# Proof 10: Parallelogram and trapezium area

A trapezium consists of a rectangle and two weird triangles at the ends.

\begin{align} A_\text{trapezium}&= A_\text{rectangle}+ A_{\text{triangle } a}+ A_{\text{triangle } b}\\ ~&=\underbrace{lw_1}_\text{rectangle}+\underbrace{\frac12 w_al}_{\text{rectangle } a}+\underbrace{\frac12 w_b l }_{\text{rectangle } b} \qquad\leftarrow A_\text{rectangle} =lw \text{ and } A_\text{triangle}=\frac12 lw \\ ~&=lw_1+\frac12(w_a+w_b)l \qquad\leftarrow \text{ The triangles have the same height l} \\ ~&=lw_1+\frac12(w_2-w_1)l \qquad\leftarrow w_a+w_b+w_1=w_2\Leftrightarrow w_a+w_b=w_2-w_1 \\ ~&=\left(w_1+\frac12(w_2-w_1)\right)l \\ ~&=\left(w_1+\frac12w_2- \frac12 w_1\right)l \\ ~&=\left(\frac12w_1+ \frac12 w_2\right)l \\ ~&=\frac12(w_1+ w_2)l\quad_\blacksquare \\ \end{align}

A parallelogram is just a trapezium with pairwise parallel sides. This is only possible if the sides are pairwise equal. Equal sides $w_1=w_2$ – let’s just call them both $w$ – means:

$$A_\text{parallelogram}=\frac12 (\underbrace{w}_{w_1}+\underbrace{w}_{w_2})l=\frac1{\cancel 2} \cancel 2wl=lw \quad_\blacksquare$$