# 11.4: The Pythagorean Theorem and its Converse

**At Grade**Created by: CK-12

One of the most important theorems in history is Pythagorean’s Theorem. Simply put, it states, “The sum of the square of each leg of a right triangle is equal to the square of the hypotenuse.”

Let’s review basic right triangle anatomy.

The two segments forming the right angle \begin{align*}(90^\circ)\end{align*}*legs* of the right triangle. The segment opposite the right angle is called the *hypotenuse*.

So, the Pythagorean Theorem states, \begin{align*}(\text{leg}_1)^2+(\text{leg}_2)^2 = (\text{hypotenuse})^2\end{align*}

\begin{align*}a^2+b^2=c^2\end{align*}

Or, to find the hypotenuse, \begin{align*}c=\sqrt{a^2+ b^2}\end{align*}

Notice this relationship is only true for **right triangles**. In later courses, you will learn how to determine relationships with non-right triangles.

Although we usually refer to the Pythagorean Theorem when determining side lengths of a right triangle, the theorem originally made a statement about areas. If we build squares on each side of a right triangle, the Pythagorean Theorem says that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares formed by the legs of the triangle.

**Multimedia Link:** For an interactive version of Pythagorean’s Theorem, use this Shockwave http://www.pbs.org/wgbh/nova/proof/puzzle/theorem.html - applet produced by NOVA and PBS.

## The Converse of Pythagorean’s Theorem

The **Converse of the Pythagorean Theorem** is also true. That is, if the lengths of three sides of a triangle make the equation \begin{align*}a^2+b^2=c^2\end{align*}

With this converse, you can use the Pythagorean Theorem to prove that a triangle is a right triangle, even if you do not know any of the triangle’s angle measurements.

Example: *Does the triangle below contain a right angle?*

Solution: This triangle does not have any right angle marks or measured angles, so you cannot assume you know whether the triangle is acute, right, or obtuse just by looking at it. Take a moment to analyze the side lengths and see how they are related. Two of the sides, 15 and 17, are relatively close in length. The third side, 8, is about half the length of the two longer sides.

To see if the triangle might be right, try substituting the side lengths into the Pythagorean Theorem to see if they makes the equation true. *The hypotenuse is always the longest side,* so 17 should be substituted for \begin{align*}c\end{align*}

\begin{align*}a^2+b^2 & = c^2\\ 8^2+15^2&=17^2\\ 64+225 &= 289\\ 289 & = 289\end{align*}

Since both sides of the equation are equal, these values satisfy the Pythagorean Theorem. Therefore, the triangle described in the problem is a right triangle.

Example: *One leg of a right triangle is 5 more than the other leg. The hypotenuse is one more than twice the size of the short leg. Find the dimensions of the triangle.*

Solution: Let \begin{align*}x=\end{align*}

The sides of the triangle must satisfy the Pythagorean Theorem.

\begin{align*}&\text{Therefore,} && x^2+(x+5)^2 = (2x+1)\\ &\text{Eliminate the parentheses.} && x^2+x^2+10x+25 = 4x^2+4x+1\\ &\text{Move all terms to the right hand side of the equation.} && 0 = 2x^2-6x-24\\ &\text{Divide all terms by} \ 2. && 0 = x^2-3x-12\\ &\text{Solve using the Quadratic Formula.} && x =\frac{3 \pm \sqrt{9+48}}{2}=\frac{3 \pm \sqrt{57}}{2}\\ &&& x \approx 5.27 \ \text{or} \ x \approx -2.27\end{align*}

The negative solution does not make sense in the context of this problem. So, use \begin{align*}x=5.27\end{align*}

## Real-World Right Triangles

Example: *Find the area of the shaded region in the following diagram.*

**Solution:**

Draw the diagonal of the square on the figure.

Notice that the diagonal of the square is also the diameter of the circle.

