In trapezoid $A B C D(A B \| C D)$, the diagonals $A C=a, B D=\frac{7}{5} a$. Find the area of the trapezoid if $\angle C A B=2 \angle D B A$.
Through vertex $C$, draw a line parallel to diagonal $B D$, until it intersects the extension of base $A B$ at point $E$. Then triangle $A C E$ is equal in area to the given trapezoid.

## Solution

Let $\angle A B D=\alpha$. Then $\angle C A B=2 \alpha$. Through vertex $C$, draw a line parallel to diagonal $B D$, until it intersects the extension of base $A B$ at point $E$. Then

$$
\angle A E C=\angle A B D=\alpha, C E=B D=\frac{7}{5} a \text {. }
$$

Applying the Law of Sines to triangle $A C E$, we get that

$$
\frac{C E}{\sin \angle C A E}=\frac{A C}{\sin \angle A E C}, \text { or } \frac{\frac{7}{5} a}{\sin 2 \alpha}=\frac{a}{\sin \alpha},
$$

or $7 \sin \alpha=5 \sin 2 \alpha$. Since $\sin \alpha \neq 0$, we find from this equation that $\cos \alpha=\frac{7}{10}$. Then $\sin \alpha=\frac{\sqrt{51}}{10}$. Therefore,

$$
\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha=\frac{12 \sqrt{51}}{125}
$$

Thus,

$$
S_{\mathrm{ABCD}}=S_{\triangle \mathrm{ACE}}=\frac{1}{2} \cdot A C \cdot C E \sin \left(180^{\circ}-3 \alpha\right)=\frac{1}{2} \cdot a \cdot \frac{7}{5} a \cdot \sin 3 \alpha=\frac{42 \sqrt{51}}{625} a^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-29.jpg?height=666&width=734&top_left_y=610&top_left_x=662)

## Answer

$\frac{42 \sqrt{51}}{625} a^{2}$