At the base of the pyramid lies a square with side $a=\sqrt{21}$. The height of the pyramid passes through the midpoint of one of the base edges and is equal to $\frac{a \sqrt{3}}{2}$. Find the radius of the sphere circumscribed around the pyramid.
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, and $R$ 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 $O1$ of the circumscribed circle of the triangle $A P B$. Let $r$ be the radius of this circle. Since

$$
\operatorname{tg} \angle P A B=\frac{P M}{A M M}=\frac{\frac{\sigma_{3}}{\frac{1}{2}}}{\frac{1}{2}}=\sqrt{3}
$$

then $\angle P B A=\angle P A B=60^{\circ}$, which means that triangle $A P B$ is equilateral, so

$$
r=O 1 P=\frac{a \sqrt{3}}{3}, O 1 M=\frac{0 \sqrt{3}}{6}
$$

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 $B Q O$, we find that

$$
R=O B=\sqrt{B Q^{2}+O Q^{2}}=\sqrt{\frac{\omega^{2}}{2}+\frac{\omega^{2}}{12}}=\frac{a \sqrt{21}}{6}=\sqrt{21} \cdot \frac{\sqrt{21}}{6}=\frac{7}{2}
$$

## Answer

3.5 .