10,11

Find the volume of a regular hexagonal pyramid with a side length of the base $a$ and the radius $R$ of the circumscribed sphere.

#
Let $O$ be the center of the sphere of radius $R$, circumscribed around a regular hexagonal pyramid PABCDEF with vertex $P$, and $M$ be the center of the base $A B C D E F$ (Fig.1). Then the point $O$ lies on the line $P M$. Denote $P M = h$. Consider the section of the pyramid and the sphere by a plane passing through points $A, D$, and $P$ (Fig.2). We obtain an isosceles triangle $A P D$ inscribed in a circle of radius $R$ with center $O$. Extend the segment $P M$ beyond point $M$ to intersect the circle at point $Q$. Since $A M$ is the altitude of the right triangle $P A Q$ drawn from the vertex of the right angle, we have

$$
A M^2 = P M \cdot M Q \text{, or } a^2 = h(2R - h)
$$

From the obtained equation, we find that

$$
h = R \pm \sqrt{R^2 - a^2}
$$

Therefore,

$$
V_{P A B C D E F} = \frac{1}{3} S_{A B C D E F} \cdot P M = \frac{1}{3} \cdot \frac{3a^2 \sqrt{3}}{2} \cdot h =
$$

$$
= \frac{1}{2} a^2 \sqrt{3} \left(R \pm \sqrt{R^2 - a^2}\right)
$$

## Answer

$\frac{1}{2} a^2 \sqrt{3} \left(R \pm \sqrt{R^2 - a^2}\right)$.