Question

Prove that the area of a rhombus is equal to half of the product of the diagonals.

Answer 1

Given ABCD is a rhombus the diagonal AC and BD cut at point O

Then $\angle AOD=\angle AOB=\angle COD=\angle BOC={90}^0$

The area of rhombus ABCD divided diagonal in four parts

So area of rhombus ABCD $=$area of triangle AOD+area of triangle AOB+area of triangle BOC+area of triangle COD

$=\frac{1}{2}\times AO\times OD+\frac{1}{2}\times AO\times OB+\frac{1}{2}\times BO\times OC+\frac{1}{2}\times OD\times OC$

$=\frac{1}{2}\times AO(OD+OB)+\frac{1}{2}OC(BO+OD)$

$=\frac{1}{2}\times AO\times BD+\frac{1}{2}\times OC\times BD$

$=\frac{1}{2}BD(AO+OC)=\frac{1}{2}\times BD\times AC$

So area of rhombus is equal to half of the product of diagonals