Prove that the area of parellelograms is the product of any its sides and the corresponding altitudes .

please provide the whole proof !!!!!!

given: let ABCD be a parallelogram. and AE be the altitude from A to side DC.

TPT ; area(parallelogram ABCD)=CD*AE

construction: join AC and from C draw an altitude to AB. 

proof:

area(ABCD)=area(ΔADC)+area(ΔABC)

the distance between two parallel lines is always constant.

and AB=CD [opposite sides of a parallelogram are equal]

therefore

area(ABCD)=CD*AE

i.e. area of parallelogram is the product of its one side and corresponding altitude.

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BEFORE LOOKING AT THE ANSWER PLZZ DRAW A FIGURE ACCORDING TO MY INSTRUCTION...

TAKE A PARALLELOGRAM PQRS . MAKE A DIAGONAL  PR IN IT  . DRAW  PM PERPENDICULAR TO QR .

WE KNOW THAT DIAGONAL DIVIDES PARALLOGRAM INTO TWO CONGRUENT TRIANGLES 

THEREFORE,  TRIANGLE PQR IS CONGRUENT TO TRAINGLE PRS . THIS MEANS,  PARALLELOGRAM PQRS = 2(PQR) . AREA OF TRIANGLE PQR = 12 PM X QR . PUTTING THIS IN PQRS = 2(PQR) . WE GET , PQRS= 2(12 PM X QR) = PM X QR . HENCE PQRS = BASE X CORRESPONDING ALTITUDE . 

HOPE THIS MAY HELP YOU .. THUMPS UP PLZ

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