Question

prove that area of a Trapezium is half the sum of The parallel sides multiplied by the distance between them ​

Answer 1

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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 respectivelyFrom figure

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

(i)Consider rectangle CDHFLength = FHBreadth = CFArea of rectangle = length × breadtharea of rectangle CDFH = FH × CF …(ii)

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Consider ΔAFC

Consider ΔAFCbase = AFheight = CF⇒ area(ΔAFC) =× AF × CF …(iii)Consider ΔDBHbase = BHheight = HD⇒ area(ΔDBH) =× BH × HD …(iv)

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

Consider ΔAFCbase = AFheight = CF⇒ area(ΔAFC) =× AF × CF …(iii)Consider ΔDBHbase = BHheight = HD⇒ area(ΔDBH) =× BH × HD …(iv)Substitute (ii), (iii) and (iv) in (i) we getArea of trapezium ABCD = FH×CF +×AF×CF +×BH×HD

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Since CDHF is rectangleCF = HD = h⇒ Area of trapezium ABCD = FH×h +×AF×h +×BH×h

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⇒ 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 rectangleFH = 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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