Right triangles (45-45-90)

A triangle with angle measurements of 45, 45, and 90 degrees is called a 45-45-90 triangle. The relative measurements of the sides and angles will always be in the same ratio as the figure below.

Regardless of the size of the triangle, these ratios will always be consistent. The two sides adjacent to the right angle are always equal, and the side opposite the right angle (the hypotenuse) is always equal to the smaller side length $x$ multiplied by $\sqrt{2}$, i.e., $x\sqrt{2}$.

Let’s consider a concrete example. Imagine if the two smaller side lengths were $2$, and we used the Pythagorean Theorem to find the length of the hypotenuse.

$$\begin{array}{rl}{a}^2+b^2& =c^2\\ 2^2+2^2& =c^2\\ 4+4& =c^2\\ 8& =c^2\\ 2\sqrt{2}& =c\end{array}$$

It’s the same! So while you could derive these equations using the Pythagorean Theorem, it’s better to simply memorize this $x:x:x\sqrt{2}$ ratio of the 3 sides of a 45-45-90 triangle.

Example 45-45-90 triangle question

Let’s try a question that involves 45-45-90 triangles.

Square A has half the area of Square B.

Quantity A: Twice the length of the diagonal of Square A
Quantity B: The length of the diagonal of Square B

Give it a try!

At a glance, it may seem likely that the two quantities are equal because Quantity A doubles the diagonal of the square with half the area. However, this logic isn’t entirely accurate, and you can check that by plugging in numbers for the area of the two squares.

Let’s imagine that Square A has an area of $1$, which means that Square B has an area of $2$.

If Square A has an area of $1$, its side lengths would also be $1$, since $1(1)=1$. The diagonal of a square is the hypotenuse of a 45-45-90 triangle, which has the side ratios of $x:x:x\sqrt{2}$. Given that the short side lengths $x$ are $1$, the length of the diagonal must be $1\sqrt{2}$, which is simply $\sqrt{2}$.

Now let’s take a look at Square B. If Square B has an area of $2$, and the area is $A=x^2$, we can plug in $2=x^2$ to get the side length of $x=\sqrt{2}$. And again, given that the side lengths of the square are also the short side lengths of a 45-45-90 triangle, the length of the diagonal must be $x\ast \sqrt{2}=\sqrt{2}\ast \sqrt{2}=2$.

We know that was a lot of steps in a row, especially written out in text, but it’s essential that you know how to break down these questions. If you’re unclear, please take a moment to slowly re-read the last two paragraphs to understand how we reached these values for the diagonals:

• Diagonal of Square A: $\sqrt{2}$
• Diagonal of Square B: $2$

Now that we’ve defined the diagonals, we just need to come back to the quantities.

• Quantity A: Twice the length of the diagonal of Square A $=2\sqrt{2}$
• Quantity B: The length of the diagonal of Square B $=2$

Clearly, $2\sqrt{2}$ is greater than $2$, so Quantity A is greater.

The final step is to ask if this is true in every case, and the answer is yes.

The length of the diagonals is always based on the ratio of the side lengths, and the ratio of the side lengths is always consistent with the constraints of the question. This means double the diagonal of Square A will always be greater than the diagonal of Square B.