Q. 74.0( 14 Votes )

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

Answer :


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


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