Show that the area of a rhombus is half the product of the lengths of its diagonals. GIVEN : A rhombus `A B C D` whose diagonals `A C` and `B D` intersect at `Odot`

Explanation

TO PROVE
: Area of rhombus`(ABCD)`=`1/2(ACxxBD)`
Given:`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=ar(triangleAOD)+ar(triangle AOB)+ar(triangleBOC)+ar( triangleCOD)`
`=1/2​xxAOxxOD+1/2​xxAOxxOB+1/2​xxBOxxOC+1/2​xxODxxOC`
`=1/2​xxAO(OD+OB)+1/2​OC(BO+OD)`
`=1/2​xxAOxxBD+1/2​xxOCxxBD`
`=1/2xxBD(AO+OC)`
`=1/2​xxBDxxAC`
So area of rhombus is equal to half of the product of diagonals.
Hence Proved.