How to find side of an equilateral triangle when area is given?

Rule

side a

A = (a^2)(√3)/4

solve for a

Let area = A units ² and side length = y units

1/2 x y x y sin 60° = A

(1/2) y² ( √3/2 ) = A

y² = 4 A / √3

y = 2 [ A/√3 ]^(1/2)

The angles of an equilateral triangle are all 60 degrees and the sides are all 2a (the choice of 2a is on purpose as per what follows). If you bisect one angle, you create two equal right-30-60 triangles with a short side of a and a hypotenuse of 2a, the bisector is sqrt(3)a. (you have to know this basic characteristic of a right-30-60 triangle to make this jump, although you can also derive it if you want; just longer to do).

The equilateral triangle will have the same area as a rectangle made by flipping one of the 30-60-90 triangles over and attaching to the other along the hypotenuse. This creates a rectangle which has an area of a*sqrt(3)*a, and of course which is also the area of the original equilateral triangle (made by moving the area around to create a different shape).

So the area of this rectangle is simply sqrt(3)*a^2. a= sqrt(area/sqrt(3)). A side of the original equilateral triangle is twice that (because of the choice of one side being 2a back at the beginning).

Side = 2(sqrt(area/sqrt(3)), and this actually does turn into The Landpirate's answer of 2*3^(1/4)*area^1/2 despite someone having marked him thumbs-down.

A = area of an equilateral Δ with side s

A = (1/2)s^2sin60 =

A = s^2*(√3)/4

s^2 = 4A/√3

s = 2√( A/√3)

Side of equilateral triangle = square root of (area*2)/(sine 60 degrees)

Let s= side.

h= √(s² - s² /4)= √3 s/2

A= √3 s²/4

s= 2√(A/√3)

In an equilateral triangle, if you draw a line down from one top down to the opposite base, you create two right triangles that are identical.

The smaller triangle's base is half that of the original, and the height of both are the same.

So if the hypotenuse is "s", then the base is "s/2". Knowing this is a right triangle we can solve for the height using Pythagorean Theorem:

a² + b² = c²

(s/2)² + b² = s²

s²/4 + b² = s²

s² + 4b² = 4s²

4b² = 3s²

b² = 3s²/4

b = √(3)s/2

That's the height of the triangle, with the base being "s" so we can now come up with an area equation:

A = bh/2

A = [s√(3)s/2] / 2

A = s²√(3)/4

If you know the area and want to solve for the side, we get:

4A = s²√(3)

4A / √3 = s²

√(4A / √3) = s

which simplifies to:

2√A / ⁴√3 = s

If we want to rationalize the denominator, multiply both halves by ⁴√(3³) to get:

2 ⁴√(3³) √A / ⁴√3⁴ = s

2 ⁴√(27) √A / 3 = s

And if we want to have the 27 and A under a single radical, we get:

2 ⁴√(27A²) / 3 = s

Kind of ugly. This would likely make more sense if we had the area to start with and just worked the numbers from there, but this will get you the side knowing the area of any equilateral triangle.

Area = one quarter of √3 times the square of the side, so side = 2√(area / √3)

Rationalise the denominator if you must.

Heron's formula tells us that the area of any triangle is:

sqrt(s * (s - a) * (s - b) * (s - c))

Where a , b , and c are side lengths and s is the semi-perimeter

a = b = c

sqrt(s * (s - a) * (s - a) * (s - a)) =>

sqrt(s * (s - a)^3) =>

(s - a) * sqrt(s * (s - a))

s = (a + b + c) / 2 = (a + a + a) / 2 = 3a/2

(3a/2 - a) * sqrt((3a/2) * (3a/2 - a)) =>

a * (3/2 - 1) * sqrt(a * (3/2) * a * (3/2 - 1)) =>

a * (1/2) * sqrt(a^2 * (3/2) * (1/2)) =>

a * (1/2) * a * (1/2) * sqrt(3) =>

(sqrt(3)/4) * a^2

A = (sqrt(3)/4) * a^2

If you have the area

A * 4 / sqrt(3) = a^2

A * 4 * sqrt(3) / 3 = a^2

2 * sqrt(A * sqrt(3) / 3) = a

(2/3) * sqrt(3 * sqrt(3) * A) = a

a is the side length.

(2/3) * sqrt(3^(3/2) * A)

(2/3) * 3^(3/4) * A^(1/2)

2 * 3^(-1/4) * A^(1/2)