The Pythagorean Theorem and Right Triangles

10 min read

Stand a ladder against a wall. The wall, the ground, and the ladder make a right triangle. You can find the ladder’s length without ever stretching a tape along it. That shortcut is the Pythagorean theorem, and it is the one rule trig leans on more than any other.

By the end of this page you can:

• Name the legs and the hypotenuse and say which side is longest.
• Find the hypotenuse when both legs are known, and find a missing leg when the hypotenuse and one leg are known.
• Test in reverse whether a triangle has a right angle.

The floor. Know $a^2 + b^2 = c^2$ with $c$ as the hypotenuse, and be able to solve it for any one missing side. Everything in right-triangle trig is built on that single equation.

Quick Reference

The right-triangle relationships in Stewart 7e, Section 6.2 (sine, cosine, and tangent as ratios of side lengths) rest on the Pythagorean relationship reviewed here.

Before You Start

This is a foundation skill with no prior skills inside MATH142. The only background it asks for:

• I can square a number, so $5^2 = 25$.
• I can take a square root, so $\sqrt{25} = 5$.

Both are reviewed in OpenStax Prealgebra 2e, Section 9.3.

The Idea

A right triangle has one square corner. That corner locks the three side lengths together: once two sides are fixed, the third is determined, with no ruler needed. This is one of the oldest results in mathematics and the bridge from plain geometry into trigonometry.

The two short sides meeting at the right angle are the legs. The long side across from the right angle is the hypotenuse, always the longest. As the legs grow, the hypotenuse grows too, but not by simple addition. Legs of $3$ and $4$ do not give a hypotenuse of $7$. The square corner ties the sides together through their squares, and that is the surprise worth understanding.

Prerequisite Hub

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Builds on: A foundation skill with no prerequisite skills inside MATH142 and no cross-course prerequisites. The reader needs only squaring and square roots, reviewed in OpenStax Prealgebra 2e, Section 9.3.

Unlocks: Once the Pythagorean relationship is secure, it feeds three later skills.

The Official Statement

Definition

Pythagorean Theorem. In a right triangle with legs of lengths $a$ and $b$ and hypotenuse of length $c$, the relationship $a^2 + b^2 = c^2$ holds. The right-triangle relationships of Section 5.4 build on this fact.

Source: OpenStax Precalculus 2e, Section 5.4 Right Triangle Trigonometry; reviewed in OpenStax Prealgebra 2e, Section 9.3.

Key Theorem

Pythagorean Theorem. For a right triangle with legs $a$ and $b$ and hypotenuse $c$, $$a^2 + b^2 = c^2.$$

Source: OpenStax Prealgebra 2e, Section 9.3 Use Properties of Angles, Triangles, and the Pythagorean Theorem.

Reading the Statement

The squares carry the meaning. The square on the hypotenuse equals the combined squares on the two legs.

One equation runs in two directions. When both legs are known, solving for $c$ gives the hypotenuse: $$c = \sqrt{a^2 + b^2}.$$ When the hypotenuse and one leg are known, the same equation rearranges to find the missing leg: $$a = \sqrt{c^2 - b^2}.$$

A Picture of Why It Is True

        +-----------+
        |           |
        |   c^2     |       The square built on the hypotenuse (area c^2)
        |           |       has the same area as the two squares built
        +-----------+       on the legs combined (area a^2 plus area b^2).

   +------+    +-----------+
   | a^2  |    |    b^2    |
   +------+    |           |
              |           |
              +-----------+

Build a square on each side. The largest square (on the hypotenuse) has the same area as the two smaller squares (on the legs) combined. The relationship is a statement about areas, which is why squaring appears, not about adding lengths.

Set the two legs and watch the areas. The square on the hypotenuse always equals the two leg squares added together. Hunt for a triangle whose three sides all come out whole.

Worked Examples

Example 1: Find the Hypotenuse (legs 3 and 4)

A right triangle has legs of length $3$ and $4$. Find the hypotenuse.

