[ Linear dependence of vectors $]$ [ Angles between lines and planes $]$

The side of the base of a regular quadrilateral pyramid is equal to $a$. The lateral face forms an angle of $45^{\circ}$ with the base plane. Find the height of the pyramid.
Let $ABCDP$ be a given regular quadrilateral pyramid with vertex $P, AB=BC=CD=AD=a, M-$ the center of the square $ABCD, K$ - the midpoint of the segment $AB$. Since $PK \perp AB$ and $MK \perp AB$, the angle $PKM$ is the linear angle of the dihedral angle between the plane of the lateral face $ABP$ and the plane of the base of the pyramid. By the condition $\angle$ $PKM=45^{\circ}$. Since the pyramid is regular, its height passes through the center of the base, so $PM$ is the height of the pyramid. From the isosceles right triangle $PKM$, we find that $PM=MK=\frac{\underline{a}}{2}$.

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