Question

Prove that area of a trapezium is equal to half the product of the sum of parallel sides and distance between the parallel sides.

Answer 1

Consider the trapezium $ABCD$ with with $AB||CD$

$\therefore AP=BQ=h\left(say\right)$ [distance between parallel lines one equal]

$ar\left(ABCD\right)=ar\left(△ADC\right)+ar\left(△ABC\right)$.....(1)

$ar\left(△ADC\right)=\frac{1}{2}\times base\times height=\frac{1}{2}\times \left(DC\right)\times h$......(2)

$ar\left(△ABC\right)=\frac{1}{2}\times base\times height=\frac{1}{2}\times \left(AB\right)\times h$.....(3)

From (1),(2)and (3)

$ar\left(ABCD\right)=\frac{1}{2}\times h\left(CD\right)+\frac{1}{2}\times h\left(AB\right)$

$ar\left(△ABCD\right)=\frac{1}{2}h[AB+CD]$

$\therefore$ Area of a trapezium is half the product of the sum of parallel sides and distance between the parallel sides.