4. The height of the trapezoid is $6 \mathrm{~cm}$, and the length of one base is $4 \mathrm{~cm}$. The interior angles of the trapezoid on that base are $120^{\circ}$ and $135^{\circ}$.

a) Construct this trapezoid.

b) Determine the perimeter and area of this trapezoid.
Solution. a) First, we construct the segment $CD=4 \text{ cm}$, and then the angles

\[
\begin{aligned}
& \measuredangle C=135^{\circ}=90^{\circ}+45^{\circ} \text{ and } \\
& \measuredangle D=120^{\circ}=2 \cdot 60^{\circ}.
\end{aligned}
\]

Then, we construct a line $p$

![](https://cdn.mathpix.com/cropped/2024_06_05_e381a8d26dc145f66c35g-2.jpg?height=322&width=548&top_left_y=704&top_left_x=916)
parallel to $CD$ which is at a distance of $6 \text{ cm}$ from $CD$. The intersection points of line $p$ with the sides of the previously constructed angles are the vertices $A$ and $B$ of the trapezoid.

b) We draw heights $CE$ and $DF$ (see the diagram). Then, since $\measuredangle BCE=45^{\circ}$, we get $BE=CE=6 \text{ cm}$ and $BC=6\sqrt{2}$, and from $\measuredangle ADF=30^{\circ}$ it follows that $DF=\frac{AD \sqrt{3}}{2}$, so $AD=4\sqrt{3}$ and $FA=\frac{1}{2} AD=2\sqrt{3}$.

Now we know the sides $CD=4 \text{ cm}, AD=4\sqrt{3} \text{ cm}, AB=10+2\sqrt{3} \text{ cm}$, and $BC=6\sqrt{2} \text{ cm}$, so $O=AB+BC+CD+DA=14+6\sqrt{2}+6\sqrt{3} \text{ cm}$ and $P=\frac{AB+BC}{2} \cdot h=42+6\sqrt{3} \text{ cm}^2$.