Right triangles (45-45-90)

A triangle with angle measurements of 45°, 45°, and 90° is called a 45-45-90 triangle. No matter how large or small the triangle is, the side lengths always stay in the same ratio.

In a 45-45-90 triangle:

• The two sides adjacent to the right angle (the legs) are equal.
• The side opposite the right angle (the hypotenuse) is the leg length multiplied by $\sqrt{2}$.

So if each leg has length $x$, the three sides are in the ratio:

• $x:x:x\sqrt{2}$

Let’s see this with a concrete example. Suppose the two legs are both $2$. Use the Pythagorean Theorem to find the hypotenuse.

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

This matches the ratio rule: if $x=2$, then the hypotenuse is $x\sqrt{2}=2\sqrt{2}$.

You can re-derive this using the Pythagorean Theorem, but on the GRE it’s usually faster to memorize the side ratio $x:x:x\sqrt{2}$ for a 45-45-90 triangle.

Example 45-45-90 triangle question

Let’s try a question that uses 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 first glance, you might think the quantities are equal: Square A has half the area, and Quantity A doubles its diagonal. But diagonal length doesn’t scale linearly with area, so it’s worth checking with numbers.

Assume:

• Area of Square A is $1$
• Area of Square B is $2$

Square A

If Square A has area $1$, then its side length is $1$ because $1\cdot 1=1$.

A square’s diagonal is the hypotenuse of a 45-45-90 triangle whose legs are the side lengths of the square. Using the ratio $x:x:x\sqrt{2}$ with $x=1$, the diagonal is:

• $1\sqrt{2}=\sqrt{2}$

Square B

If Square B has area $2$, and the area formula is $A=x^2$, then:

• $2=x^2\Rightarrow x=\sqrt{2}$

So the side length of Square B is $\sqrt{2}$. Again using the 45-45-90 ratio, the diagonal is:

• $x\sqrt{2}=\sqrt{2}\cdot \sqrt{2}=2$

So the diagonals are:

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

Now compare 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 $=2$

Since $2\sqrt{2}>2$, Quantity A is greater.

This relationship holds in every case. When one square has half the area of another, its side length (and therefore its diagonal) is smaller by a factor of $\sqrt{\frac{1}{2}}$, not by a factor of $\frac{1}{2}$. Because diagonals scale with side length, doubling Square A’s diagonal will always produce a value greater than Square B’s diagonal.