Recognize special right triangles

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Difficulty: Intermediate
Estimated time: 26 minutes
Domains: Geometry and Trigonometry
Techniques: Special-right-triangles45-45-9030-60-90Side-roles

What you’ll learn

Tell whether a right triangle is a $45^\circ$ - $45^\circ$ - $90^\circ$ or a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle before you pick a ratio.
Name each side as a leg, short leg, long leg or hypotenuse.
Scale $x:x:x\sqrt2$ and $x:x\sqrt3:2x$ to find exact lengths.
Find the special triangles hiding inside a square, a rectangle or an equilateral triangle.
Keep an exact radical unless the question asks for a decimal.

Prerequisites

Why this matters on the SAT

Spot the special triangle first

The SAT reference sheet shows both special right triangles with their side lengths. The hard part is spotting one when it’s hidden in a square’s diagonal, an equilateral triangle’s height, a ladder against a wall, or a rectangle. Once you see it, you name each side’s role and scale the ratio. Here’s a typical question.

Solution to the example

The diagonal cuts the square into two right triangles. In triangle $ABC$ , the legs are two sides of the square, so they’re equal. That makes it a $45^\circ$ - $45^\circ$ - $90^\circ$ triangle, where

\text{hypotenuse}=(\text{leg})\sqrt2.

Call the side length $s$ . The diagonal is the hypotenuse, so

s\sqrt2=18\sqrt2,

and $s=18$ . The perimeter is four sides:

4s=4(18)=72.

The answer is C. Notice that the diagonal wasn’t what the question asked for. It was your way to the side length. Choice D is what you get if you skip that step and treat the diagonal as a side.

SAT example

A square diagonal divides the square into two

45^\circ

45^\circ

90^\circ

triangles.

Square $ABCD$ has diagonal $\overline{AC}$ of length $18\sqrt2$ . What is the perimeter of the square?

A
$36$
B
$36\sqrt2$
C
$72$
D
$72\sqrt2$

Recognize first, then choose the ratio

Name each side by the angle across from it, not by where it sits on the page.

Before you multiply or divide anything, make two quick decisions.

Decision 1: Which special triangle is it?

A right angle alone isn’t enough. You need one more clue:

It’s a $45^\circ$ - $45^\circ$ - $90^\circ$ triangle if it has a $45^\circ$ angle, if its two legs are equal, or if it’s half of a square cut along the diagonal.
It’s a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle if it has a $30^\circ$ or $60^\circ$ angle, or if it’s half of an equilateral triangle cut by its height.

If you find neither clue, the special ratios don’t apply. When you know two sides, use the Pythagorean theorem. When some other angle, like $25^\circ$ , sets the sides, use trigonometry.

Decision 2: Which role does each side play?

The hypotenuse is across from the $90^\circ$ angle.
In a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, the short leg is across from $30^\circ$ .
The long leg is across from $60^\circ$ .

Each ratio is a side map. It gives every side’s length from one number, $x$ :

\begin{aligned} 45^\circ\text{-}45^\circ\text{-}90^\circ &: x,\ x,\ x\sqrt2,\\[1.4em] 30^\circ\text{-}60^\circ\text{-}90^\circ &: x,\ x\sqrt3,\ 2x. \end{aligned}

The second line runs short leg, long leg, hypotenuse. Turn or flip the triangle, and the roles stay the same.

Try it yourself:

A $30^\circ$ - $60^\circ$ - $90^\circ$ triangle has a hypotenuse of $26$ . Before you calculate, decide which part of the map $26$ is. Then find the short leg and the long leg.

Answer: The hypotenuse is $2x$ , so $2x=26$ and $x=13$ . That makes the short leg $13$ and the long leg $13\sqrt3$ .

Check your understanding:

A right triangle has legs of $9$ and $14$ . No acute angle is given, and it isn’t part of a square or an equilateral triangle. Should you use a special-right-triangle ratio?

Where the ratios come from

You don’t have to take either map on faith. Each one comes from geometry you already know, so you can rebuild it in a few lines.

The $45^\circ$ - $45^\circ$ - $90^\circ$ ratio

The two $45^\circ$ angles are equal, so the legs across from them are equal too. Call each leg $x$ and the hypotenuse $h$ . The Pythagorean theorem gives

\begin{aligned} h^2&=x^2+x^2\\[1.4em] &=2x^2,\\[1.4em] h&=x\sqrt2. \end{aligned}

That’s the map $x:x:x\sqrt2$ .

