[ Isosceles, Inscribed, and Circumscribed Trapezoids ]

The bases of an isosceles trapezoid are equal to $a$ and $b(a>b)$, and the lateral side is equal to $l$. Find the radius of the circle circumscribed around this trapezoid.
Let $A D=a$ and $B C=b-$ be the bases of the isosceles trapezoid $A B C D, C H-$ the height of the trapezoid. Then

$$
D H=\frac{1}{2}(A D-B C)=\frac{a-b}{2}, A H=\frac{1}{2}(A D+B C)=\frac{a+b}{2}
$$

Let $\angle C D A=\alpha$. From the right triangles $C D H$ and $A C H$, we find that

$$
\begin{gathered}
C H=\sqrt{C D^{2}-D H^{2}}=\sqrt{l^{2}-\left(\frac{a-b}{2}\right)^{2}}, \sin \alpha=\frac{C H}{C D}=\frac{\sqrt{l^{2}-\left(\frac{a-b}{2}\right)^{2}}}{l} \\
A C=\sqrt{C H^{2}+A H^{2}}=\sqrt{l^{2}-\left(\frac{a-b}{2}\right)^{2}+\left(\frac{a-b}{2}\right)^{2}}=\sqrt{l^{2}+a b}
\end{gathered}
$$

The circle of radius $R$, circumscribed around the trapezoid $A B C D$ is the circle circumscribed around the triangle $A C D$. By the Law of Sines,

$$
R=\frac{A C}{2 \sin \alpha}=\frac{\frac{\sqrt{l^{2}+a b}}{2 \sqrt{l^{2}-\left(\frac{a-b}{2}\right)^{2}}}}{1}=\frac{\sqrt{l^{2}+a b}}{\sqrt{4 l^{2}-(a-b)^{2}}}
$$

## Answer

$\frac{\sqrt{l^{2}+a b}}{\sqrt{4 l^{2}-(a-b)^{2}}}$