How do I remember which triangle uses sqrt(2) and which uses sqrt(3)?

Match the count. The 45^-45^-90^ has two equal angles and comes from cutting a square (a 2-sided idea) — it uses sqrt(2). The 30^-60^-90^ comes from cutting an equilateral triangle, a 3-sided figure — it uses sqrt(3). You can also re-derive either one in about fifteen seconds with the Pythagorean Theorem if you forget.

Can I just use the Pythagorean Theorem instead of memorizing the ratios?

Yes, if you are given two sides. But these problems usually give you only one side plus the angles, and the Pythagorean Theorem alone cannot finish that — you would need trigonometry. The ratios let you get exact radical answers from a single side with no calculator, and they are much faster on multi-step problems involving areas, perimeters, and polygons.

Do I always have to rationalize the denominator?

Mathematically, 8sqrt(3) and 8sqrt(3)3 are the same number. Most geometry classes require the rationalized form in simplest radical form, so unless your teacher says otherwise, clear the radical out of the denominator by multiplying the top and bottom by that radical.

How can I tell whether a triangle in a problem is actually a special right triangle?

Look for a guarantee, not an appearance. Marked angles of 30^, 45^, or 60^ with a right angle qualify. So do constructions: a diagonal of a square, an altitude of an equilateral triangle, a radius-and-apothem pair in a regular hexagon, or an isosceles triangle with a right angle. If the figure only looks like a half-square, you cannot assume it.

GEOM-7.2

Special Right Triangles: 45-45-90 & 30-60-90

Learn to derive and use the 45-45-90 ratio x : x : x√2 and the 30-60-90 ratio x : x√3 : 2x to find every side of a special right triangle from just one.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Special Right Triangles: 45-45-90 & 30-60-90, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Special Right Triangles: 45-45-90 & 30-60-90 free

What this lesson covers

Every isosceles right triangle in the world has exactly the same shape, and so does every triangle you get by slicing an equilateral triangle in half. Because the shapes are fixed, so are the ratios of their sides — which means you can find all three sides from just one, with no calculator and no trig. These two triangles, the

45^\circ

45^\circ

90^\circ

and the

30^\circ

60^\circ

90^\circ

, show up constantly in geometry: in squares and their diagonals, in equilateral triangles and their heights, in regular hexagons, and later in the unit circle.

In this lesson you will derive both ratios from the Pythagorean Theorem (so you never have to trust your memory blindly), practice going forward from a leg and backward from a hypotenuse, and learn to spot the mistakes that trip most students up — especially mixing up which leg is which in a

30^\circ

60^\circ

90^\circ

Deriving the 45-45-90 Ratio from a Square

Draw a square with side length

x

and cut it along a diagonal. Each half is a right triangle whose two legs are the sides of the square, so both legs equal

x

. Since the two legs are congruent, the base angles are congruent, and because they must add to

90^\circ

, each is

45^\circ

. That is why this triangle is also called an isosceles right triangle.

Now apply the Pythagorean Theorem to find the hypotenuse

h

x^2 + x^2 = h^2 \quad\Rightarrow\quad 2x^2 = h^2 \quad\Rightarrow\quad h = x\sqrt{2}

(We take the positive root because a length cannot be negative.) So the side ratio is

\text{leg} : \text{leg} : \text{hypotenuse} = x : x : x\sqrt{2}

Two rules follow, and they are inverses of each other. To go from a leg to the hypotenuse, multiply by

\sqrt{2}

. To go from the hypotenuse back to a leg, divide by

\sqrt{2}

.

Since

\sqrt{2} \approx 1.414

, the hypotenuse is always about 1.4 times a leg — a quick sanity check. If you start from a leg of 6 and get 4.2, you divided when you should have multiplied.

When you divide by

\sqrt{2}

, most classes ask you to rationalize the denominator:

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

A hypotenuse of 10 gives legs of

5\sqrt{2} \approx 7.07

, which is comfortably less than 10. Good.

The most common error here is calling both non-right sides "legs" when one of them is drawn as the slanted side. Identify the hypotenuse by the right angle it faces, not by how the picture is tilted.

Deriving the 30-60-90 Ratio from an Equilateral Triangle

Start with an equilateral triangle whose sides are all

2x

and whose angles are all

60^\circ

. Drop an altitude from one vertex to the opposite side. That altitude is also a median and an angle bisector, so it cuts the base into two segments of length

x

and splits the

60^\circ

vertex angle into two

30^\circ

angles.

