Question

Prove that the area of a trapezium ABCD $=\frac{1}{2}\times$ (sum of parallel sides) × distance between parallel sides.

Answer 1

Draw DF and CE perpendiculars from D and C on AB.
Area of a trapezium ABCD = area ($△$ DFA) + area (rectangle DFEC) + area ($△$ CEB)

$=(\frac{1}{2}\times AF\times DF)+(FE\times DF)+(\frac{1}{2}\times EB\times CE)$

$=(\frac{1}{2}\times AF\times h)+(FE\times h)+(\frac{1}{2}\times EB\times h)$

$=\frac{1}{2}\times h\times (AF+2FE+EB)$

$=\frac{1}{2}\times h\times (AF+FE+EB+FE)$

$=\frac{1}{2}\times h\times (AB+FE)$

$=\frac{1}{2}\times h\times (AB+DC)$ square units.

$=\frac{1}{2}\times$ (sum of parallel sides) $\times$ (distance between them)