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` TO PROVE : `a r(r hom b u sA B C D)=1/2(A CxB D)`

Explanation

Given: A rhombus `ABCD`
whose diagonals `AC` and `BD`
intersect at `O`.
To prove: `ar(rh om busABCD)=1/2 (AC\xx\BD)`
Proof: Diagonals of a rhombus are perpendicular to each other.
Area of rhombus `ABCD` is split into area of `triangleABD` and area of `triangleBCD`
area of triangle `=1/2xxbasexxheight`
area of `△ABD=1/2×BD×OA`
area of `△CBD=1/2xxBDxxOC`
area of rhombus`=1/2xxBDxxOA+1/2xxBDxxOC`
`=1/2xxBD[AD+OC]=1/2xxBDxxAC`

`therefore` Area of rhombus=half the product of length of diagonals.