Problem 704

Let $A B C D P$ be the given regular quadrilateral pyramid with vertex $P, A B=B C=C D=A D=a, M-$ the center of the square $A B C D, K$ - the midpoint of the segment $A B$. Since $P K \perp A B$ and $M K \perp A B$, the angle $P K M$ is the linear angle of the dihedral angle between the plane of the lateral face $A B P$ and the plane of the base of the pyramid. By the condition $\angle$ $P K M=45^{\circ}$. Since the pyramid is regular, its height passes through the center of the base, so $P M$ is the height of the pyramid. From the isosceles right triangle $P K M$ we find that $P M=M K=\frac{a}{2}$. Therefore,

$$V_{A B C D P}=\frac{1}{3} S_{A B C D} \cdot P M=\frac{1}{3} a^2 \cdot \frac{a}{2}=\frac{a^{3}}{6}$$

## Answer

$\frac{\mathbf{a}^{3}}{6}$.