Question

If each side of a rhombus is doubled, how much will its area increase?

Options

• 1.5 times

• 2 times

• 3 times

• 4 times

Answer 1

3 times

Explanation:

Let the given rhombus be ABCD with side length a unit and length of diagonals AC and BD are d₁ and d₂ respectively.

In ΔAOD,

(AD)² = (OA)² + (OD)²  ...[Using pythagoras theorem]

`a^2 = (d_1/2)^2 + (d_2/2)^2`

`\implies a = sqrt(d_1^2/4 + d_2^2/4)`

`\implies a = 1/2 sqrt(d_1^2 + d_2^2)`

`\implies 2a = sqrt(d_1^2 + d_2^2)`  ...(1)

Now, side length of rhombus is doubled.

i.e., 2a and let length of new diagonals are d’₁ and d’₂·

`\implies 2a = sqrt((d_1^’/2)^2 + (d_2^’/2)^2`

Using (1), we get

`d_1^’ = 2d_1` and `d_2^’ = 2d_2`

Area of rhombus ABCD = `1/2 xx d_1 xx d_2 = 1/2d_1d_2`

After doubling the side length, area of rhombus = `1/2 xx 2d_1 xx 2d_2 = 2d_1d_2`

∴ Increase in area = `2d_1d_2 - 1/2 d_1d_2`

= `3/2 d_1d_2`

= 3 × (Area of rhombus ABCD)