}=\left(m_{1}+m_{2}\right)^{2}+B C^{\prime 2}$, where $C^{\prime}$ is the perpendicular projection of $C$ onto $A B$.

$$
\begin{aligned}
B C^{2} & =\left(\frac{5 \sqrt{7}+10 \sqrt{3}}{2}\right)^{2}+\left(\frac{5}{2}\right)^{2}=\frac{500+100 \sqrt{21}}{4} \approx 239.56 \\
B C & \approx 15.48
\end{aligned}
$$

In the second case, $m=m_{2}-m_{1}$.

$$
\begin{aligned}
& m=\frac{10 \sqrt{3}-5 \sqrt{7}}{2} \\
& T=\frac{10+15}{2} \cdot \frac{10 \sqrt{3}-5 \sqrt{7}}{2}=\frac{250 \sqrt{3}-125 \sqrt{7}}{4} \approx 25.57
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_9f17add37fa2ed6803ebg-1.jpg?height=407&width=415&top_left_y=1919&top_left_x=844)

The leg of the trapezoid:

$$
\begin{aligned}
D A^{2} & =\left(\frac{10 \sqrt{3}-5 \sqrt{7}}{2}\right)^{2}+\left(\frac{5}{2}\right)^{2}=\frac{500-100 \sqrt{21}}{4} \\
D A & =\sqrt{\frac{500-100 \sqrt{21}}{4}} \approx 3.23
\end{aligned}
$$