Define variables. Let \begin{align*}c=\end{align*}

\begin{align*}&& 2^2+2^2 &= c^2\\ && 4+4&=c^2\\ \text{Write the formula and solve}. && c^2& =8 \Rightarrow c = \sqrt{8} \Rightarrow c = 2\sqrt{2}\end{align*}

The diameter of the circle is \begin{align*}2\sqrt{2}\end{align*}

Area of a circle is \begin{align*}A=\pi r^2 = \pi \left ( \sqrt{2} \right )^2 = 2 \pi\end{align*}

Area of the shaded region is therefore \begin{align*}2\pi -4 \approx 2.28\end{align*}.

## Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.

CK-12 Basic Algebra: Pythagorean Theorem (13:03)

Verify that each triangle is a right triangle.

- \begin{align*}a=12, \ b=9, c=15\end{align*}
- \begin{align*}a=6, \ b=6, \ c=6\sqrt{2}\end{align*}
- \begin{align*}a=8, \ b=8\sqrt{3}, \ c=16\end{align*}

Find the missing length of each right triangle.

- \begin{align*}a=12, \ b=16, \ c=?\end{align*}
- \begin{align*}a=?, \ b=20, \ c=30\end{align*}
- \begin{align*}a=4, \ b=?, \ c=11\end{align*}
- One leg of a right triangle is 4 feet less than the hypotenuse. The other leg is 12 feet. Find the lengths of the three sides of the triangle.
- One leg of a right triangle is 3 more than twice the length of the other. The hypotenuse is 3 times the length of the short leg. Find the lengths of the three legs of the triangle.
- A regulation baseball diamond is a square with 90 feet between bases. How far is second base from home plate?
- Emanuel has a cardboard box that measures \begin{align*}20 \ cm \times 10 \ cm \times 8 \ cm (\text {length} \times \text{width} \times \text{height})\end{align*}. What is the length of the diagonal from a bottom corner to the opposite top corner?
- Samuel places a ladder against his house. The base of the ladder is 6 feet from the house and the ladder is 10 feet long. How high above the ground does the ladder touch the wall of the house?
- Find the area of the triangle if area of a triangle is defined as \begin{align*}A=\frac{1}{2} \text{base} \times \text{height}\end{align*}.
- Instead of walking along the two sides of a rectangular field, Mario decided to cut across the diagonal. He saves a distance that is half of the long side of the field. Find the length of the long side of the field given that the short side is 123 feet.
- Marcus sails due north and Sandra sails due east from the same starting point. In two hours, Marcus’s boat is 35 miles from the starting point and Sandra’s boat is 28 miles from the starting point. How far are the boats from each other?
- Determine the area of the circle.
- In a right triangle, one leg is twice as long as the other and the perimeter is 28. What are the measures of the sides of the triangle?
- Maria has a rectangular cookie sheet that measures \begin{align*}10 \ inches \times 14 \ inches\end{align*}. Find the length of the diagonal of the cookie sheet.
- Mike is loading a moving van by walking up a ramp. The ramp is 10 feet long and the bed of the van is 2.5 feet above the ground. How far does the ramp extend past the back of the van?

**Mixed Review**

- A population increases by 1.2% annually. The current population is 121,000.
- What will the population be in 13 years?
- Assuming this rate continues, when will the population reach 200,000?

- Write \begin{align*}1.29651843 \cdot 10^5\end{align*} in standard form.
- Is \begin{align*}4,2,1,\frac{1}{2},\frac{1}{6},\frac{1}{8},\ldots\end{align*} an example of a geometric sequence? Explain your answer.
- Simplify \begin{align*}6x^3 (4xy^2+y^3 z)\end{align*}.
- Suppose \begin{align*}0=(x-2)(x+1)(x-3)\end{align*}. What are the \begin{align*}x-\end{align*}intercepts?
- Simplify \begin{align*}\sqrt{300}\end{align*}.

## Quick Quiz

- Identify the origin of \begin{align*}h(x)=\sqrt{x-2}+5\end{align*}, then graph the function.
- Simplify \begin{align*}\frac{6}{\sqrt[3]{2}}\end{align*} by rationalizing the denominator.
- Simplify: \begin{align*}\sqrt[4]{-32}\end{align*}. If the answer is not possible, explain why.
- What is an extraneous solution? In what situations do such solutions occur?
- Can 3, 4, 6 form a right triangle?
- Solve \begin{align*}5=\sqrt[3]{y+6}\end{align*}.