Question

prove that the area of rhombus is equal to half of the product of the diagonals

Answer 1

Hey,

Since a property of rhombus is that its all sides are equal

And AO = OC ; OB = OD

Now. ar.( AOD ) = ( 1 / 2 ) * AO * OD it is same for ar.( BOC ) and further ar.( AOB ) = ( 1 / 2 ) * AO * OB and it is also the same for ar.( COD )

from the figure, ar.( ABCD ) = ar.( AOD ) + ar.( BOC ) + ar.( AOB ) + ar.( COD )

=> ( 1 / 2 ) * AO * OD + ( 1 / 2 ) * BO * OC  + ( 1 / 2 ) * AO * OB + ( 1 / 2 ) * OC * OD

=> { ( 1 / 2 ) * AO * OD + ( 1 / 2 ) * OC * OD } + { ( 1 / 2 ) * BO * AO + ( 1 / 2 ) * BO * OC } proceed by adding AO to OC

=> Finally , ar.( ABCD ) = half the product of diagonals    

HOPE IT HELPS:-))

Answer 2

Hey! See the attached pic

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ABCD is a rhombus :-

(Diagonal of rhombus are perpendicular to each other)

ΔABD

ar(ABD)=1/2×b×h

=1/2×BD×AO

ΔBDC

=Base×height/2

=1/2×BD×OC

ar of rhombus

=ar(ΔABD)+ar(BDC)

=1/2×BD×AO+1/2×BD×OC

=1/2×BD×(AO+OC)

=1/2×BD×AC

Therefore :-

(AO+OC=AC) Proved

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Regards :)

Yash Raj ✌