8,9

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

Find $AH$ and $DH$, if the bases of the trapezoid are 15 and 35.
Let $Q$ be the base of the perpendicular dropped from vertex $B$ to $AD$ ($AD=25, BC=15$). From the equality of right triangles $ABQ$ and $DCH$, it follows that $AQ=DH$, and since $BCHQ$ is a rectangle, then $HQ=BC=15$. Therefore,

$$
\begin{gathered}
DH=\frac{AD-HQ}{2}=\frac{35-15}{2}=10 \\
AH=AD-DH=35-10=25
\end{gathered}
$$

## Answer

$25 \cdot 10$.