There are a couple of facts that make it easy to find the height of an equilateral triangle.
- First of all, all the side lengths of an equilateral triangle are equal to each other.
- Secondly, all the angles of an equilateral triangle equal \(60^\circ\).
Because of these two facts, we can apply the Pythagorean Theorem or the Law of Sines to an equilateral triangle to find this formula:
Height of an Equilateral Triangle Formula
In this formula, the h represents the height of the equilateral triangle and the x represents the side length of the equilateral triangle.
\[h=\frac{\sqrt{3}}{2}x\]
How to Find the Height of an Equilateral Triangle
- Identify the side length of the equilateral triangle.
- Evaluate the formula \(h=\frac{\sqrt{3}}{2}x\)
Example
What is the side length of the equilateral triangle?
The side length is \(4\, in\).
What is the evaluated height formula?
\[h=\frac{\sqrt{3}}{2}(4 \, in)\]
Four times the square root of 3 is \(4\sqrt{3}\).
\[h=\frac{4\sqrt 3}{2}\, in\]
Four and two are both divisible by two, so the fraction can be reduced.
\[h=2\sqrt 3 \, in\]
The height of this equilateral triangle is \(2\sqrt 3 \, in\).
In decimal form, this is approximately 3.46410161514 in.
Why It Works
The dotted line representing the height of an equilateral triangle creates two congruent right triangles that are each exactly half of the equilateral triangle.
The reason we know the two right triangles are congruent is because they share a leg (h) and their hypotenuses (x) are congruent, so the Hypotenuse Leg Theorem proves their congruence.
We can find the height of the triangle by applying the Pythagorean Theorem or the Law of Sines to these right triangles. And when we apply them to an equilateral triangle with side lengths of x, we end up with the formula \(h=\frac{\sqrt 3}{2}x\).
The Pythagorean Theorem says that \({\yellow a}^2+{\red b}^2={\blue c}^2\) for every right triangle where \(\yellow a\) and \(\red b\) are the legs of the triangle and \(\blue c\) is the hypotenuse.
The height of the equilateral triangle (\(\yellow h\)) is one of the legs of the right triangle.
The other leg of the right triangle is one half of the base of the triangle (\({\red \frac{1}{2}x}\)) because the dotted height line splits the base of the equilateral triangle exactly in half.
The hypotenuse of the right triangle is the side length of the equilateral triangle (\(\blue x\)).
When we plug these values into the Pythagorean Theorem, we get…
\[({\yellow h})^2+({\red \frac{1}{2}x})^2=({\blue x})^2\]
\[({\yellow h})^2+({\red \frac{1}{2}x})^2=({\blue x})^2\]
When we square \(\yellow h\) and \(\blue x\), we get \(\yellow h^2\) and \(\blue x^2\).
When we square \(\red \frac{1}{2}x\), we have to square each part individually, which gives us \(\red \frac{1}{4}x^2\).
So, the simplified Pythagorean Theorem is…
\[{\yellow h^2}+{\red \frac{1}{4}x^2}={\blue x^2}\]
\[{\yellow h^2}+{\red \frac{1}{4}x^2}={\blue x^2}\]
To get h by itself, we first need to subtract \(\red \frac{1}{4}x^2\) from both sides of the equation.
\[{\yellow h^2}={\blue x^2}-{\red \frac{1}{4}x^2}\]
Next, we can combine the like terms on the left side of the equation (\(\blue x^2\) and \(- \red \frac{1}{4}x^2\)) by subtracting \(\frac{1}{4}\) from the “invisible” coefficient of 1.
\[{\yellow h^2}=\frac{3}{4}x^2\]
Lastly, we will square root both sides of the equation to get h by itself.
\[{\yellow h}=\sqrt{\frac{3}{4}x^2}\]
The square root of \(x^2\) is x. The square root of 4 is 2. And the square root of 3 cannot be simplified.
So, the simplified expression is…
\[{\yellow h}=\frac{\sqrt 3}{2}x\]
The height of an equilateral triangle is \(\yellow \frac{\sqrt 3}{2}x\) where \(\blue x\) is the side length of the equilateral triangle.
The Law of Sines says that \(\frac{sin {\yellow A}}{\yellow a}=\frac{sin{\blue B}}{\blue b}\) where A and B are the angles opposite of side lengths of a and b.
In the right triangle formed by the height of the equilateral triangle, the height is opposite of the \(\yellow 60^\circ\) angle. And the hypotenuse is opposite of the right angle.
So, we can write the Law of Sines as…
\[\frac{sin \,{\yellow 60^\circ}}{\yellow h}=\frac{sin \,{\blue 90^\circ}}{\blue x}\]
\[\frac{sin \,{\yellow 60^\circ}}{\yellow h}=\frac{sin \,{\blue 90^\circ}}{\blue x}\]
The sine of \(\yellow 60^\circ\) is \(\yellow \frac{\sqrt{3}{2}\). And the sine of \(\blue 90^\circ\) is \(\blue 1\). So, the simplified Law of Sines is…
\[\frac{\yellow \frac{\sqrt{3}}{2}}{\yellow h}=\frac{\blue 1}{\blue x}\]
\[\frac{\yellow \frac{\sqrt{3}}{2}}{\yellow h}=\frac{\blue 1}{\blue x}\]
To solve for h, we can cross-multiply, which gives us…
\[({\yellow h})({\blue 1})=({\yellow\frac{\sqrt{3}}{2}})({\blue x})\]
Anything times 1 will equal itself, so the simplified equation is…
\[{\yellow h}={\yellow\frac{\sqrt{3}}{2}}{\blue x}\]
The height of an equilateral triangle is \(\yellow \frac{\sqrt 3}{2}x\) where \(\blue x\) is the side length of the equilateral triangle.
