8,9 |  |

In parallelogram $A B C D$, the diagonals $A C=15, B D=9$ are known. The radius of the circle circumscribed around triangle $A D C$ is 10. Find the radius of the circle circumscribed around triangle $A B D$.
Apply the formula $a=2 R \sin \boldsymbol{\alpha}$.

## Solution

Let $R=10$ be the radius of the circumcircle of triangle $A D C$. Then

$$
\sin \angle A D C=\frac{A C}{2 R}=\frac{15}{20}=\frac{3}{4}
$$

Since $\angle B A D=180^{\circ}-\angle A D C$, then

$$
\sin \angle B A D=\sin \angle A D C=\frac{3}{4}
$$

If $R_{1}$ is the radius of the circumcircle of triangle $A B D$, then

$$
R_{1}=\frac{B D}{2 \sin \angle B A D}=\frac{B D}{2 \sin \angle A D C}=\frac{9 \cdot 2}{3}=6
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5ddd4eab4fe1a12edeb9g-17.jpg?height=489&width=732&top_left_y=1719&top_left_x=663)

## Answer

6.