Special Right Triangles: 45-45-90

Two right triangles show up so often that geometry gives them their own shortcuts: the 45-45-90 triangle and the 30-60-90 triangle, named by their angle measures. In each one, the sides always come in the same fixed ratio — so if you know any one side, you can find the other two with a single multiplication or division. No Pythagorean theorem required.

The 45-45-90 sides are in the ratio $x : x : x\sqrt{2}$ , and the 30-60-90 sides are in the ratio $x : x\sqrt{3} : 2x$ . These patterns are worth memorizing cold — they power half of right-triangle trig, they explain the exact values of $\sin 30^\circ$ , $\cos 45^\circ$ , and friends, and they turn many two-step Pythagorean problems into one-step problems.

The 45-45-90 triangle: x, x, x√2

A 45-45-90 triangle is half of a square, cut along the diagonal. Its two base angles are both $45^\circ$ , which makes it isosceles — the two legs are always equal. The hypotenuse is the leg times $\sqrt{2}$ .

So the rules are: leg to hypotenuse, multiply by $\sqrt{2}$ ; hypotenuse to leg, divide by $\sqrt{2}$ . Dividing leaves a radical in the denominator, so rationalize: $\dfrac{10}{\sqrt{2}} = \dfrac{10\sqrt{2}}{2} = 5\sqrt{2}$ .

You can confirm the pattern with the Pythagorean theorem: legs of $x$ and $x$ give $x^2 + x^2 = 2x^2$ , and $\sqrt{2x^2} = x\sqrt{2}$ .

The 30-60-90 triangle: x, x√3, 2x

A 30-60-90 triangle is half of an equilateral triangle, cut along an altitude. Its three sides are all different, so keeping track of which is which matters: the short leg sits opposite the $30^\circ$ angle, the long leg sits opposite the $60^\circ$ angle, and the hypotenuse sits opposite the right angle.

Everything runs through the short leg $x$ : the hypotenuse is twice the short leg ( $2x$ ), and the long leg is the short leg times $\sqrt{3}$ ( $x\sqrt{3}$ ). Given any side, get to $x$ first, then build the others.

The classic mistake is doubling the short leg to get the long leg. Doubling gives the hypotenuse — the long leg comes from multiplying by $\sqrt{3}$ , and since $\sqrt{3} \approx 1.73$ , the long leg is always shorter than the hypotenuse.

Working backward from a bigger side

Problems love to hand you the hypotenuse or the long leg and ask for something smaller. The moves are the reverse of the forward rules: from a 45-45-90 hypotenuse, divide by $\sqrt{2}$ to get a leg. From a 30-60-90 hypotenuse, divide by $2$ to get the short leg. From a 30-60-90 long leg, divide by $\sqrt{3}$ to get the short leg.

Whenever a division leaves a square root in the denominator, rationalize by multiplying top and bottom by that root: $\dfrac{9}{\sqrt{3}} = \dfrac{9\sqrt{3}}{3} = 3\sqrt{3}$ . And watch for gift-wrapped answers — if a 45-45-90 hypotenuse is given as $8\sqrt{2}$ , the leg is just $8$ , because the $\sqrt{2}$ is already built in.

Worked examples

Example 1: 45-45-90, finding the hypotenuse

Each leg of a 45-45-90 triangle measures $7$ . Find the hypotenuse.

Write the 45-45-90 ratio

x : x : x\sqrt{2}

The legs are

x = 7

, so multiply by

\sqrt{2}

for the hypotenuse

c = 7\sqrt{2}

Check with the Pythagorean theorem:

7^2 + 7^2 = 98

and

(7\sqrt{2})^2 = 49 \cdot 2 = 98

✓

Answer: $c = 7\sqrt{2}$

Example 2: 45-45-90, working backward

The hypotenuse of a 45-45-90 triangle measures $10$ . Find each leg.

Hypotenuse equals leg times

\sqrt{2}

x\sqrt{2} = 10

Divide both sides by

\sqrt{2}

x = \dfrac{10}{\sqrt{2}}

Rationalize the denominator

x = \dfrac{10\sqrt{2}}{2} = 5\sqrt{2}

Check:

5\sqrt{2} \cdot \sqrt{2} = 5 \cdot 2 = 10

✓

Answer: Each leg is $5\sqrt{2}$

Example 3: 30-60-90 from the short leg

The short leg of a 30-60-90 triangle measures $5$ . Find the other two sides.

Write the 30-60-90 ratio with the short leg first

x : x\sqrt{3} : 2x

The short leg is

x = 5

The hypotenuse is twice the short leg

2x = 10

The long leg is the short leg times

\sqrt{3}

x\sqrt{3} = 5\sqrt{3}

Answer: Long leg $= 5\sqrt{3}$ , hypotenuse $= 10$

Example 4: 30-60-90 from the long leg

The long leg of a 30-60-90 triangle measures $9$ . Find the short leg and the hypotenuse.

The long leg is the short leg times

\sqrt{3}

x\sqrt{3} = 9

Divide both sides by

\sqrt{3}

x = \dfrac{9}{\sqrt{3}}

Rationalize

x = \dfrac{9\sqrt{3}}{3} = 3\sqrt{3}

The hypotenuse is twice the short leg

2x = 6\sqrt{3}

Check the long leg:

3\sqrt{3} \cdot \sqrt{3} = 3 \cdot 3 = 9

✓

Answer: Short leg $= 3\sqrt{3}$ , hypotenuse $= 6\sqrt{3}$

Try one yourself

Common questions

How do I remember which leg is opposite which angle in a 30-60-90 triangle?

Smaller angle, smaller side: the shortest side always faces the smallest angle. So the short leg is opposite the $30^\circ$ angle, the long leg is opposite the $60^\circ$ angle, and the hypotenuse — always the longest side — is opposite the $90^\circ$ angle.

Can I just use the Pythagorean theorem instead?

For a 45-45-90 triangle with both legs known, yes — it gives the same answer with more arithmetic. But for a 30-60-90 triangle with only one side known, the Pythagorean theorem alone isn't enough, because it needs two sides. The ratio patterns work from a single side, which is exactly why they're worth memorizing.

Why do the answers have square roots in them?

Because the ratios themselves contain $\sqrt{2}$ and $\sqrt{3}$ , which are irrational. Exact answers like $5\sqrt{2}$ are preferred in geometry; only round to a decimal like $7.07$ if the problem asks for it. If a radical ends up in a denominator, rationalize it.

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