Predict first. The hypotenuse beats either leg but stays under their sum $3 + 4 = 7$, so the answer sits between $4$ and $7$.

Compute. $$c^2 = 3^2 + 4^2 = 9 + 16 = 25, \qquad c = \sqrt{25} = 5.$$

Compare. The result $5$ lands between $4$ and $7$ and confirms the grid-paper measurement. Legs $3$ and $4$ always give a hypotenuse of exactly $5$.

Example 2: Find a Missing Leg (hypotenuse 13, one leg 5)

A right triangle has a hypotenuse of $13$ and one leg of $5$. Find the other leg.

Predict first. The missing side is a leg, so it must be shorter than the hypotenuse $13$.

Compute. Rearrange to isolate the missing leg: $$b^2 = 13^2 - 5^2 = 169 - 25 = 144, \qquad b = \sqrt{144} = 12.$$

Compare. The result $12$ is less than $13$, as a leg must be. Check the triple: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$.

Example 3: A Non-Whole-Number Answer (legs 1 and 1)

A right triangle has both legs equal to $1$. Find the hypotenuse.

Predict first. Equal, small legs give a hypotenuse just over $1$ and well under $2$.

Compute. $$c^2 = 1^2 + 1^2 = 2, \qquad c = \sqrt{2} \approx 1.414.$$

Compare. The value sits between $1$ and $2$. Not every right triangle has whole-number sides. The exact answer is $\sqrt{2}$, an irrational number, and the decimal is an approximation.

Example 4: A Real Context (a ladder against a wall)

A ladder $10$ feet long leans against a vertical wall with its base $6$ feet from the wall. How high up the wall does it reach?

Predict first. The ladder is the hypotenuse (longest side, leaning across the right angle of wall and ground). The height reached is a leg, so it is under $10$.

Compute. With hypotenuse $c = 10$ and one leg $a = 6$, the height $b$ is the missing leg: $$b^2 = 10^2 - 6^2 = 100 - 36 = 64, \qquad b = \sqrt{64} = 8.$$

Compare. The height $8$ feet is less than the $10$-foot ladder. The base, wall, and ladder form a $6$-$8$-$10$ right triangle.

Common Misconceptions

The hypotenuse equals the sum of the legs. This treats the relationship as additive, so legs $3$ and $4$ would give $7$. The true hypotenuse is $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$, not $7$. The relationship is multiplicative through squares, and the hypotenuse is always strictly less than the sum of the legs.

$a^2 + b^2 = c^2$ applies to any triangle. It holds only when the triangle has a right angle, and only with $c$ as the side across from that angle. Sides $4$, $5$, $6$ give $4^2 + 5^2 = 41$ while $6^2 = 36$, so the equation fails. Confirm a square corner and identify the side facing it before applying the theorem.

$\sqrt{a^2 + b^2}$ simplifies to $a + b$. A square root does not distribute across addition. For $a = 3$, $b = 4$: $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$, while $a + b = 7$. Carry out the squaring and the square root in order before simplifying.

Practice Problems

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Mastery Checklist

Novice (Level 1-2):

• I can identify the legs and the hypotenuse in a right triangle
• I can state $a^2 + b^2 = c^2$ from memory
• I can find the hypotenuse when both legs are given

Competent (Level 3-4):

• I can find a missing leg when the hypotenuse and one leg are given
• I can handle irrational answers, giving both the exact form and a decimal
• I can predict the rough size of the answer before computing, as a check

Proficient (Level 5):

• I can use the relationship in reverse to test whether a triangle is right
• I can explain why the relationship is about areas, not added lengths
• I can connect a side-length problem back to its meaning in a real context

Connections

Looking back:

• The background is squaring and square roots, reviewed in OpenStax Prealgebra 2e, Section 9.3.

Looking ahead:

• Coordinate Plane and Distance builds the distance formula as a hypotenuse computation
• Right Triangle Trigonometry defines sine, cosine, and tangent as ratios of these side lengths
• Special Right Triangles derives the exact ratios of the 45-45-90 and 30-60-90 triangles

Last updated: 2026-06-24