Each half is a right triangle with angles

30^\circ

60^\circ

, and

90^\circ

. Its short leg is

x

(half the original side), its hypotenuse is

2x

(a full original side), and its long leg

a

is the altitude:

x^2 + a^2 = (2x)^2 \quad\Rightarrow\quad a^2 = 4x^2 - x^2 = 3x^2 \quad\Rightarrow\quad a = x\sqrt{3}

So the ratio, written from shortest side to longest, is

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

The key to using this correctly is remembering that sides are ordered the same way as the angles they face. The short leg is across from

30^\circ

, the long leg is across from

60^\circ

, and the hypotenuse is across from

90^\circ

. Never assign sides by position on the page.

Think of the short leg as the hub. Everything routes through it:

You know	To get the short leg	Then
Short leg $x$	already have it	long leg $=x\sqrt{3}$ , hyp $=2x$
Long leg $\ell$	$x = \ell/\sqrt{3}$	hyp $=2x$
Hypotenuse $h$	$x = h/2$	long leg $=x\sqrt{3}$

Since

\sqrt{3} \approx 1.732

, a triangle with short leg 5 has sides about 5, 8.66, and 10 — check that your answers grow in that order.

Working Backward: From Hypotenuse or Long Leg

Going forward is easy; going backward is where errors pile up. The safe method is always the same: first solve for $x$ , then build the other sides from $x$ . Do not try to jump directly from the hypotenuse to the long leg.

Suppose a

30^\circ

60^\circ

90^\circ

triangle has hypotenuse 18. Set

2x = 18

, so

x = 9

. The short leg is 9 and the long leg is

9\sqrt{3} \approx 15.6

. A very common wrong answer is 18 divided by

\sqrt{3}

, which comes from grabbing the wrong relationship.

Now suppose the long leg is 12. Set

x\sqrt{3} = 12

, so

x = \frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}

The short leg is

4\sqrt{3} \approx 6.93

and the hypotenuse is

8\sqrt{3} \approx 13.86

. Notice

6.93 < 12 < 13.86

: the ordering confirms the answer.

Radical arithmetic that shows up constantly:

\sqrt{2}\cdot\sqrt{2} = 2, \qquad \sqrt{3}\cdot\sqrt{3} = 3, \qquad \frac{a}{\sqrt{b}} = \frac{a\sqrt{b}}{b}

So if a

45^\circ

45^\circ

90^\circ

triangle has a leg of

7\sqrt{2}

, the hypotenuse is

7\sqrt{2}\cdot\sqrt{2} = 14

— a clean whole number, which is exactly why problems are written that way.

One more habit worth building: after every answer, do a decimal reality check. In a

45^\circ

45^\circ

90^\circ

, the hypotenuse must be between one leg and two legs. In a

30^\circ

60^\circ

90^\circ

, the hypotenuse must be exactly double the short leg and a bit more than the long leg. If your numbers violate that, you multiplied where you should have divided.

Recognizing Special Triangles Inside Bigger Figures

Problems rarely hand you a labeled triangle. More often, a special right triangle is hiding inside a square, a rectangle, an equilateral triangle, a rhombus, or a hexagon, and your job is to find it.

A square's diagonal creates two

45^\circ

45^\circ

90^\circ

triangles, so a square with side

s

has diagonal

s\sqrt{2}

. Reversing that, a square with diagonal 20 has side

10\sqrt{2}

, and therefore area

(10\sqrt{2})^2 = 200

square units.

An equilateral triangle's altitude creates two

30^\circ

60^\circ

90^\circ

triangles. For side length

s

, the altitude is

\frac{s\sqrt{3}}{2}

, so the area is

A = \frac{1}{2}\cdot s \cdot \frac{s\sqrt{3}}{2} = \frac{s^2\sqrt{3}}{4}

A regular hexagon splits into six equilateral triangles, so its apothem is a

30^\circ

60^\circ

90^\circ

long leg. The diagonal of a rhombus, the height of an isosceles trapezoid with

45^\circ

base angles, the height of a ramp at a

30^\circ

incline — all of these reduce to the same two ratios.

Why use ratios instead of the Pythagorean Theorem or trigonometry? All three give the same answers. The special ratios are faster, they give exact radical answers instead of rounded decimals, and they are the foundation for the sine and cosine values you will meet in the next lessons:

\sin 30^\circ = \frac{1}{2}

and

\sin 45^\circ = \frac{\sqrt{2}}{2}

come straight from these triangles.

Where students go wrong: assuming a triangle is special because it looks like it. You need marked angles, an isosceles right triangle, or a construction (an altitude of an equilateral triangle, a diagonal of a square) that guarantees it.

