The base of the pyramid is an equilateral triangle with side $a$. The height of the pyramid passes through the midpoint of one of the sides of the base and is equal to $\frac{3 a}{2}$. Find the radius of the sphere circumscribed around the pyramid.
Let $O$ be the center of the sphere circumscribed around the given pyramid $ABCD$ with height $DK = \frac{3a}{2}$, where $K$ is the midpoint of $BC$, and $R$ is the radius of the sphere. Let $M$ be the center of the equilateral triangle $ABC$. Then the point $O$ lies on the perpendicular to the plane $ABC$ passing through the point $M$. Denote $OM = x$. From the right triangle $OMA$, we find that

$$
R^2 = AM^2 + OM^2 = \left(\frac{a}{\sqrt{3}}\right)^2 + x^2.
$$

The lines $OM$ and $DK$ are perpendicular to the plane $ABC$, so they are parallel and therefore lie in the same plane. In this plane, consider the right trapezoid $ODKM$, where

$$
DK = \frac{3a}{2}, \quad OD = R, \quad OM = x, \quad KM = \frac{1}{3}AK = \frac{a\sqrt{3}}{6}.
$$

Drop a perpendicular $ON$ to $KD$. In the right triangle $DON$,

$$
\begin{aligned}
R^2 = OD^2 &= ON^2 + DN^2 = KM^2 + (DK - KN)^2 = \\
&= \left(\frac{a\sqrt{3}}{6}\right)^2 + \left(\frac{3a}{2} - x\right)^2.
\end{aligned}
$$

Thus, we have the equation

$$
\left(\frac{a}{\sqrt{3}}\right)^2 + x^2 = \left(\frac{a\sqrt{3}}{6}\right)^2 + \left(\frac{3a}{2} - x\right)^2
$$

from which $x = \frac{2a}{3}$. Therefore,

$$
R = \sqrt{\left(\frac{a}{\sqrt{3}}\right)^2 + x^2} = \sqrt{\frac{a^2}{3} + \frac{4a^2}{9}} = \frac{a\sqrt{7}}{3}
$$

## Answer

$\frac{a\sqrt{7}}{3}$.