Point $H$ lies on the larger base $AD$ of the isosceles trapezoid $ABCD$, and $CH$ is the height of the trapezoid.

Find the bases of the trapezoid if $AH=20$ and $DH=8$.

#
Let $A D=x, B C=y$. Let $Q$ be the foot of the perpendicular dropped from vertex $B$ to $A D$. From the equality of right triangles $A B Q$ and $D C H$, it follows that $A Q=D H$. Since $B C H Q$ is a rectangle, $H Q=B C=y$. Therefore,

$$
\begin{gathered}
D H=\frac{A D-H Q}{2}=\frac{x-y}{2} \\
A H=A D-D H=x-\frac{x-y}{2}=\frac{x+y}{2}
\end{gathered}
$$

From the system

$$
\left\{\begin{array}{l}
\frac{x-y}{2}=8 \\
\frac{x+y}{2}=20
\end{array}\right.
$$

we find that $x=28, y=12$.

## Answer

$12 \cdot 28$