A = [math]bh[/math]

The area of the parallelogram we created would be the base [math]\left(a+b\right)[/math] times the height. Since the parallelogram is half the height of the trapezoid, we would multiple the height by [math]\frac{1}{2}[/math].[br]So, the area would be [math]\left(\frac{1}{2}\right)h\left(a+b\right)[/math] which is also the formula for the area of a trapezoid!

A = [math]bh[/math]

A = [math]\left(\frac{1}{2}\right)bh[/math]

a or [math]a_{_{_1}}[/math]

(base-a)

The area of the rectangle would be [math]\left(ah\right)[/math] and the area of the triangle would be [math]\left(\frac{1}{2}\right)\left(b-a\right)h[/math].[br]If we add them, we get [math]\left(ah\right)+\left(\frac{1}{2}\right)\left(b-a\right)h[/math]. Let's simplify this:[br][math]ah+\frac{1}{2}bh-\frac{1}{2}ah[/math] [math]\longleftarrow[/math]distribute h and [math]\frac{1}{2}[/math] to (b-a)[br][br][math]h\left(a+\frac{1}{2}b-\frac{1}{2}a\right)[/math][math]\leftarrow[/math] factor out h[br][br][math]h\left(\frac{1}{2}b+\frac{1}{2}a\right)[/math][math]\leftarrow[/math] simplify inside the parenthesis[br][br][math]\frac{1}{2}h\left(b+a\right)[/math][math]\leftarrow[/math] factor out [math]\frac{1}{2}[/math][br][br]Now we are left with [math]\frac{1}{2}h\left(b+a\right)[/math] or [math]\frac{1}{2}h\left(a+b\right)[/math] which is the area of a trapezoid![br][br][br]