The area of a rhombus is 36 cm2. One of the diagonals is twice as long as the other. Find the lengths of the

The area of a rhombus is 36 cm2
One of the diagonals is twice as long as the other.
Find the lengths of the diagonals.

I have no idea how to do this D:
Can you please give me the steps needed to do this kind of question, also if you can please explain why you're doing what you're doing, thank you :)

The area of a rhombus is:
A = (1/2) ⋅ d₁ ⋅ d₂, where d₁ and d₂ are the diagonals of the rhombus.

So, one of the diagonals is twice as long as the other. We can write it this way:
d₂ = 2 ⋅ d₁

So, we can substitute it into the formula, now with only one variable:
A = (1/2) ⋅ d₁ ⋅ d₂
(36 cm²) = (1/2) ⋅ d₁ ⋅ 2 ⋅ d₁
(36 cm²) = d₁²
d₁ = √(36 cm²)
d₁ = 6 cm

Since the other diagonal is twice as long:
d₂ = 2 ⋅ d₁ = 2 ⋅ (6 cm) = 12 cm

So, the lengths of the diagonals are: 6 cm, and 12 cm.

Area of the Rhombus = Product of the diagonals/2
One diagonal = x other diagonal =2x
Hence 2x²/2 =36
→x= 6 cm
Other diagonals= 12 cm

the correct answer is 6cm and 12 cm

since,
Area=d1d2/2

substitude d2=2d1 in the formula of area and simplify to get the length of d1=6cm
then d2=2 * 6cm =12 cm