Question

Find the area of a rhombus whose perimeter is 200m and one of the diagnols is 80m.

Answer 1

Let side of the rhombus be 's' and diagonals be d and d'
Given , perimeter of rhombus=200m
➡4s= 200
➡s=50 m
Also given , d'=80m
We know , diagonals of a rhombus are perpendicular bisectors.
➡(d/2)²+(d'/2)²=s²
➡(d/2)²=900
➡d=2*30=60 m
Area of the rhombus = d*d'/2 = 60*80/2 = 2400 m²

Answer 2

Answer:

perimeter of rhombus = 4a

⇒4a = 200m

⇒a = 200/4 = 50m

In a rhombus diagonals bisect each other at right angles.

In ΔABD,

s = (a+b+c)/2

⇒s = (50+ 50+ 80)/2 = 180/2 = 90 m

By herons formula,

Area of ΔABD = $\sqrt{s(s-a)(s-b)(s-c)}$

⇒ $\sqrt{90(90-50)(90-50)(90-80)}$

⇒$\sqrt{90 * 40 * 40 * 10}$

⇒40√(9*10*10)

⇒40 * 3 * 10 = 1200 m²

Area of ΔABD can also be written as

1/2*b*h = 1200

1/2 * 80 * h = 1200

h = 1200 * 2/ 80 = 30 m

AO = CO = 30 m [∵ In a rhombus diagonals bisect each other at right angles]

⇒ AC = 30 *2 = 60 m

Area of rhombus = product of diagonals/2

⇒60 * 80/2 = 2400 m²