Prove that the area of a trapezium is half the sum of the parallel sides multiplied by the distance between them.

ABCD be trapezium with CD || AB

CF and DH are perpendiculars to segment AB from C and D respectively

From figure

Area of trapezium ABCD = area(ΔAFC) + area of rectangle CDFH + area(ΔBHD) …(i)

Consider rectangle CDHF

Length = FH

Breadth = CF

Area of rectangle = length × breadth

area of rectangle CDFH = FH × CF …(ii)

Consider ΔAFC

base = AF

height = CF

⇒ area(ΔAFC) = × AF × CF …(iii)

Consider ΔDBH

base = BH

height = HD

⇒ area(ΔDBH) = × BH × HD …(iv)

Substitute (ii), (iii) and (iv) in (i) we get

Area of trapezium ABCD = FH×CF + ×AF×CF + ×BH×HD

Since CDHF is rectangle

CF = HD = h

⇒ Area of trapezium ABCD = FH×h + ×AF×h + ×BH×h

⇒ Area of trapezium ABCD = h × (FH + ×AF + ×BH)

= h × [FH + × (AF + BH)]

= h × [FH + × (AB – FH)]

= h × (FH + ×AB - ×FH)

= h × (×FH + ×AB)

= × h × (FH + AB)

Since CDHF is rectangle

FH = CD

⇒ Area of trapezium ABCD = × h × (CD + AB)

h is the distance between parallel sides AB and CD

Therefore, area of a trapezium is half the sum of the parallel sides multiplied by the distance between them