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prove that area of the rhombus is half the prodcut of the 2 diagnols

since the diagonals of a rhombus bisect each other at 90 degree.

since ABCD is a rhombus. OB=OD and OA=OC.

and ∠AOD=∠AOB=∠BOC=∠COD=90 degree.

area of ABCD can be divided in four parts.

area(ABCD)=area(ΔAOD)+area(ΔAOB)+area(ΔBOC)+area(COD)

=1/2(product of diagonals)

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Given: Rhombus ABCD
/| To prove: Area of Rhombus ABCD = DB�AC/2
/ | 
D/ E| B
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| /
|/ 
C

Area of rhombus ABCD = area of triangle ABD + area of 
triangle CBD

Triangles ABD and CBD are congruent by SSS

Area of rhombus ABCD = 2�(Area of triangle ABD)

AE is perpendicular to DB because the diagonals 
of a rhombus are perpendicular bisectors of each other.

Area of triangle ABD = DB�AE/2 because a triangle 's area
is one-half the product of a side and the altitude drawn
to that side.

Area of rhombus ABCD = 2�(Area of triangle ABD)

So area of rhombus ABCD = 2�(DB�AE/2) = DB�AE

AE = AC/2 because the diagonals of a rhombus are perpendicular
bisectors of each other.

So area of rhombus DB�AE = DB�(AC/2) = DB�AC/2

  • -1

 Given: Rhombus ABCD
/| To prove: Area of Rhombus ABCD = DB�AC/2
/ | 
D/ E| B
��|��/
| /
|/ 
C

Area of rhombus ABCD = area of triangle ABD + area of 
triangle CBD

Triangles ABD and CBD are congruent by SSS

Area of rhombus ABCD = 2�(Area of triangle ABD)

AE is perpendicular to DB because the diagonals 
of a rhombus are perpendicular bisectors of each other.

Area of triangle ABD = DB�AE/2 because a triangle 's area
is one-half the product of a side and the altitude drawn
to that side.

Area of rhombus ABCD = 2�(Area of triangle ABD)

So area of rhombus ABCD = 2�(DB�AE/2) = DB�AE

AE = AC/2 because the diagonals of a rhombus are perpendicular
bisectors of each other.

So area of rhombus DB�AE = DB�(AC/2) = DB�AC/2

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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    

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