10.319. The diagonals of an isosceles trapezoid are perpendicular to each other, and its area is $a^{2}$. Determine the height of the trapezoid.
Solution.

Let in trapezoid $A B C D$ (Fig. 10.108) $A B=C D, A C \perp B D, O$ - the intersection point of $A C$ and $B D, C K$ - the height of the trapezoid. Since the trapezoid is isosceles, then $A K=\frac{A D+B C}{2}, A O=D O$, and since $\angle A O D=90^{\circ}$, then $\angle O A D=45^{\circ}$. Therefore, in the right $\triangle A K C \quad A K=C K$. The area of the trapezoid $S=\frac{A D+B C}{2} \cdot C K=A K \cdot C K=C K^{2}=a^{2}$. Thus, $C K=a$.

Answer: $a$.