This is an easy exercise! Indeed, it's not difficult to see that $(BCM)=\frac{1}{2}(ABCD)$. But $(BCM)=\frac{1}{2}BC \cdot CM \sin 150^{\circ}=\frac{1}{4}ab$. Hence $(ABCD)=\frac{1}{8}ab$.

I'm sorry, the conclusion of this solution should be: Hence $(ABCD)=\frac{1}{2}ab$.

Jutaro wrote: This is an easy exercise! Of course, with trig. all is easy.

Anyway, I think a solution that avoids the use of the trigonometric formula for the area of a triangle is still easy.

Let $N$ be midpoint of $BC$.Then, $A_{MNC}$=$\frac{\frac{a}{2}\cdot{b}\cdot{sin150^{\circ}}}{2}=\frac{{a}\cdot{b}}{8}$ and also $A_{MNC}=A_{BMN}=\frac{a\cdot{b}}{8}$.Then $A_{BMC}$=$\frac{a\cdot{b}}{4}$ and also we know that, $A_{BMC}$=$A_{AMB}$+$A_{MDC}$ so $A_{ABCD}$=$A_{BMC}$+$A_{AMB}$+$A_{MDC}$=${2}\cdot{A_{BMC}}$ =$2\cdot{\frac{ab}{4}}$=$\frac{ab}{2}$

Sorry to revive such an old thread but I don't understand why $ [BCM] = \frac{1}{2} [ABCD]$..

Quote: Let $ ABCD$ be a trapezoid with $ AB\parallel CD$ . Denote the length $ h$ of the altitude of the trapezoid. Consider the points $ M\in (AD)\ ,\ N\in (BC)$ so that $ MN\parallel AB$ . Prove that $ [MBC] = \frac h2\cdot MN$ . In the particular case when $ M$ is the midpoint of the side $ [AD]$ obtain that $ [ABCD] = 2\cdot [MBC]$ .