A rhombus has side of length 60 units each and the lengths of its perpendicular diagonals are in the ratio of 2 to 1. What is the area of the rhombus, in square units?

2880 square units

Step-by-step explanation:

Since the area of a rhombus, A = d₁ × d₂/2 where d₁ and d₂ are its diagonals. Given that d₁:d₂ = 2:1, d₁/d₂ = 2/1, so d₁ = 2d₂. Also, the length of side of the rhombus L = 60 units.

We find the length of the diagonal by Pythagoras' theorem since the diagonals are perpendicular to each other and bisect each other at their mid point.

The mid point of d₁ = d₁/2 and the midpoint of d₂ = d₂/2.

By Pythagoras' theorem,

L² = (d₁/2)² + (d₂/2)²

L² = d₁²/4 + d₂²/4

Now, d₁ = 2d₂

So, L² = (2d₂)²/4 + d₂²/4

L² = 4d₂²/4 + d₂²/4

L² = d₂² + d₂²/4

L² = 5d₂²/4

d₂² = 4L²/5

taking square root of both sides,

√d₂² = √(4L²/5)

d₂ = 2L/√5

Since L = 60 units, then

d₂ = 2L/√5

d₂ = 2(60)/√5

d₂ = 120/√5

rationalizing the denominator by multiplying the numerator and denominator by √5, we have

d₂ = 120/√5 × √5/√5

d₂ = 120√5/5

d₂ = 24√5 units

So, our area A = d₁ × d₂/2

Since d₁ = 2d₂, our area is

A = 2d₂ × d₂/2

A = d₂ × d₂

A = d₂²

substituting d₂ = 24√5, we have

A = d₂²

A = (24√5)²

A = (24)²(√5)²

A = 576 × 5

A = 2880 square units