Math Core

You already know the Pythagorean theorem as a tool for finding a missing side. In geometry you can do more with it: prove it using similar triangles, run it backwards to decide whether a triangle has a right angle, and even tell whether a triangle is acute or obtuse from its side lengths alone.

The theorem, restated

In a right triangle, the two sides that form the right angle are the legs, and the side across from the right angle is the hypotenuse, the longest side.

The Pythagorean theorem

If a triangle is a right triangle with legs $a$ and $b$ and hypotenuse $c$ , then

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

A proof with similar triangles

There are hundreds of proofs of this theorem. Here is one that uses the similarity you just studied.

Start with right triangle $ABC$ , with the right angle at $C$ . Draw the altitude from $C$ to the hypotenuse, meeting it at $D$ . That altitude splits the hypotenuse $c$ into two pieces, $x = AD$ and $y = DB$ , so $x + y = c$ .

Right triangle ABC with hypotenuse c = AB. The altitude CD splits c into pieces x and y.

The altitude creates two smaller right triangles, and all three triangles are similar:

$\triangle ACD$ and $\triangle ABC$ share angle $A$ , and each has a right angle, so they are similar by AA.
$\triangle CBD$ and $\triangle ABC$ share angle $B$ , and each has a right angle, so they are similar by AA.

Corresponding sides of similar triangles are proportional. Compare the small triangle $ACD$ with the big triangle $ABC$ . In each, pair the side next to angle $A$ with the hypotenuse:

$\frac{x}{b} = \frac{b}{c} \quad\Longrightarrow\quad b^2 = cx.$

Do the same with triangle $CBD$ and angle $B$ :

$\frac{y}{a} = \frac{a}{c} \quad\Longrightarrow\quad a^2 = cy.$

Now add the two results:

$a^2 + b^2 = cy + cx = c(x + y) = c \cdot c = c^2.$

That's the whole proof. Notice that it works for every right triangle, because the similarity argument never used any particular side lengths.

Using the theorem with exact answers

In geometry you'll often be asked for an exact answer, which means a simplified radical instead of a rounded decimal. To simplify $\sqrt{n}$ , pull out the largest perfect-square factor: $\sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5}$ .

Worked example: Legs 4 and 8

Find the exact length of the hypotenuse.

\begin{aligned} c^2 &= 4^2 + 8^2 = 16 + 64 = 80 \\ c &= \sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5} \end{aligned}

The hypotenuse is $4\sqrt{5} \approx 8.94$ .

Some side lengths come out whole. A set of three positive whole numbers with $a^2 + b^2 = c^2$ is called a Pythagorean triple. The most common are $3, 4, 5$ ; $5, 12, 13$ ; $8, 15, 17$ ; and $7, 24, 25$ . Any whole-number multiple of a triple is also a triple, so $6, 8, 10$ and $15, 20, 25$ work too. Recognizing triples saves time, but you can always fall back on the equation.

The converse

The theorem says: if a triangle is right, then $a^2 + b^2 = c^2$ . The converse swaps the two parts. It is also true.

Converse of the Pythagorean theorem

If the side lengths of a triangle satisfy $a^2 + b^2 = c^2$ , where $c$ is the longest side, then the triangle is a right triangle, and the right angle is opposite side $c$ .

Why is it true? Suppose a triangle has sides $a$ , $b$ , $c$ with $a^2 + b^2 = c^2$ . Build a second triangle with legs $a$ and $b$ and a genuine right angle between them. By the theorem, its hypotenuse is $\sqrt{a^2 + b^2} = c$ . The two triangles have the same three side lengths, so they are congruent by SSS. So the original triangle has a right angle too.

Builders use the converse all the time: to check that a corner is square, measure $3$ feet along one wall, $4$ feet along the other, and make sure the diagonal between the marks is exactly $5$ feet.

Worked example: Is it a right triangle?

A triangle has sides $20$ , $21$ and $29$ . Is it a right triangle?

The longest side is $29$ , so it plays the role of $c$ .

$20^2 + 21^2 = 400 + 441 = 841 \qquad 29^2 = 841$

The two sides match, so by the converse the triangle is right, with the right angle opposite the side of length $29$ .

Acute or obtuse?

What if $a^2 + b^2$ and $c^2$ don't match? Picture two sides $a$ and $b$ joined at a hinge. At $90^\circ$ the third side is exactly $\sqrt{a^2 + b^2}$ . Close the hinge and the third side shrinks; open it and the third side grows. That gives a test for any triangle.

Classifying a triangle by its sides

Let $c$ be the longest side of a triangle with sides $a$ , $b$ , $c$ .

If	then the triangle is
$c^2 = a^2 + b^2$	right
$c^2 < a^2 + b^2$	acute
$c^2 > a^2 + b^2$	obtuse

Before you classify, make sure the three lengths make a triangle at all. By the triangle inequality, the two shorter sides must add up to more than the longest side.

Worked example: Classify the triangle

Classify each triangle as acute, right or obtuse.

Sides $7$ , $9$ , $12$
Sides $6$ , $7$ , $8$

Solutions.

First check it's a triangle: $7 + 9 = 16 > 12$ . Then compare: $7^2 + 9^2 = 49 + 81 = 130$ and $12^2 = 144$ . Since $144 > 130$ , the longest side is too long for a right angle, so the triangle is obtuse.
$6 + 7 = 13 > 8$ , so it's a triangle. $6^2 + 7^2 = 36 + 49 = 85$ and $8^2 = 64$ . Since $64 < 85$ , the triangle is acute.

Common mistake

Always compare the square of the longest side with the sum of the squares of the other two. For sides $5$ , $12$ , $13$ , checking $5^2 + 13^2$ against $12^2$ gives nonsense. The right check is $5^2 + 12^2 = 169 = 13^2$ .

Tip

If you forget which inequality means obtuse, think of an extreme case: a very flat triangle with sides $1$ , $1$ and almost $2$ has a huge angle, and $2^2 = 4$ is much bigger than $1^2 + 1^2 = 2$ . Big $c^2$ means big angle.

Practice

Practice 1

A right triangle has legs $9$ and $12$ . How long is the hypotenuse?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

A right triangle has a hypotenuse of $26$ and one leg of $10$ . Find the other leg.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

A right triangle has two legs of length $6$ . Find the exact length of the hypotenuse. (Type a square root as sqrt(…), for example 2sqrt(3).)

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Which set of side lengths forms a right triangle?

Practice 5

A triangle has sides $8$ , $15$ and $16$ . How should it be classified?

Practice 6

A triangle is supposed to have sides $5$ , $7$ and $13$ . How should it be classified?

Practice 7

A soccer field is a rectangle $100$ yards long and $60$ yards wide. How far is it from one corner to the opposite corner, to the nearest tenth of a yard?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

The sides of a right triangle are $x$ , $x + 7$ and $x + 8$ . Find $x$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.