3 [ Perpendicularity of a line and a plane (other).]

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 $Q$ be the center of the base $ABC$ of the triangular pyramid $ABCD$, where the lateral edge $AD$ is perpendicular to the plane of the base, and $AB = BC = AC = a = 6$, $AD = h = 4$. The center $O$ of the sphere with radius $R$, circumscribed around the pyramid $ABCD$, lies on the line perpendicular to the plane of the base $ABC$ passing through the point $Q$, and also in the plane perpendicular to the line $AD$ and passing through the midpoint $P$ of the segment $AD$. The lines $AP$ and $OQ$ are parallel (since they are perpendicular to the same plane), and the quadrilateral $APOQ$ is a rectangle. In the right triangle $AOQ$, it is known that

$$
AQ = \frac{a \sqrt{3}}{3}, OQ = AP = \frac{h}{2}
$$

Therefore,

$$
R = OA = \sqrt{AQ^2 + OQ^2} = \sqrt{\left(\frac{a \sqrt{3}}{3}\right)^2 + \left(\frac{h}{2}\right)^2} = \sqrt{12 + 4} = 4
$$

## Answer

4.00