10.351. Let $B D$ be the height of triangle $A B C$, and point $E$ be the midpoint of $B C$. Calculate the radius of the circle circumscribed around triangle $B D E$, if $A B=30 \text{ cm}, B C=26 \text{ cm}$ and $A C=28 \text{ cm}$.
Solution.

Let's find the area of the given triangle using Heron's formula:

$S_{\triangle A B C}=336$ cm $^{2}$. Then $B D=\frac{2 S_{\triangle A B C}}{A C}=24$ cm (Fig. 10.133). $D E$ is the median of the right $\triangle B D C$, drawn to the hypotenuse $B C$. Then $D E=\frac{1}{2} B C=$ $=13$ cm, $\cos \angle C B D=\frac{B D}{B C}=\frac{12}{13}$. Then $\sin \angle C B D=\sqrt{1-\cos ^{2} \angle C B D}=\frac{5}{13}$.

The radius of the circumcircle of $\triangle B E D, R=\frac{D E}{2 \sin \angle C B D}=16.9$ cm.

Answer: 16.9 cm.