The base of the pyramid is an equilateral triangle with a side length of 6. One of the lateral edges is perpendicular to

the base plane and equals 4. Find the radius of the sphere circumscribed around the pyramid.

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Let $A B C D$ be a given triangular pyramid, the base of which is an equilateral triangle $A B C$, and the lateral edge $C D$ is perpendicular to the plane of the base. $O$ is the center of the sphere circumscribed around the pyramid $A B C D$, and $R$ is the radius of the sphere. Since point $O$ is equidistant from points $A, B$, and $C$, it lies on the line perpendicular to the plane of triangle $A B C$ passing through the center $M$ of the circumscribed circle of the triangle. On the other hand, point $O$ is equidistant from points $C$ and $D$, so it lies in the plane perpendicular to edge $C D$ and passing through its midpoint $P$. Since planes perpendicular to the same line are parallel, the quadrilateral $M O P C$ is a rectangle. We then find:

$$
\begin{gathered}
O P=M C=A B^{\frac{\sqrt{3}}{3}}=2 \sqrt{3}, C P=\frac{1}{2} C D=2, \\
R=O C=\sqrt{O P^{2}+C P^{2}}=\sqrt{12+4}=4
\end{gathered}
$$

## Answer

4.00