The $30^\circ$ - $60^\circ$ - $90^\circ$ ratio

Start with an equilateral triangle with sides of length $2x$ . All three of its angles are $60^\circ$ . Now draw its altitude, the height that runs from the top corner straight down to the base.

The altitude splits the triangle into two right triangles. Their hypotenuses are equal, because they’re sides of the equilateral triangle, and they share a leg, the altitude itself. So the two halves are congruent, meaning the same size and shape, by the hypotenuse-leg rule. Matching parts of congruent triangles are equal, so the altitude cuts the base into two pieces of length $x$ and cuts the top $60^\circ$ angle into two $30^\circ$ angles.

In either half, the short leg is $x$ and the hypotenuse is $2x$ . The Pythagorean theorem gives the long leg:

\sqrt{(2x)^2-x^2} =\sqrt{3x^2} =x\sqrt3.

That’s the map $x:x\sqrt3:2x$ .

Common mistake:

Since $60^\circ$ is twice $30^\circ$ , it’s tempting to make the long leg twice the short leg. Angles and sides don’t grow together like that. The long leg is $x\sqrt3$ , only about $1.7$ times the short leg. The side that’s twice the short leg is the hypotenuse, $2x$ .

Scale from the side you know

You don’t need a new formula for every pair of sides. One routine covers them all: match the side you know to its place in the map, find $x$ , then build the side you need.

$45^\circ$ - $45^\circ$ - $90^\circ$

If you know a leg, $L$ , the other leg is also $L$ , and the hypotenuse is $L\sqrt2$ .
If you know the hypotenuse, $H$ , solve $x\sqrt2=H$ . Multiplying the top and bottom by $\sqrt2$ moves the radical to the top:

x=\frac{H}{\sqrt2}=\frac{H\sqrt2}{2}.

For example, if the hypotenuse is $14$ , each leg is

\frac{14\sqrt2}{2}=7\sqrt2.

$30^\circ$ - $60^\circ$ - $90^\circ$

If you know the short leg, $S$ , the long leg is $S\sqrt3$ , and the hypotenuse is $2S$ .
If you know the hypotenuse, $H$ , halve it to get the short leg, $\frac H2$ . The long leg is then $\frac{H\sqrt3}{2}$ .
If you know the long leg, $L$ , solve $x\sqrt3=L$ first. Then $x$ is the short leg, and $2x$ is the hypotenuse.

Check your understanding:

In a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, the long leg is $15\sqrt3$ . What are the short leg and the hypotenuse?

Example: Turn an equilateral side into area

Worked example

The altitude splits the equilateral triangle into two congruent

30^\circ

60^\circ

90^\circ

triangles.

In equilateral triangle $ABC$ , $AB=12$ . The altitude $\overline{CM}$ meets $\overline{AB}$ at $M$ . What is the area of triangle $ABC$ ?

Step 1

Find the hidden special triangle

Every angle of an equilateral triangle is $60^\circ$ . Just as when we built the map, altitude $CM$ splits triangle $ABC$ into two congruent halves and cuts the top $60^\circ$ angle into two $30^\circ$ angles.

So triangle $ACM$ has angles of $30^\circ$ , $60^\circ$ and $90^\circ$ . It’s a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle.

Step 2

Name each side’s role

The altitude also cuts the base in half. Since $AB=12$ ,

AM=MB=6.

Now name each side of triangle $ACM$ by the angle across from it:

$AM=6$ is across from $30^\circ$ , so it’s the short leg, $x$ .
$CM=h$ is across from $60^\circ$ , so it’s the long leg, $x\sqrt3$ .
$AC=12$ is across from the right angle, so it’s the hypotenuse, $2x$ .

Step 3

Scale the ratio

The short leg is $x=6$ , so the long leg is

h=x\sqrt3=6\sqrt3.

As a check, $2x=12$ matches the hypotenuse $AC$ . Leave the height as $6\sqrt3$ . It’s exact, and you don’t need a decimal.

Step 4

Finish with the area

The full triangle has base $12$ and height $6\sqrt3$ :

\begin{aligned} \text{Area} &=\frac12 bh\\[1.4em] &=\frac12(12)(6\sqrt3)\\[1.4em] &=36\sqrt3. \end{aligned}

So the area of triangle $ABC$ is

\boxed{36\sqrt3}.

The special triangle only gave you the height. The area formula finished the job.

