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The bases of the trapezoid are equal to $a$ and $b$, and the angles at the larger base are $30^{\circ}$ and $45^{\circ}$. Find the area of the trapezoid.

#
Draw a line through the vertex of the smaller base parallel to the lateral side of the trapezoid.

## Solution

Let $a>b$. Draw a line through the vertex $C$ of the smaller base $BC$ of trapezoid $ABCD$, parallel to the lateral side $AB$, until it intersects the base $AD$ at point $M$, and draw a perpendicular $CK$ to $AD$.

Let $\angle A=45^{\circ}, \angle D=30^{\circ}$. Denote $CK=h$. Then

\[
\begin{gathered}
MK=KC=h, KD=\sqrt{3}KC=h\sqrt{3} \\
MD=AD-AM=AD-BC=a-b
\end{gathered}
\]

On the other hand,

\[
MD=MK+KD=h+h\sqrt{3}=h(\sqrt{3}+1)=a-b
\]

From this, we find that $h=\frac{a-b}{\sqrt{3}+1}$. Therefore,

\[
\begin{gathered}
S_{\mathrm{ABCD}}=\frac{1}{2}(AD+BC)CK= \\
=\frac{1}{2} \cdot \frac{(a+b)(a-b)}{\sqrt{3}+1}=\frac{1}{4}\left(a^{2}-b^{2}\right)(\sqrt{3}-1)
\end{gathered}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-46.jpg?height=346&width=740&top_left_y=2094&top_left_x=659)

Answer

$\frac{1}{4}\left(a^{2}-b^{2}\right)(\sqrt{3}-1)$