Apply the Pythagorean theorem

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Difficulty: Beginner
Estimated time: 32 minutes
Domains: Geometry and Trigonometry
Techniques: Pythagorean-theoremHypotenuseDiagonalsDistance-formula

What you’ll learn

Check that a triangle has a right angle before you use the Pythagorean theorem.
Find the hypotenuse from the right angle, not from how the drawing looks.
Find a missing leg or a missing hypotenuse.
Spot right triangles hiding in rectangles, squares, coordinate planes, and word problems.
Use Desmos distance(A,B) when the coordinates are messy decimals.
Keep an exact radical, or round to a decimal only when the question asks.
Check that a length you found makes sense.

Prerequisites

Rewrite radicals and rational exponents0%
Simplify square roots and keep radical forms exact.

Why this matters on the SAT

Find the hidden right triangle.

An SAT question may never say, “Use the Pythagorean theorem.” Instead, it might give you a rectangle’s diagonal, two points on a grid, a vertical pole on level ground, or a diameter across a circle. Your first job is to notice the right angle.

Once you’ve found a right triangle, you can use the formula on the SAT reference sheet:

a^2+b^2=c^2.

So you don’t need to memorize it. What the sheet can’t tell you is whether the triangle really has a right angle, which side is $c$ , and what the question wants you to do with the length you find. Those decisions are yours.

Pick your route from the answer form and the numbers. Work by hand when the answer is an exact radical or the sides are a familiar whole-number triple like $7$ - $24$ - $25$ . For the distance between two points with messy decimal coordinates, define the points in Desmos and enter distance(A,B). Either way, first make sure you know which segment the question is asking about.

Solution to the example

$AB$ is across from the right angle at $C$ , so it’s the hypotenuse, and it goes alone on the $c^2$ side:

\begin{aligned} BC^2+7^2&=25^2\\[1.4em] BC^2&=576\\[1.4em] BC&=24. \end{aligned}

The answer is B. The formula was on the reference sheet all along. Spotting the hypotenuse is what told you to subtract.

Each wrong choice comes from a common slip. A subtracts the sides without squaring them ( $25-7$ ), D adds them ( $25+7$ ), and C adds the squares ( $\sqrt{625+49}$ ), as if $25$ were a leg.

SAT example

The right-angle marker identifies AB as the hypotenuse.

In right triangle $ABC$ , $\angle C$ is a right angle, $AB=25$ , and $AC=7$ . What is the length of $\overline{BC}$ ?

A
$18$
B
$24$
C
$\sqrt{674}$
D
$32$

Example: turn a diagonal into an area

Worked example

A rectangle’s diagonal is the hypotenuse of a right triangle whose legs are the side lengths.

A rectangle has a perimeter of $56$ units and a diagonal length of $20$ units. What is the area, in square units, of the rectangle?

The diagonal cuts the rectangle into two right triangles. Call the side lengths $\ell$ and $w$ . The perimeter gives

2\ell+2w=56,

\ell+w=28.

The diagonal is the hypotenuse, so the Pythagorean theorem gives

\ell^2+w^2=20^2=400.

Here’s the twist: the question asks for the area, $\ell w$ , not the side lengths. Square the sum you know, and $\ell w$ shows up in the middle:

\begin{aligned} (\ell+w)^2&=\ell^2+2\ell w+w^2\\[1.4em] 28^2&=400+2\ell w\\[1.4em] 784-400&=2\ell w\\[1.4em] 192&=\ell w. \end{aligned}

The area is $\boxed{192}$ square units.

Follow the target: The theorem gave you $\ell^2+w^2$ , not the area. Squaring $\ell+w$ bridged the gap, so you never had to find $\ell$ and $w$ on their own.

Check your understanding:

Suppose you learn the sides are really $12$ and $16$ . How could you quickly check both the diagonal and the area?

Find the hypotenuse before you write an equation

A right triangle has two legs and one hypotenuse. The legs are the two sides that meet to make the right angle. The hypotenuse is the side across from the right angle, and it’s always the longest side.

The right-angle marker decides which side is which, not the way the drawing looks. A slanted side isn’t automatically the hypotenuse, and a side named $c$ doesn’t belong in the $c^2$ spot unless it’s across from the right angle.

