The lateral side, the smaller base, and the diagonal of an isosceles trapezoid are 10, 6, and 14, respectively. Find the larger base.

#
The projection of the lateral side of an isosceles trapezoid onto the larger base is equal to half the difference of the bases, and the projection of the diagonal is half the sum of the bases.

## Solution

Let $H$ be the projection of vertex $C$ of the smaller base of trapezoid $ABCD$ onto the larger base $AD$.

Let $AD = x$. Then

$$
\begin{aligned}
& DH = \frac{x - 6}{2} = \frac{AD + BC}{2} \\
& AH = AD - HD = AD - \frac{x - 6}{2} = \\
& \frac{x + 6}{2} = \frac{(x + 6)^2}{4}
\end{aligned}
$$

Expressing the square of the leg $C$ from the right triangles $CHA$ and $CHD$, we get the equation

$$
196 - \frac{(x - 6)^2}{4} = 100 -
$$

from which we find that $x = 16$.

## Answer

16.