Choosing a Strategy and Avoiding the Classic Mistakes

Before computing, run a three-question check. Which angles are given or forced? Which side am I given — leg, short leg, long leg, or hypotenuse? What is

x

Situation	First step	Result
$45$ - $45$ - $90$ , leg $=8$	$x=8$	hyp $=8\sqrt{2}$
$45$ - $45$ - $90$ , hyp $=8$	$x\sqrt{2}=8$	legs $=4\sqrt{2}$
$30$ - $60$ - $90$ , short leg $=8$	$x=8$	long $=8\sqrt{3}$ , hyp $=16$
$30$ - $60$ - $90$ , long leg $=8$	$x\sqrt{3}=8$	short $=\frac{8\sqrt{3}}{3}$ , hyp $=\frac{16\sqrt{3}}{3}$
$30$ - $60$ - $90$ , hyp $=8$	$2x=8$	short $=4$ , long $=4\sqrt{3}$

Four mistakes account for most lost work.

First, using

\sqrt{2}

in a

30^\circ

60^\circ

90^\circ

\sqrt{3}

in a

45^\circ

45^\circ

90^\circ

. Anchor it: two equal angles goes with

\sqrt{2}

; three distinct angles goes with

\sqrt{3}

.

Second, doubling the long leg to get the hypotenuse. Only the short leg doubles. If the long leg is 10, the hypotenuse is not 20 — it is about 11.5.

Third, forgetting to rationalize when your class requires it, or rationalizing incorrectly:

\frac{5}{\sqrt{3}}

becomes

\frac{5\sqrt{3}}{3}

, not

\frac{5\sqrt{3}}{\sqrt{3}}

and not

\frac{\sqrt{15}}{3}

.

Fourth, rounding too early. Keep radicals exact until the final line; if the problem asks for a decimal, round once at the end.

A final habit: label the picture as you go. Write

x

next to the short leg or the shared leg, then write the other two sides in terms of that same

x

. Solving one equation is much safer than juggling three unlabeled lengths.

Key terms

45-45-90 triangle.: An isosceles right triangle; half of a square cut along its diagonal. Its sides are in the ratio $x : x : x\sqrt{2}$ , with the hypotenuse equal to a leg times $\sqrt{2}$ .
30-60-90 triangle.: Half of an equilateral triangle, formed by an altitude. Its sides are in the ratio $x : x\sqrt{3} : 2x$ for the short leg, long leg, and hypotenuse.
Short leg.: In a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, the side opposite the $30^\circ$ angle. It equals half the hypotenuse and serves as the $x$ that all other sides are built from.
Long leg.: In a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, the side opposite the $60^\circ$ angle; it equals the short leg times $\sqrt{3}$ .
Hypotenuse.: The side opposite the right angle, always the longest side of a right triangle.
Rationalizing the denominator.: Rewriting a fraction so no radical remains in the denominator, by multiplying numerator and denominator by that radical, as in $\frac{6}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$ .
Altitude of an equilateral triangle.: A segment from a vertex perpendicular to the opposite side; it bisects both that side and the vertex angle, and has length $\frac{s\sqrt{3}}{2}$ for side length $s$ .
Exact value.: An answer left in radical or fraction form, such as $5\sqrt{3}$ , rather than a rounded decimal like 8.66.

Worked example

An equilateral triangle has an altitude of 9 centimeters. Find the side length of the equilateral triangle and the perimeter, giving exact answers in simplest radical form, then a decimal rounded to the nearest tenth.

Draw the equilateral triangle and its altitude. The altitude splits it into two congruent

30^\circ

60^\circ

90^\circ

triangles, because it bisects the

60^\circ

vertex angle into two

30^\circ

angles and meets the base at a right angle.

Identify the parts. In each half, the altitude is the long leg (it is opposite the

60^\circ

base angle), half of the base is the short leg, and a full side of the equilateral triangle is the hypotenuse.

Solve for

x

, the short leg. Since the long leg is

x\sqrt{3}

x\sqrt{3} = 9 \quad\Rightarrow\quad x = \frac{9}{\sqrt{3}} = \frac{9\sqrt{3}}{3} = 3\sqrt{3}

Build the hypotenuse. The hypotenuse is

2x

, so the side of the equilateral triangle is

2(3\sqrt{3}) = 6\sqrt{3} \text{ centimeters}

Find the perimeter. All three sides are equal, so

P = 3(6\sqrt{3}) = 18\sqrt{3} \text{ centimeters}

Convert and check. Since

\sqrt{3} \approx 1.732

, the side is about

10.4

centimeters and the perimeter is about

31.2

centimeters. Sanity check: the altitude (9) must be shorter than a side (10.4), and it is. A common wrong answer is a side of

9\sqrt{3} \approx 15.6

, which comes from treating the altitude as the short leg instead of the long leg — the altitude is always opposite the

60^\circ

angle.

