[ Lines and Planes in Space (Miscellaneous).]

The base of the pyramid $S A B C D$ is an isosceles trapezoid $A B C D$, where $A B=B C=a, A D=2 a$. The planes of the faces $S A B$ and $S C D$ are perpendicular to the plane of the base of the pyramid. Find the height of the pyramid if the height of the face $S A D$, drawn from the vertex $S$, is $2 a$.
Note that $A D$ and $B C$ are the bases of trapezoid $A B C D$, and $A B$ and $C D$ are its lateral sides. Since each of the planes $S A B$ and $S C D$ is perpendicular to the plane $A B C D$, the line of intersection of planes $S A B$ and $S C D$ is perpendicular to the plane $A B C D$. Therefore, if $H$ is the point of intersection of this line with the plane $A B C D$, then $S H$ is the height of the given pyramid. Let $S M$ be the height of triangle $A S D$. Then, by the theorem of three perpendiculars, $H M \perp A D$, i.e., $H M$ is the height of the isosceles triangle $A H D$. Since $B C \| A D$ and $B C=\frac{1}{2} A D$, $B C$ is the midline of triangle $A H D$, so $A H=D H=2 a=A D$. Triangle $A H D$ is equilateral, therefore

$$
H M=A H \sin 60^{\circ}=a \sqrt{3}
$$

Therefore,

$$
S H=\sqrt{A H^{2}-H M^{2}}=\sqrt{4 a^{2}-3 a^{2}}=a
$$

## Answer

$a$.