Math Core

Two right triangles show up so often in geometry, trigonometry and design that it pays to know their side ratios by heart: the 45°-45°-90° triangle and the 30°-60°-90° triangle. Once you know the patterns, you can find every side from just one, with exact answers and no calculator.

The 45°-45°-90° triangle

Cut a square along a diagonal and you get two copies of an isosceles right triangle. Its two legs are equal, and its acute angles are both $45^\circ$ .

Call each leg $s$ . The Pythagorean theorem gives the hypotenuse:

$c^2 = s^2 + s^2 = 2s^2 \quad\Longrightarrow\quad c = \sqrt{2s^2} = s\sqrt{2}.$

A 45°-45°-90° triangle: legs s and s, hypotenuse s√2.

45°-45°-90° triangle

The legs are equal, and the hypotenuse is $\sqrt{2}$ times a leg.

$\text{leg} : \text{leg} : \text{hypotenuse} = s : s : s\sqrt{2}$

Leg to hypotenuse: multiply by $\sqrt{2}$ .
Hypotenuse to leg: divide by $\sqrt{2}$ .

Worked example: From leg to hypotenuse

A 45°-45°-90° triangle has a leg of length $7$ . Find the other sides.

The other leg is also $7$ . The hypotenuse is $7\sqrt{2} \approx 9.90$ .

Going the other way means dividing by $\sqrt{2}$ . You'll usually want to rationalize the denominator: multiply the top and bottom by $\sqrt{2}$ so that no radical is left in the denominator.

Worked example: From hypotenuse to leg

The hypotenuse of a 45°-45°-90° triangle is $10$ . Find the length of each leg.

$s = \frac{10}{\sqrt{2}} = \frac{10}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$

Each leg is $5\sqrt{2} \approx 7.07$ . Check: $(5\sqrt{2})^2 + (5\sqrt{2})^2 = 50 + 50 = 100 = 10^2$ . ✓

A handy shortcut: the leg is always half the hypotenuse times $\sqrt{2}$ . So a hypotenuse of $10$ gives legs of $5\sqrt{2}$ , and a hypotenuse of $8$ gives legs of $4\sqrt{2}$ .

The 30°-60°-90° triangle

Start with an equilateral triangle with side length $2s$ . All its angles are $60^\circ$ . Draw an altitude from the top vertex. It lands at the midpoint of the base and cuts the triangle into two congruent right triangles.

An equilateral triangle with side 2s. The altitude h cuts it into two 30°-60°-90° triangles.

Look at one half. Its angles are $60^\circ$ (an original corner), $90^\circ$ (where the altitude meets the base) and $30^\circ$ (half of the top angle). Its sides are:

the hypotenuse, an original side: $2s$
the short leg, half the base: $s$
the long leg, the altitude $h$ : by the Pythagorean theorem, $h^2 = (2s)^2 - s^2 = 3s^2$ , so $h = s\sqrt{3}$ .

A 30°-60°-90° triangle: short leg s, hypotenuse 2s, long leg s√3.

30°-60°-90° triangle

$\text{short leg} : \text{long leg} : \text{hypotenuse} = s : s\sqrt{3} : 2s$

The short leg is opposite the $30^\circ$ angle. Always find it first.
The hypotenuse is twice the short leg.
The long leg (opposite $60^\circ$ ) is the short leg times $\sqrt{3}$ .

Worked example: Starting from the short leg or the hypotenuse

The short leg of a 30°-60°-90° triangle is $4$ . Find the other two sides.
The hypotenuse of a 30°-60°-90° triangle is $18$ . Find the other two sides.

Solutions.

Hypotenuse $= 2 \cdot 4 = 8$ . Long leg $= 4\sqrt{3} \approx 6.93$ .
Short leg $= 18 \div 2 = 9$ . Long leg $= 9\sqrt{3} \approx 15.59$ .

The trickiest case is when you're given the long leg. Divide by $\sqrt{3}$ to get back to the short leg, then double for the hypotenuse.

Worked example: Starting from the long leg

In the triangle below, the side of length $9$ is opposite the $60^\circ$ angle. Find $x$ and $y$ .

The side of length $9$ is the long leg, so the short leg is

$x = \frac{9}{\sqrt{3}} = \frac{9\sqrt{3}}{3} = 3\sqrt{3}.$

The hypotenuse is twice the short leg: $y = 2 \cdot 3\sqrt{3} = 6\sqrt{3}$ .

Check with the Pythagorean theorem: $(3\sqrt{3})^2 + 9^2 = 27 + 81 = 108$ and $(6\sqrt{3})^2 = 36 \cdot 3 = 108$ . ✓

Common mistake

Don't mix up the two patterns. The $\sqrt{2}$ belongs only to the 45°-45°-90° triangle; the 30°-60°-90° triangle uses $2$ and $\sqrt{3}$ . And in a 30°-60°-90° triangle, the hypotenuse is twice the short leg, not twice the long one.

Tip

Sanity check with sizes: the longest side is always opposite the biggest angle. In a 30°-60°-90° triangle, $s < s\sqrt{3} < 2s$ because $1 < 1.73 < 2$ . If your "long leg" came out longer than the hypotenuse, something went wrong.

Practice

Practice 1

A 45°-45°-90° triangle has legs of length $5$ . Find the exact length of the hypotenuse. (Type a square root as sqrt(…), for example 3sqrt(2).)