And according to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides (called legs). So, for the triangle above –

In this tutorial, we’ll learn how to work with and solve certain special right triangles. Let’s start with a quick recap of what right triangles are.

A right triangle (or right-angled triangle) is a triangle in which one of the three internal angles is a right angle ($\hspace{0.2em} 90 \degree$).

The longest side in a right triangle, known as the hypotenuse, is the side opposite the right angle.

And according to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides (called legs). So, for the triangle above –

$BC^2 + AC^2 = {\color{Red} AB} ^2$

Also, in every right triangle, the two non-right angle are acute (less than 90o) and their sum must be $\hspace{0.2em} 90 \degree$. Otherwise, the sum of the angles wouldn’t be $\hspace{0.2em} 180 \degree$ – something impossible.

Here are two examples. Notice how in each case the two acute angles add to give a right angle.

And that brings us to the two special right triangles we want to talk about.

Certain special cases of right triangles present themselves so frequently in the study of triangles and geometry that it makes sense to study them separately and make a note of some important conclusions.

The first special right triangle we have is one in which one angle is $\hspace{0.2em} 90 \degree$ (of course) and the other two are $\hspace{0.2em} 45 \degree$ each.

As you can see in the figure above, each leg of a $\hspace{0.2em} \text{45-45-90} \hspace{0.2em}$ triangle has the same length – sides opposite equal angles ($\hspace{0.2em} 45 \degree$ each).

And using the Pythagorean theorem, we can see that the length of the hypotenuse would be $\hspace{0.2em} \sqrt{2} \hspace{0.2em}$ the length of a leg.

So, if the length of one leg is $\hspace{0.2em} k \hspace{0.2em}$, that of the other leg will also be $\hspace{0.2em} k \hspace{0.2em}$, and the hypotenuse will have a length of $\hspace{0.2em} \sqrt{2}k \hspace{0.2em}$.

A right isosceles triangle is a $\hspace{0.2em} \text{45-45-90} \hspace{0.2em}$ triangle.

The second special right triangle on our list is one in which one angle is $\hspace{0.2em} 90 \degree$ (again, of course) and the other two are $\hspace{0.2em} 30 \degree$ and $\hspace{0.2em} 60 \degree$.

As you can see in the figure above, in a $\hspace{0.2em} \text{30-60-90} \hspace{0.2em}$ triangle -

If the length of the shorter leg is $\hspace{0.2em} k \hspace{0.2em}$, that of the longer leg would be $\hspace{0.2em} \sqrt{3}k \hspace{0.2em}$, and the hypotenuse will have a length of $\hspace{0.2em} \sqrt{2}k \hspace{0.2em}$.

Consider the $\hspace{0.2em} \text{30-60-90} \hspace{0.2em}$ triangle below. The length of its shorter leg is $\hspace{0.2em} k \hspace{0.2em}$.

If we make a copy of the triangle and join the two as shown below, we will end up with an equilateral triangle (each angle would be equal to $\hspace{0.2em} 60 \degree$).

Now, in this equilateral triangle, the side $\hspace{0.2em} BD \hspace{0.2em}$ would have a length of$\hspace{0.2em} k + k \hspace{0.2em}$, or $\hspace{0.2em} 2k \hspace{0.2em}$. And since the sides of an equilateral triangle are all equal, the lengths of sides $\hspace{0.2em} AB \hspace{0.2em}$ and $\hspace{0.2em} AD \hspace{0.2em}$ would also be $\hspace{0.2em} 2k \hspace{0.2em}$.

Using the Pythagorean theorem in $\hspace{0.2em} \triangle ABC \hspace{0.2em}$ (or $\hspace{0.2em} \triangle ACD \hspace{0.2em}$), we can see that the length of $\hspace{0.2em} AC \hspace{0.2em}$ (the longer leg of our original triangle) would be $\hspace{0.2em} \sqrt{3}k \hspace{0.2em}$.

Example

Find the lengths of BC and AC in the triangle below.

Solution

Here, we have a right triangle in which one acute angle is $\hspace{0.2em} 60 \degree$. That means we have a $\hspace{0.2em} \text{30-60-90} \hspace{0.2em}$ triangle.

And we know that in such a triangle the sides are $\hspace{0.2em} k \hspace{0.2em}$ (shorter leg), $\hspace{0.2em} \sqrt{3}k \hspace{0.2em}$ (longer leg), and $\hspace{0.2em} 2k \hspace{0.2em}$ (hypotenuse) — where $\hspace{0.2em} k \hspace{0.2em}$ is some constant.

Now, the question tells us that the length of the hypotenuse is $\hspace{0.2em} 8 \hspace{0.2em}$. So,

$\begin{align*} 2k &= 8 \\[1em] k &= 4 \end{align*}$

Having calculated the value of $\hspace{0.2em} k \hspace{0.2em}$, we can find the values of the two legs.

$\begin{align*} \text{shorter leg} &= k \\[1em] &= 4 \end{align*}$

$\begin{align*} \text{longer leg} &= \sqrt{3}k \\[1em] &= 4 \sqrt{3} \\[1em] & \approx 6.93 \end{align*}$

So, the length of $\hspace{0.2em} AC \hspace{0.2em}$ is $\hspace{0.2em} 4 \hspace{0.2em}$ and that of $\hspace{0.2em} BC \hspace{0.2em}$ is $\hspace{0.2em} 6.93 \hspace{0.2em}$.

Example

If one leg of a right isosceles triangle measures $\hspace{0.2em} 5 \hspace{0.2em}$ inches, what is the length of the hypotenuse?

Solution

If you remember from earlier, a right isosceles triangle is a $\hspace{0.2em} \text{45-45-90} \hspace{0.2em}$ triangle. And such a triangle has side lengths $\hspace{0.2em} k \hspace{0.2em}$ (each leg) and $\hspace{0.2em} \sqrt{2}k \hspace{0.2em}$ (hypotenuse).

According to the question, one leg of the triangle has a length of $\hspace{0.2em} 5 \hspace{0.2em}$ in. So,

$k = 5$

Therefore, the length of the hypotenuse would be -

$\begin{align*} \sqrt{2} k &= 5 \sqrt{2} \\[1em] &\approx 7.07 \end{align*}$

So, the length of the hypotenuse is $\hspace{0.2em} 7.07 \text{ in} \hspace{0.2em}$.

And with that, we come to the end of this tutorial on special right triangles.

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