2.55. The base of the pyramid is an equilateral triangle with a side length of $a$. One of the lateral edges is perpendicular to the base plane and equals $b$. Find the radius of the sphere circumscribed around the pyramid.
2.55. Let $O$ be the center of the sphere circumscribed around the pyramid $ABCD$ (Fig. 2.45). Then $OA = OB = OC = OD$. Drop the perpendicular $OK$ to the plane $ABC$ and draw $OE \perp DB$. Since point $O$ is equidistant from the vertices of triangle $ABC$, point $K$ is the

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-484.jpg?height=341&width=343&top_left_y=1310&top_left_x=811)

Fig. 2.45

center of the triangle and $BK = \frac{a}{\sqrt{3}}$. Furthermore, since $OB = OD$, we have $EB = ED = \frac{b}{2}$. Therefore,

$$
OB = \sqrt{OK^2 + BK^2} = \sqrt{\frac{b^2}{4} + \frac{a^2}{3}} = \frac{\sqrt{12a^2 + 9b^2}}{6}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-485.jpg?height=525&width=486&top_left_y=351&top_left_x=53)

Fig. 2.46

Om. $m: \frac{\sqrt{12a^2 + 9b^2}}{6}$.