The theorem only works once the problem gives you a $90^\circ$ angle or lets you show there is one. Good signs to look for:

a small square marking a corner in a figure
a statement that an angle measures $90^\circ$
perpendicular lines
something vertical standing on level ground
the corner of a rectangle or square
a horizontal change and a vertical change on a coordinate grid

Without a right angle, the theorem doesn’t apply, even when a problem gives you three side lengths.

Check your understanding:

In triangle $ABC$ , $AB=9$ , $BC=12$ , and $\angle B=90^\circ$ . Which side is the hypotenuse, and how long is it?

Common mistake:

Writing $9^2+AC^2=12^2$ treats $BC$ as the hypotenuse, but $BC$ touches the right angle, so it’s a leg. Find the right angle first, look straight across the triangle to the opposite side, and put that side alone on the $c^2$ side of the equation.

Add for the hypotenuse, subtract for a leg

It’s one theorem either way. What changes is which side you’re missing.

Missing hypotenuse

If the legs are $7$ and $11$ , add their squares:

\begin{aligned} c^2&=7^2+11^2\\[1.4em] &=170,\\[1.4em] c&=\sqrt{170}. \end{aligned}

A length is positive, so take the positive square root. $170$ has no perfect-square factor bigger than $1$ , so $\sqrt{170}$ is already as simple as it gets.

If simplifying square roots feels shaky, review Rewrite radicals and rational exponents.

Missing leg

If the hypotenuse is $13$ and one leg is $8$ , call the other leg $x$ :

\begin{aligned} x^2+8^2&=13^2\\[1.4em] x^2&=169-64\\[1.4em] &=105,\\[1.4em] x&=\sqrt{105}. \end{aligned}

There’s no separate subtraction formula to learn. The subtraction comes from getting $x^2$ alone in $x^2+8^2=13^2$ .

Check your understanding:

A right triangle has hypotenuse $25$ and one leg $7$ . Do you add or subtract squares to find the other leg, and how long is it?

Common mistake:

If your work ends at $x^2=105$ , you’ve found the square of the length, not the length. Take the square root, and keep the positive one, since a length can’t be negative.

Turn diagonals into right triangles

A diagonal connects two corners that aren’t next to each other. Every corner of a rectangle or square is $90^\circ$ , so a diagonal and two sides always form a right triangle.

For a rectangle with sides $\ell$ and $w$ and diagonal $d$ ,

d^2=\ell^2+w^2.

That’s the Pythagorean theorem again, not a new diagonal formula.

A box, or rectangular prism, also has a space diagonal, which runs through the inside from one corner to the opposite corner. With length $\ell$ , width $w$ , and height $h$ , use the theorem twice, first across the base and then up:

\begin{aligned} \text{base diagonal}^2&=\ell^2+w^2,\\[1.4em] d^2&=(\text{base diagonal})^2+h^2\\[1.4em] &=\ell^2+w^2+h^2. \end{aligned}

The second line is the theorem again. The height stands straight up from the base, so it’s perpendicular to the base diagonal, and those two are the legs.

Check your understanding:

A rectangular prism has dimensions $3$ , $4$ , and $12$ . How long is its space diagonal?

Use coordinate changes as the legs

To find the distance between $P(x_1,y_1)$ and $Q(x_2,y_2)$ , picture a right triangle:

the horizontal leg is $\lvert x_2-x_1\rvert$ long
the vertical leg is $\lvert y_2-y_1\rvert$ long
the segment from $P$ to $Q$ is the hypotenuse

PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.

That’s the distance formula, and it’s the Pythagorean theorem written with coordinates.

Let’s try it with $P(-4,2)$ and $Q(2,9)$ . The horizontal change is

2-(-4)=6,

and the vertical change is

9-2=7.

PQ=\sqrt{6^2+7^2}=\sqrt{85}.

An exact radical like this one is quicker to keep by hand. Messy decimal coordinates are the Desmos case from earlier: when the question wants a decimal, define the points and let Desmos do the arithmetic.

A=(-3.7,2.4)
B=(4.6,-1.9)
distance(A,B)

Desmos gives the full decimal distance. Round it only if the question tells you to.

