# 1.2: Pythagorean Theorem

Difficulty Level: At Grade Created by: CK-12

The Pythagorean Theorem allows you to find the lengths of the sides of a right triangle, which is a triangle with one \begin{align*}90^\circ\end{align*} angle (known as the right angle). An example of a right triangle is depicted below.

A right triangle is composed of three sides: two legs, which are labeled in the diagram as \begin{align*}leg_{1}\end{align*} and \begin{align*}leg_{2}\end{align*}, and a hypotenuse, which is the side opposite to the right angle. The hypotenuse is always the longest of the three sides. Typically, we denote the right angle with a small square, as shown above, but this is not required.

The Pythagorean Theorem states that the length of the hypotenuse squared equals the sum of the squares of the two legs. This is written mathematically as:

\begin{align*}({\color{green}leg}_{\color{green}1})^2 + ({\color{blue}leg}_{\color{blue}2})^2 = ({\color{red}hypotenuse})^2 \end{align*}

To verify this statement, first explicitly expressed by Pythagoreans so many years ago, let’s look at an example.

### Example 1

Consider the right triangle below. Does the Pythagorean Theorem hold for this triangle?

Solution

As labeled, this right triangle has sides with lengths 3, 4, and 5. The side with length 5, the longest side, is the hypotenuse because it is opposite to the right angle. Let’s say the side of length 4 is \begin{align*}leg_{1}\end{align*} and the side of length 3 is \begin{align*}leg_{2}\end{align*}.

Recall that the Pythagorean Theorem states:

\begin{align*}(leg_1)^2 + (leg_2)^2 &= (hypothenuse)^2 \intertext{If we plug the values for the side lengths of this right triangle into the mathematical expression of the Pythagorean Theorem, we can verify that the theorem holds:} (4)^2+(3)^2 &=(5)^2 \\ 16+9 &=25 \\ 25 &=25\end{align*}

Although it is clear that the theorem holds for this specific triangle, we have not yet proved that the theorem will hold for all right triangles. A simple proof, however, will demonstrate that the Pythagorean Theorem is universally valid.

### Proof Based on Similar Triangles

The diagram below depicts a large right triangle (triangle \begin{align*}{\color{red}ABC}\end{align*}) with an altitude (labeled \begin{align*}h\end{align*}) drawn from one of its vertices. An altitude is a line drawn from a vertex to the side opposite it, intersecting the side perpendicularly and forming a \begin{align*}90^\circ\end{align*} angle.

In this example, the altitude hits side \begin{align*}AB\end{align*} at point \begin{align*}D\end{align*} and creates two smaller right triangles within the larger right triangle. In this case, triangle \begin{align*}{\color{red}ABC}\end{align*} is similar to triangles \begin{align*}{\color{green}CBD}\end{align*} and \begin{align*}{\color{blue}ACD}\end{align*}. When a triangle is similar to another triangle, corresponding sides are proportional in lengths and corresponding angles are equal. In other words, in a set of similar triangles, one triangle is simply an enlarged version of the other.

Similar triangles are often used in proving the Pythagorean Theorem, as they will be in this proof. In this proof, we will first compare similar triangles \begin{align*}{\color{red}ABC}\end{align*} and \begin{align*}{\color{green}CBD}\end{align*}, then triangles \begin{align*}{\color{red}ABC}\end{align*} and \begin{align*}{\color{blue}ACD}\end{align*}.

Comparing Triangles \begin{align*}{\color{red}ABC}\end{align*} and \begin{align*}{\color{green}CBD}\end{align*}

In the diagram above, side \begin{align*}{\color{red}AB}\end{align*} corresponds to side \begin{align*}{\color{green}CB}\end{align*}. Similarly, side \begin{align*}{\color{red}BC}\end{align*} corresponds to side \begin{align*}{\color{green}BD}\end{align*}, and side \begin{align*}{\color{red}CA}\end{align*} corresponds to side \begin{align*}{\color{green}DC}\end{align*}. It is possible to tell which side corresponds to the appropriate side on a similar triangle by using angles; for example, corresponding sides \begin{align*}{\color{red}AB}\end{align*} and \begin{align*}{\color{green}CB}\end{align*} are both opposite a right angle.

Because corresponding sides are proportional and have the same ratio, we can set the ratios of their lengths equal to one another. For example, the ratio of side \begin{align*}{\color{red}AB}\end{align*} to side \begin{align*}{\color{red}BC}\end{align*} in triangle \begin{align*}{\color{red}ABC}\end{align*} is equal to the ratio of side \begin{align*}{\color{green}CB}\end{align*} to corresponding side \begin{align*}{\color{green}BD}\end{align*} in triangle \begin{align*}{\color{green}CBD}\end{align*}:

\begin{align*}\frac{\text{length of } {\color{red}AB}}{\text{length of } {\color{red}BC}} &= \frac{\text{length of } {\color{green}CB}}{\text{length of } {\color{green}BD}} \intertext{Written with variables, this becomes:} \frac{c}{a} &= \frac{a}{x} \intertext{Next, we can simplify this equation by multiplying both sides of the equation by $a$ and $x$:} {x} \times {a} \times \frac{c}{a} &= \frac{a}{x} \times {x} \times {a} \intertext{With simplification, we obtain:} {cx} &=a^2\end{align*}

Comparing Triangles \begin{align*}{\color{red}ABC}\end{align*} and \begin{align*}{\color{blue}ACD}\end{align*}

Triangle \begin{align*}{\color{red}ABC}\end{align*} is also similar to triangle \begin{align*}{\color{blue}ACD}\end{align*}. Side \begin{align*}{\color{red}AB}\end{align*} corresponds to side \begin{align*}{\color{blue}CA}\end{align*}, side \begin{align*}{\color{red}BC}\end{align*} corresponds to side \begin{align*}{\color{blue}CD}\end{align*}, and side \begin{align*}{\color{red}AC}\end{align*} corresponds to side \begin{align*}{\color{blue}DA}\end{align*}.

Using this set of similar triangles, we can say that:

\begin{align*}\frac{\text{length of } {\color{blue}CA}}{\text{length of } {\color{blue}DA}} &= \frac{\text{length of } {\color{red}AB}}{\text{length of } {\color{red}AC}} \intertext{Written with variables, this becomes:} \frac{b}{c-x} &= \frac{c}{b} \intertext{Similar to before, we can multiply both sides of the equation by $c-x$ and $b$:} {b} \times {(c-x)} \times \frac{b}{c-x} &= \frac{c}{b} \times {b} \times {(c-x)} \\ b^2 &= c(c-x) \\ c^2 &= cx + b^2\end{align*}

Earlier, we found that \begin{align*}cx = a^2\end{align*}. If we replace \begin{align*}cx\end{align*} with \begin{align*}a^2\end{align*}, we obtain \begin{align*}c^2 = a^2 + b^2\end{align*}. This is just another way to express the Pythagorean Theorem. In the triangle \begin{align*}{\color{red}ABC}\end{align*}, side \begin{align*}c\end{align*} is the hypotenuse, while sides \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are the two legs of the triangle.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: