Question

Prove that the area of a trapezium is half the sum of the parallel sides x the distance between them

Answer 1

Please refer to the image below

Answer 2

Question:-Prove that the area of a trapezium is half the sum of the parallel sides×the distance between them

Solution;-

Given:

A trapezium ABCD in which AB||DC and CM perpendicular AB.

Let CM=h

To prove:

$area \: of \: trapezium = \frac{1}{2} \times (ab + dc) \times h$

Construction:

Draw AL perpendicular DC,meeting CD produced at L.join AC

proof:-

ar(trap.ABCD)=ar(traingleABC)+ar(traingleACD)

$(\frac{1}{2} \times ab \times cm) + ( \frac{1}{2} \times dc \times al)$

$\frac{1}{2} \times(ab + dc) \times h \: \: \: {(as \: cm = al = h)}$

__________

Therefore

$area \: of \: trapezium = \frac{1}{2} \times sum \: of \: parallel \: sides \times distance \: between \: them$