Problem 4. In the isosceles trapezoid $ABCD$ with leg length 10, one of the bases is twice as large as the other. If one of the angles of the trapezoid is $75^{\circ}$, calculate its area.
Solution. Let E be the midpoint of the larger base $AB$. Then $\overline{AE} = \overline{BE} = \overline{CD}$. Since $AB \| CD$, we get that $AEC D$ and $EBC D$ are parallelograms. Therefore, $\overline{AD} = \overline{ED} = \overline{EC} = \overline{BC}$, from which we get $\triangle AED \cong \triangle BEC \cong \triangle CDE \quad(\mathrm{CCC}) . \quad$ Then

![](https://cdn.mathpix.com/cropped/2024_06_05_0c08f972e294b2535969g-11.jpg?height=382&width=601&top_left_y=439&top_left_x=896)
$P_{ABCD} = 3 P_{BEC}$. Let $EG$ be the height dropped from $E$ to $BC$. Since $\quad \measuredangle BEC = \measuredangle EBC = 75^{\circ}$, we get $\measuredangle ECB = 30^{\circ}$, from which it follows that $\overline{EG} = \frac{1}{2} \overline{EC} = 5$. Finally, $P_{BEC} = \frac{10 \cdot 5}{2} = 25$ and thus $P_{ABCD} = 75$.