The base of the pyramid is a square with side $a$, the height of the pyramid passes through the midpoint of one of the sides of the base and is equal to $\frac{a \sqrt{3}}{2}$. Find the radius of the circumscribed sphere.
Let $P A B C D$ be a quadrilateral pyramid, the base of which is a square $A B C D$ with side $a$, and the height $P M$ passes through the midpoint $M$ of edge $A B ; P M=\frac{a \sqrt{3}}{2} ; Q$ is the center of the square $A B C D ; O$ is the center of the circumscribed sphere, $R-$ is the radius of this sphere. The point $O$ is equidistant from the vertices of the square $A B C D$, so it lies on the perpendicular to the plane of the base of the pyramid passing through the center $Q$ of the square. On the other hand, the point $O$ is equidistant from the vertices of the triangle $A P B$, so it lies on the perpendicular to the plane $A P B$ passing through the center $O 1$ of the circumscribed circle of triangle $A P B$. Let $r$ be the radius of this circle. Since

$$
\operatorname{tg} \angle P A B=\frac{\Gamma P^{\prime} M}{A M}=\frac{\frac{a \sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}
$$

$\angle P B A=\angle P A B=60^{\circ}$, hence, triangle $A P B$ is equilateral, so

$$
r=O 1 P=\frac{a \sqrt{3}}{3}, O 1 M=\frac{a \sqrt{3}}{6}
$$

The lines $O Q$ and $P M$ are parallel, as they are perpendicular to the same plane. Similarly, $O O 1 \| Q M$. Therefore,

$$
O Q=O 1 M=\frac{a \sqrt{3}}{6}
$$

From the right triangle $A Q O$, we find that

$$
R=O A=\sqrt{A Q^{2}+O Q^{2}}=\sqrt{\frac{\alpha^{2}}{2}+\frac{\alpha^{2}}{12}}=\frac{a \sqrt{21}}{6}
$$

Answer

$\frac{a \sqrt{21}}{6}$