Pythagoras Theorem Notes

Published on: September 29, 2026

This Pythagoras Theorem Calculator finds the hypotenuse of a right triangle from its two shorter sides, and shows the working line by line. The square drawn on the hypotenuse has the same area as the two squares on the shorter sides put together, so each shorter side is squared, the squares are added, and the square root of that total is the hypotenuse. Following the same pattern every time is what makes the theorem easy to check and repeat.

Right triangle with legs a, b and hypotenuse c

In the diagram, the shorter sides are \(a\) and \(b\), and the hypotenuse is \(c\).

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Substitute the two shorter sides into the formula.
  4. Square each of the shorter sides.
  5. Add the two squares under the root.
  6. Take the square root to get the hypotenuse.

Formula:

\(c = \sqrt{a^{2} + b^{2}}\)
See Example 1 Hide Example 1

Problem

\(a = 9, b = 12\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Substitute the two shorter sides into the formula.
  4. Square each of the shorter sides.
  5. Add the two squares under the root.
  6. Take the square root to get the hypotenuse.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the hypotenuse of a right triangle whose shorter sides are \(a = 9\) and \(b = 12\).

\(a = 9,\; b = 12\)

Step 2

Step 2 - Write the formula.

In this problem: In a right triangle the square drawn on the hypotenuse has the same area as the two squares on the shorter sides together, so the hypotenuse is the square root of their sum.

\(c = \sqrt{a^{2} + b^{2}}\)

Step 3

Step 3 - Substitute the two shorter sides into the formula.

In this problem: Replacing \(a\) with \(9\) and \(b\) with \(12\) in \(c = \sqrt{a^{2} + b^{2}}\) gives \(c = \sqrt{9^{2} + 12^{2}}\).

\(c = \sqrt{9^{2} + 12^{2}}\)

Step 4

Step 4 - Square each of the shorter sides.

In this problem: Under the root the two shorter sides are still written as squares, so each one is worked out in turn.

\(c = \sqrt{9^{2} + 12^{2}}\)

Step 4a

Step 4a - Square each of the shorter sides.

In this problem: Squaring side \(a\) gives \(9^{2} = 81\).

\(c = \sqrt{81 + 12^{2}}\)

Step 4b

Step 4b - Square each of the shorter sides.

In this problem: Squaring side \(b\) gives \(12^{2} = 144\).

\(c = \sqrt{81 + 144}\)

Step 5

Step 5 - Add the two squares under the root.

In this problem: Adding the two squares gives \(81 + 144 = 225\).

\(c = \sqrt{225}\)

Step 6

Step 6 - Take the square root to get the hypotenuse.

In this problem: Taking the square root undoes the squaring and gives \(\sqrt{225} = 15\).

\(c = \sqrt{225} = 15\)

Final Answer

\(c = 15\)
See Example 2 Hide Example 2

Problem

\(a = 7, b = 24\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Substitute the two shorter sides into the formula.
  4. Square each of the shorter sides.
  5. Add the two squares under the root.
  6. Take the square root to get the hypotenuse.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the hypotenuse of a right triangle whose shorter sides are \(a = 7\) and \(b = 24\).

\(a = 7,\; b = 24\)

Step 2

Step 2 - Write the formula.

In this problem: In a right triangle the square drawn on the hypotenuse has the same area as the two squares on the shorter sides together, so the hypotenuse is the square root of their sum.

\(c = \sqrt{a^{2} + b^{2}}\)

Step 3

Step 3 - Substitute the two shorter sides into the formula.

In this problem: Replacing \(a\) with \(7\) and \(b\) with \(24\) in \(c = \sqrt{a^{2} + b^{2}}\) gives \(c = \sqrt{7^{2} + 24^{2}}\).

\(c = \sqrt{7^{2} + 24^{2}}\)

Step 4

Step 4 - Square each of the shorter sides.

In this problem: Under the root the two shorter sides are still written as squares, so each one is worked out in turn.

\(c = \sqrt{7^{2} + 24^{2}}\)

Step 4a

Step 4a - Square each of the shorter sides.

In this problem: Squaring side \(a\) gives \(7^{2} = 49\).

\(c = \sqrt{49 + 24^{2}}\)

Step 4b

Step 4b - Square each of the shorter sides.

In this problem: Squaring side \(b\) gives \(24^{2} = 576\).