Practice questions

In a

45^\circ

45^\circ

90^\circ

triangle, the hypotenuse measures 14. What is the length of each leg in simplest radical form?

$7$
$7\sqrt{2}$
$14\sqrt{2}$
$7\sqrt{3}$

Answer: $7\sqrt{2}$

The ratio is

x : x : x\sqrt{2}

, so set

x\sqrt{2} = 14

and solve:

x = \frac{14}{\sqrt{2}} = \frac{14\sqrt{2}}{2} = 7\sqrt{2} \approx 9.9

. Check the size: each leg must be less than 14 but more than half of 14, and 9.9 fits. Choosing 7 means you halved the hypotenuse, which is the

30^\circ

60^\circ

90^\circ

rule, not this one. Choosing

14\sqrt{2}

means you multiplied by

\sqrt{2}

instead of dividing, giving a leg longer than the hypotenuse — impossible. And

\sqrt{3}

never appears in an isosceles right triangle.

30^\circ

60^\circ

90^\circ

triangle has a long leg of 15 inches. Find the short leg and the hypotenuse in simplest radical form, and explain how you know which side is which.

Answer: Short leg $= 5\sqrt{3}$ inches; hypotenuse $= 10\sqrt{3}$ inches.

Sides are ordered like the angles they face, so the long leg is opposite the

60^\circ

angle and equals

x\sqrt{3}

, where

x

is the short leg opposite the

30^\circ

angle. Set

x\sqrt{3} = 15

, giving

x = \frac{15}{\sqrt{3}} = \frac{15\sqrt{3}}{3} = 5\sqrt{3} \approx 8.66

inches. The hypotenuse is

2x = 10\sqrt{3} \approx 17.3

inches. Verify with the Pythagorean Theorem:

(5\sqrt{3})^2 + 15^2 = 75 + 225 = 300

, and

(10\sqrt{3})^2 = 100 \cdot 3 = 300

. The lengths order correctly as

8.66 < 15 < 17.3

. Doubling 15 to get a hypotenuse of 30 is the classic error — only the short leg doubles.

A square has a diagonal of 12 meters. Find the exact area of the square.

Answer: 72 square meters

The diagonal splits the square into two

45^\circ

45^\circ

90^\circ

triangles whose legs are sides of the square. With side

s

, the diagonal is

s\sqrt{2}

, so

s\sqrt{2} = 12

and

s = \frac{12}{\sqrt{2}} = 6\sqrt{2}

meters. The area is

s^2 = (6\sqrt{2})^2 = 36 \cdot 2 = 72

square meters. A useful shortcut you can now justify: for any square, area equals half the diagonal squared, since

\frac{12^2}{2} = 72

. Squaring the diagonal without halving gives 144, the area of a much larger square.

FAQ

How do I remember which triangle uses $\sqrt{2}$ and which uses $\sqrt{3}$ ?: Match the count. The $45^\circ$ - $45^\circ$ - $90^\circ$ has two equal angles and comes from cutting a square (a 2-sided idea) — it uses $\sqrt{2}$ . The $30^\circ$ - $60^\circ$ - $90^\circ$ comes from cutting an equilateral triangle, a 3-sided figure — it uses $\sqrt{3}$ . You can also re-derive either one in about fifteen seconds with the Pythagorean Theorem if you forget.
Can I just use the Pythagorean Theorem instead of memorizing the ratios?: Yes, if you are given two sides. But these problems usually give you only one side plus the angles, and the Pythagorean Theorem alone cannot finish that — you would need trigonometry. The ratios let you get exact radical answers from a single side with no calculator, and they are much faster on multi-step problems involving areas, perimeters, and polygons.
Do I always have to rationalize the denominator?: Mathematically, $\frac{8}{\sqrt{3}}$ and $\frac{8\sqrt{3}}{3}$ are the same number. Most geometry classes require the rationalized form in simplest radical form, so unless your teacher says otherwise, clear the radical out of the denominator by multiplying the top and bottom by that radical.
How can I tell whether a triangle in a problem is actually a special right triangle?: Look for a guarantee, not an appearance. Marked angles of $30^\circ$ , $45^\circ$ , or $60^\circ$ with a right angle qualify. So do constructions: a diagonal of a square, an altitude of an equilateral triangle, a radius-and-apothem pair in a regular hexagon, or an isosceles triangle with a right angle. If the figure only looks like a half-square, you cannot assume it.

Learn this with a teacher, not a page

The Crimsora tutor teaches Special Right Triangles: 45-45-90 & 30-60-90 live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

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