Check your understanding:

Suppose the equilateral triangle’s side were $24$ instead of $12$ . How would the altitude and the area change?

Keep radicals exact

Once you’ve spotted the triangle, working by hand is usually the shortest reliable way, so what’s left to watch is the form of your answer. When the choices or the answer box expect an exact value, keep $\sqrt2$ and $\sqrt3$ as they are. For example,

10\sqrt3

is exact, while $17.32$ is only close: $10\sqrt3$ is $17.3205\ldots$ , and the digits never end.

Reach for the calculator only when the question asks for a decimal or when you want to check a final value. You could also check these triangles with trigonometry, typing in $\sin 30^\circ$ , $\cos45^\circ$ or $\tan60^\circ$ , but once the exact ratio is in front of you, that’s extra setup you don’t need.

Common mistake:

If you swap $\sqrt3$ for $1.73$ halfway through, your answer won’t match any exact choice. Carry the radical through the whole solution, and round only if the question names a rounding place or asks for a decimal.

Practice problems

Before you calculate, name the special triangle and the role of the side you’re given. The last problem links two special triangles, so take it one shape at a time. Each calculator starts blank, and using it is optional.

Scale a 45-45-90 triangle

Practice problem

In right triangle $JKL$ , $\angle K=90^\circ$ and $\angle J=45^\circ$ . If $JK=11$ , which choice gives the length of hypotenuse $\overline{JL}$ ?

Calculator loads as you approach

This one is faster by hand. Use Desmos only if you want to check the value.

Find a rectangle’s perimeter

Practice problem

A rectangle has a diagonal of length $20$ centimeters. The angle between the diagonal and one longer side of the rectangle is $30^\circ$ . Which choice gives the perimeter of the rectangle, in centimeters?

Calculator loads as you approach

The diagonal is the hypotenuse. Keep the side lengths exact.

Connect two special triangles

Practice problem

An equilateral triangle has height $18\sqrt3$ centimeters. A square has a diagonal equal in length to one side of the equilateral triangle. What is the perimeter, in centimeters, of the square?

Your answer

Calculator loads as you approach

Work this one by hand. Use Desmos only to check a decimal approximation.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

A right angle alone isn’t enough. Confirm the $45^\circ$ - $45^\circ$ - $90^\circ$ or $30^\circ$ - $60^\circ$ - $90^\circ$ pattern before you use a ratio.
In a $45^\circ$ - $45^\circ$ - $90^\circ$ triangle, the sides are $x:x:x\sqrt2$ .
In a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, the short leg, long leg and hypotenuse are $x:x\sqrt3:2x$ .
Name each side by the angle across from it, not by how the triangle is turned.
A square’s diagonal makes two $45^\circ$ - $45^\circ$ - $90^\circ$ triangles. An equilateral triangle’s altitude makes two $30^\circ$ - $60^\circ$ - $90^\circ$ triangles.
Find $x$ , build the side you need, then finish what the question asks for, like a perimeter or an area.
Keep $\sqrt2$ and $\sqrt3$ exact unless the question asks for a decimal.

Next lesson

Use sine, cosine, and tangent

Handle right triangles whose acute angles do not create one of the two special exact ratios.

Start next lesson

Practice

Practice this lesson

116 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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What you’ll learn

Prerequisites

Spot the special triangle first

Recognize first, then choose the ratio

Decision 1: Which special triangle is it?

Decision 2: Which role does each side play?

Where the ratios come from

The 45∘45^\circ45∘-45∘45^\circ45∘-90∘90^\circ90∘ ratio

The 30∘30^\circ30∘-60∘60^\circ60∘-90∘90^\circ90∘ ratio

Scale from the side you know

45∘45^\circ45∘-45∘45^\circ45∘-90∘90^\circ90∘

30∘30^\circ30∘-60∘60^\circ60∘-90∘90^\circ90∘

Example: Turn an equilateral side into area

Find the hidden special triangle

Name each side’s role

Scale the ratio

Finish with the area

Keep radicals exact

Practice problems

Scale a 45-45-90 triangle

Find a rectangle’s perimeter

Connect two special triangles

3 practice examples left

Quick recap

Use sine, cosine, and tangent

Practice this lesson

The $45^\circ$ - $45^\circ$ - $90^\circ$ ratio

The $30^\circ$ - $60^\circ$ - $90^\circ$ ratio

$45^\circ$ - $45^\circ$ - $90^\circ$

$30^\circ$ - $60^\circ$ - $90^\circ$