Let ABCD be the rhombus of perimeter 80 m and diagonal AC = 24 m

We have,

AB + BC + CD + DA = 80

4 AB = 80 [AB=BC=CD=DA sides of Rhombus]

AB = 20 m

In ΔABC, we have

AB = a = 20 cm, BC = b = 20 cm, AC = c = 24 cm

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 32

A=

A = = cm²

Hence, area of rhombus ABCD = 2 ×192 m²

= 384 m²