10.323. The diagonal of an isosceles trapezoid is $10 \mathrm{~cm}$, and the area is $48 \mathrm{~cm}^{2}$. Find the height of the trapezoid.
Solution.

In trapezoid $ABCD$ (Fig. 10.112) $AB=CD, AC=10$ cm, $CE$ is the height. Let $AE=x$ cm, $CE=y$ cm. Since $AE=\frac{AD+BC}{2}$, then by the condition $xy=48$ cm. From $\triangle AEC\left(\angle AEC=90^{\circ}\right): AE^{2}+CE^{2}=AC^{2} ; x^{2}+y^{2}=100$. We solve the system of equations: $\left\{\begin{array}{l}x^{2}+y^{2}=100 ; \\ xy=48 ;\end{array} ;\left\{\begin{array}{l}(x+y)^{2}-2xy=100 \\ xy=48 ;\end{array}\right.\right.$ $\left\{\begin{array}{l}(x+y)^{2}=196 ; \\ xy=48 .\end{array}\right.$ Since $x>0, y>0$, then $x+y=14$. Then $\left\{\begin{array}{l}x=8, \\ y=6,\end{array}\right.$ or $\left\{\begin{array}{l}x=6, \\ y=8 .\end{array}\right.$ Therefore, the height of the trapezoid is 6 cm or 8 cm.

Answer: 6 cm, 8 cm.