In a rectangular trapezoid $A B C D$, the base $A B$ is 1.5 times the diagonal $A C$. Angles $B A D$ and $A D C$ are right angles. Angle $D C A$ is equal to angle $B C A$. The lateral side $A D$ is equal to 4. Find the area of trapezoid $A B C D$.
Find the cosine of the angle at the base of the isosceles triangle \(ABC\).

## Solution

Let \(AC = x\). Then \(AB = \frac{3x}{2}\). Since

\[
\angle CAB = \angle DCA = \angle ACB,
\]

triangle \(ABC\) is isosceles. If \(BM\) is its height, then

\[
AM = CM = \frac{x}{2}, \quad BC = AB = \frac{3x}{2}
\]

From the right triangle \(AMB\), we find that

\[
\cos \alpha = \cos \angle MAB = \frac{AM}{AB} = \frac{\frac{x}{2}}{\frac{3x}{2}} = \frac{1}{3}
\]

From the right triangle \(ADC\), we find that

\[
\begin{gathered}
x = AC = \frac{AD}{\sin \angle DCA} = \frac{4}{\sin \alpha} = \frac{4}{\frac{2\sqrt{2}}{3}} = 3\sqrt{2} \\
DC = AC \cos \alpha = \frac{3\sqrt{2}}{3} = \sqrt{2}
\end{gathered}
\]

Therefore,

\[
S_{\mathrm{ABCD}} = \frac{AB + CD}{2} \cdot AD = \frac{\frac{9\sqrt{2}}{2} + \sqrt{2}}{2} \cdot 4 = 11\sqrt{2}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-58.jpg?height=375&width=638&top_left_y=1862&top_left_x=710)

Answer

\(11\sqrt{2}\).