If the perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is the area of the quadrilateral?

We know that perimeter of rhombus = 4 × side of the rhombus

Given perimeter of rhombus = 100cm

Side AB of rhombus = 100/4 = 25cm

Let BD be the diagonal given = 48cm

We know that diagonals of a rhombus bisect each other

E is the midpoint of BD

BE = 24 cm

Now, ∆ABE is the right angle triangle at E

∴using Pythagoras theorem,

AE² + BE²= AB²

AE = 7cm

Area of rhombus = 4×area of ∆ABE

= 2 × 24 × 7

= 336 sq cm