The horizontal and vertical changes are the legs, and the segment you want is the hypotenuse.

Check your understanding:

Why is the horizontal change from $x=-4$ to $x=2$ equal to $6$ , not $-2$ or $2$ ?

Common mistake:

Don’t add the coordinates, and don’t subtract an $x$ -coordinate from a $y$ -coordinate. Find the $x$ -difference and the $y$ -difference separately, square each one, add, and then take the square root.

Give the answer form the question asks for

If the question asks for an exact value, keep the square root and simplify it. If it asks for a decimal, or says “to the nearest tenth,” approximate only at the very end.

Suppose a right triangle has hypotenuse $26$ and one leg $17$ . The other leg $x$ satisfies

\begin{aligned} x^2&=26^2-17^2\\[1.4em] &=676-289\\[1.4em] &=387\\[1.4em] &=9\cdot43. \end{aligned}

Pull out the perfect square $9$ , and the exact length is

x=\sqrt{387}=3\sqrt{43}.

For the nearest tenth, work from the unrounded value:

3\sqrt{43}\approx19.7.

Don’t round $\sqrt{43}$ first. It’s about $6.557$ . Round that to $6.6$ , multiply by $3$ , and you get $19.8$ , which is off by a tenth.

Check your understanding:

Your work gives $d=\sqrt{72}$ . What do you submit if the question asks (a) for an exact value, or (b) for the nearest tenth?

Calculator loads as you approach

Both lines give the same length, unrounded.

Check that your answer makes sense

Three quick questions catch most Pythagorean mistakes:

Is the hypotenuse longer than each leg?
Do the sides fit $a^2+b^2=c^2$ when you plug them in?
Did you answer what the question actually asked?

Say the legs are $9$ and $14$ , so the hypotenuse is $\sqrt{277}$ . Is that a sensible size? Since

16^2=256<277<289=17^2,

the hypotenuse is between $16$ and $17$ . That’s longer than both legs, so it passes.

If a missing leg ever comes out longer than the hypotenuse, the equation was set up wrong.

Common mistake:

After you find a side, reread the last line of the question. The SAT may want an area, a perimeter, a radius, a ratio, or a leftover piece. Use the length you found to get there.

Practice problems

The first and last problems are quickest by hand. The middle one has messy decimal coordinates, so it’s a Desmos job.

Find a missing leg

Practice problem

A right triangle has a hypotenuse of length $17$ meters and one leg of length $8$ meters. What is the length of the other leg, in meters?

Your answer

Calculator loads as you approach

Use Desmos if it helps you solve or check this problem.

Find an awkward coordinate distance

Practice problem

In the $xy$ -plane, point $A$ has coordinates $(-3.7,2.4)$ and point $B$ has coordinates $(4.6,-1.9)$ . What is the length of $\overline{AB}$ , rounded to the nearest tenth?

Your answer

Calculator loads as you approach

Define both points, then enter distance(A,B).

Model a broken flagpole

Practice problem

A $25$ -foot vertical flagpole snaps during a storm. The top section remains straight and touches the level ground $15$ feet from the base of the pole. At what height above the ground, in feet, did the flagpole break?

Your answer

Calculator loads as you approach

Use Desmos if it helps you solve or check this problem.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Use the Pythagorean theorem only when a right angle is given or can be shown.
The hypotenuse is across from the right angle, and it’s always the longest side.
Add the squares of the legs to find the hypotenuse. To find a leg, get its square alone, which means subtracting.
Rectangle diagonals, segments between points, and space diagonals all hide right triangles.
Work by hand for exact radicals and simple triples. Use Desmos distance(A,B) for messy decimal coordinates.
Keep an exact radical unless the question asks for a decimal, and round only at the end.
Once you have the length, finish the job: an area, a perimeter, a ratio, or whatever the question wants.

Related lessons

If a square root needs simplifying, Rewrite radicals and rational exponents shows you how. Later, Use sine, cosine, and tangent takes on problems built from acute angles and side ratios.

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Recognize special right triangles

Use the side ratios of 45-45-90 and 30-60-90 triangles to find exact lengths.

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