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The base $AB$ of trapezoid $ABCD$ is twice the length of base $CD$ and twice the length of side $AD$. Diagonal $AC$ is equal to $a$, and side $BC$ is equal to $b$. Find the area of the trapezoid.

#
Draw a line through point $D$ parallel to $B C$.

## Solution

Draw a line through vertex $D$ parallel to $B C$. Let $K$ be the point of intersection of the drawn line with the base $A B$. Then $A D C K$ is a rhombus, and $D C B K$ is a parallelogram. Therefore,

$$
\begin{gathered}
D K=B C=b, S_{\mathrm{ADCK}}=\frac{1}{2} D K \cdot A C=\frac{1}{2} a b, \\
S_{\triangle \mathrm{KCB}}=S_{\triangle \mathrm{AKC}}=\frac{1}{2} S_{\mathrm{ADCK}}=\frac{1}{4} a b .
\end{gathered}
$$

Thus, $S_{\mathrm{ABCD}}=\frac{3}{4} a b$.

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-22.jpg?height=340&width=744&top_left_y=928&top_left_x=657)

Answer

$\frac{3}{4} a b$.