\(c = \sqrt{49 + 576}\)

Step 5

Step 5 - Add the two squares under the root.

In this problem: Adding the two squares gives \(49 + 576 = 625\).

\(c = \sqrt{625}\)

Step 6

Step 6 - Take the square root to get the hypotenuse.

In this problem: Taking the square root undoes the squaring and gives \(\sqrt{625} = 25\).

\(c = \sqrt{625} = 25\)

Final Answer

\(c = 25\)
See Example 3 Hide Example 3

Problem

\(a = 6, b = 8\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Substitute the two shorter sides into the formula.
  4. Square each of the shorter sides.
  5. Add the two squares under the root.
  6. Take the square root to get the hypotenuse.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the hypotenuse of a right triangle whose shorter sides are \(a = 6\) and \(b = 8\).

\(a = 6,\; b = 8\)

Step 2

Step 2 - Write the formula.

In this problem: In a right triangle the square drawn on the hypotenuse has the same area as the two squares on the shorter sides together, so the hypotenuse is the square root of their sum.

\(c = \sqrt{a^{2} + b^{2}}\)

Step 3

Step 3 - Substitute the two shorter sides into the formula.

In this problem: Replacing \(a\) with \(6\) and \(b\) with \(8\) in \(c = \sqrt{a^{2} + b^{2}}\) gives \(c = \sqrt{6^{2} + 8^{2}}\).

\(c = \sqrt{6^{2} + 8^{2}}\)

Step 4

Step 4 - Square each of the shorter sides.

In this problem: Under the root the two shorter sides are still written as squares, so each one is worked out in turn.

\(c = \sqrt{6^{2} + 8^{2}}\)

Step 4a

Step 4a - Square each of the shorter sides.

In this problem: Squaring side \(a\) gives \(6^{2} = 36\).

\(c = \sqrt{36 + 8^{2}}\)

Step 4b

Step 4b - Square each of the shorter sides.

In this problem: Squaring side \(b\) gives \(8^{2} = 64\).

\(c = \sqrt{36 + 64}\)

Step 5

Step 5 - Add the two squares under the root.

In this problem: Adding the two squares gives \(36 + 64 = 100\).

\(c = \sqrt{100}\)

Step 6

Step 6 - Take the square root to get the hypotenuse.

In this problem: Taking the square root undoes the squaring and gives \(\sqrt{100} = 10\).

\(c = \sqrt{100} = 10\)

Final Answer

\(c = 10\)
See Example 4 Hide Example 4

Problem

\(a = 5, b = 12\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Substitute the two shorter sides into the formula.
  4. Square each of the shorter sides.
  5. Add the two squares under the root.
  6. Take the square root to get the hypotenuse.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the hypotenuse of a right triangle whose shorter sides are \(a = 5\) and \(b = 12\).

\(a = 5,\; b = 12\)

Step 2

Step 2 - Write the formula.

In this problem: In a right triangle the square drawn on the hypotenuse has the same area as the two squares on the shorter sides together, so the hypotenuse is the square root of their sum.

\(c = \sqrt{a^{2} + b^{2}}\)

Step 3

Step 3 - Substitute the two shorter sides into the formula.

In this problem: Replacing \(a\) with \(5\) and \(b\) with \(12\) in \(c = \sqrt{a^{2} + b^{2}}\) gives \(c = \sqrt{5^{2} + 12^{2}}\).

\(c = \sqrt{5^{2} + 12^{2}}\)

Step 4

Step 4 - Square each of the shorter sides.

In this problem: Under the root the two shorter sides are still written as squares, so each one is worked out in turn.

\(c = \sqrt{5^{2} + 12^{2}}\)

Step 4a

Step 4a - Square each of the shorter sides.

In this problem: Squaring side \(a\) gives \(5^{2} = 25\).

\(c = \sqrt{25 + 12^{2}}\)

Step 4b

Step 4b - Square each of the shorter sides.

In this problem: Squaring side \(b\) gives \(12^{2} = 144\).

\(c = \sqrt{25 + 144}\)

Step 5

Step 5 - Add the two squares under the root.

In this problem: Adding the two squares gives \(25 + 144 = 169\).

\(c = \sqrt{169}\)

Step 6

Step 6 - Take the square root to get the hypotenuse.

In this problem: Taking the square root undoes the squaring and gives \(\sqrt{169} = 13\).

\(c = \sqrt{169} = 13\)

Final Answer

\(c = 13\)