[ Regular Pyramid ]

Find the volume of a regular hexagonal pyramid with a lateral edge $b$ and the area $Q$ of a lateral face.

#
Let $M$ be the center of the base $A B C D E F$ of a regular hexagonal pyramid $P A B C D E F$ with lateral edges $P A = P B = P C = P D = P E = P F = b$, and $K$ be the midpoint of $A B$. Denote $A B = B C = C D = D E = E F = A F = a$. Then

$$
\begin{gathered}
A M = a, P K = \sqrt{A P^{2} - A K^{2}} = \sqrt{b^{2} - \frac{a^{2}}{4}}, \\
Q = S_{\triangle A P B} = \frac{1}{2} A B \cdot P K = \frac{1}{2} a \sqrt{b^{2} - \frac{a^{2}}{4}} = \frac{1}{2} a \sqrt{4 b^{2} - a^{2}}, \\
4 Q = a \sqrt{4 b^{2} - a^{2}}, 16 Q^2 = 4 a^2 b^2 - a^4, a^4 - 4 a^2 b^2 + 16 Q^2 = 0.
\end{gathered}
$$

From the obtained equation, we find that

$$
a^2 = 2 b^2 - \sqrt{4 b^{2} - 16 Q^{2}}
$$

(the second root does not satisfy the condition of the problem, since $b > a$). Therefore,

$$
\begin{gathered}
P M = \sqrt{A P^{2} - A M^{2}} = \sqrt{b^{2} - a^{2}} = \sqrt{b^{2} - 2 b^{2} + \sqrt{4 b^{2} - 16 Q^{2}}} = \\
= \sqrt{\sqrt{4 b^{2} - 16 Q^{2}} - b^{2}}
\end{gathered}
$$

Thus,

$$
\begin{gathered}
V_{P A B C D E F} = \frac{1}{3} S_{A B C D E F} \cdot P M = \frac{1}{3} \cdot \frac{6 a^{2} \sqrt{3}}{4} \cdot P M = \\
= \frac{1}{3} \cdot \frac{3 \sqrt{3}}{2} \cdot \left(2 b^2 - \sqrt{4 b^{2} - 16 Q^{2}}\right) \cdot \sqrt{\sqrt{4 b^{2} - 16 Q^{2}} - b^{2}} = \\
= \frac{\sqrt{3}}{2} \left(2 b^2 - \sqrt{4 b^{2} - 16 Q^{2}}\right) \sqrt{\sqrt{4 b^{2} - 16 Q^{2}} - b^{2}}
\end{gathered}
$$