Prove that the area of a rhombus is equal to half the rectangle contained by its diagonals. Given: A rhombus `A B C D` such that its diagonals `A C\ a n d\ B D` intersect at `Odot` To Prove: `a r\ (r hom b u s\ A B C D)=1/2` (area of the rectangle contained by its diagonals `=1/2(A C\ x\ B D)`

Explanation

Proof:
`ABCD` is a rhombus the diagonal `AC` and `BD` cut at point `O`.
Then `∠AOD=∠AOB=∠COD=∠BOC=90^@`
The area of rhombus `ABCD` divided diagonal in four parts.
So area of rhombus `ABCD=`area of `triangleAOD+`area of `triangleAOB+`area of `triangleBOC+`area of `triangleCOD`
`=1/2×AO×OD+1/2×AO×OB+1/2×BO×OC+1/2×OD×OC`
`1/2×AO(OD+OB)+1/2OC(BO+OD)`
`1/2×AO×BD+1/2×OC×BD`
`1/2BD(AO+OC)=1/2×BD×AC`
So area of rhombus is equal to half of the product of diagonals.
Hence Proved.