Pythagoras' theorem

Discover, geometrically prove, and apply the Pythagorean Theorem.

Imagine you are helping an architect build a scale model of a sculpture. One triangular frame is a right triangle with legs 9 ft and 12 ft. How long is the third side?

Pythagorean Theorem

In any right triangle with legs \(a\) and \(b\), and hypotenuse \(c\):

\[ a^2 + b^2 = c^2 \]

The hypotenuse is opposite the right angle, so it is always the longest side.

Right triangle labels

We usually name vertices with uppercase letters and opposite sides with matching lowercase letters.

Finding side lengths

Find the hypotenuse

Example. If \(a=9\) and \(b=12\), find \(c\).

\[ \begin{aligned} c &= \sqrt{9^2 + 12^2} \\ &= \sqrt{81 + 144} \\ &= \sqrt{225} \\ &= 15 \end{aligned} \]

The third side is 15 ft.

Find a missing leg

Example. If \(c=13\) and \(a=5\), find \(b\).

\[ \begin{aligned} b &= \sqrt{c^2 - a^2} \\ &= \sqrt{13^2 - 5^2} \\ &= \sqrt{169 - 25} \\ &= \sqrt{144} \\ &= 12 \end{aligned} \]

Leave answers in simplest radical form

Example. If \(c=10\) and \(a=2\), find \(b\).

\[ \begin{aligned} b &= \sqrt{10^2 - 2^2} \\ &= \sqrt{100 - 4} \\ &= \sqrt{96} \\ &= \sqrt{16 \cdot 6} \\ &= 4\sqrt{6} \end{aligned} \]

Common pitfall

When solving \(c^2=289\), the equation has two numerical roots (\(\pm 17\)), but side lengths in geometry are nonnegative. So the triangle side is \(17\), not \(-17\).

Converse of the theorem

If three side lengths satisfy \(a^2+b^2=c^2\), then the triangle is right-angled.

Example: \(7,24,25\) because \(7^2+24^2=49+576=625=25^2\).

Coordinate geometry use

Distance between points \(A(x_1,y_1)\) and \(B(x_2,y_2)\):

\[ AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \]

This is Pythagoras applied to horizontal and vertical change.

Applications

• Diagonal of squares and rectangles.
• Ladders, ramps, and roof braces in construction.
• Map/grid displacement and shortest straight-line paths.
• Checking if a corner is exactly \(90^\circ\).

Vocabulary

History

The theorem is associated with Pythagoras, but right-triangle relationships appeared in earlier Babylonian, Egyptian, and Indian mathematics. In Euclidean geometry, the theorem became a central tool for proof, measurement, and construction.

Geometric proof by area

Build a large square of side length \(a+b\). Place four identical right triangles (legs \(a,b\), hypotenuse \(c\)) inside it so the uncovered center is a square of side \(c\).

Area of big square: \((a+b)^2\). Area of the same figure as pieces: \(4\left(\frac{1}{2}ab\right)+c^2=2ab+c^2\). Set them equal:

\[ (a+b)^2 = 2ab + c^2 \]

\[ a^2 + 2ab + b^2 = 2ab + c^2 \]

\[ a^2 + b^2 = c^2 \]

Checkpoints

If the legs are 3 and 4, what is the hypotenuse?

Answer: \(c=\sqrt{3^2+4^2}=\sqrt{25}=5\).

If the legs are 6 and 8, what is the hypotenuse?

Answer: \(c=\sqrt{6^2+8^2}=\sqrt{100}=10\).

If the legs are 5 and 12, what is the hypotenuse?

Answer: \(c=\sqrt{5^2+12^2}=\sqrt{169}=13\).

A square has side length 6. Find its diagonal.

Answer: \(d=\sqrt{6^2+6^2}=\sqrt{72}=6\sqrt{2}\).

A square has side length 9. Find its diagonal.

Answer: \(d=\sqrt{9^2+9^2}=\sqrt{162}=9\sqrt{2}\).

If one leg is 4 and the hypotenuse is 8, find the other leg.

Answer: \(\sqrt{8^2-4^2}=\sqrt{64-16}=\sqrt{48}=4\sqrt{3}\).

If one leg is 10 and the hypotenuse is 15, find the other leg.

Answer: \(\sqrt{15^2-10^2}=\sqrt{225-100}=\sqrt{125}=5\sqrt{5}\).

For legs \(x\) and \(y\), write the hypotenuse.

Answer: \(c=\sqrt{x^2+y^2}\).

Using the proof figure, explain why \((a+b)^2\) and \(2ab+c^2\) must be equal.

Answer: They are two area expressions for the same large square: one as a full square of side \(a+b\), and one as four right triangles plus the center square of side \(c\).