4. In an isosceles trapezoid, the midline is $s$, and the diagonal is twice as long as the midline. Determine the area of this trapezoid.
Solution. Let $ABCD$ be the given trapezoid and let $MN$ be its midline (see the diagram). We have $MN = s$ and $AC = 2s$. Now, since $AC' = a - \frac{a - b}{2} = \frac{a + b}{2} = MN = s$, we get

$$
\begin{aligned}
CC' & = \sqrt{AC^2 - AC'^2} \\
& = \sqrt{4s^2 - s^2} = s \sqrt{3}
\end{aligned}
$$

Finally,

$$
\begin{aligned}
P & = \frac{AB + CD}{2} \cdot CC' = MN \cdot CC' \\
& = s^2 \sqrt{3}.
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_b58a8127523cd0cd5361g-2.jpg?height=344&width=641&top_left_y=256&top_left_x=838)