# 11.4: The Pythagorean Theorem and Its Converse

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

## Learning Objectives

- Use the Pythagorean Theorem.
- Use the converse of the Pythagorean Theorem.
- Solve real-world problems using the Pythagorean Theorem and its converse.

## Introduction

Teresa wants to string a clothesline across her backyard, from one corner to the opposite corner. If the yard measures 22 feet by 34 feet, how many feet of clothesline does she need?

The **Pythagorean Theorem** is a statement of how the lengths of the sides of a right triangle are related to each other. A right triangle is one that contains a 90 degree angle. The side of the triangle opposite the 90 degree angle is called the **hypotenuse** and the sides of the triangle adjacent to the 90 degree angle are called the **legs**.

If we let \begin{align*}a\end{align*} and \begin{align*}b\end{align*} represent the legs of the right triangle and \begin{align*}c\end{align*} represent the hypotenuse then the Pythagorean Theorem can be stated as:

In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. That is: \begin{align*}a^2+b^2=c^2\end{align*}.

This theorem is very useful because if we know the lengths of the legs of a right triangle, we can find the length of the hypotenuse. Also, if we know the length of the hypotenuse and the length of a leg, we can calculate the length of the missing leg of the triangle. When you use the Pythagorean Theorem, it does not matter which leg you call \begin{align*}a\end{align*} and which leg you call \begin{align*}b\end{align*}, but the hypotenuse is always called \begin{align*}c\end{align*}.

Although nowadays we use the Pythagorean Theorem as a statement about the relationship between distances and lengths, originally the theorem 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.

## Use the Pythagorean Theorem and Its Converse

The Pythagorean Theorem can be used to verify that a triangle is a right triangle. If you can show that the three sides of a triangle make the equation \begin{align*}a^2+b^2=c^2\end{align*} true, then you know that the triangle is a right triangle. This is called the **Converse of the Pythagorean Theorem.**

**Note:** When you use the Converse of the Pythagorean Theorem, you must make sure that you substitute the correct values for the legs and the hypotenuse. The hypotenuse must be the longest side. The other two sides are the legs, and the order in which you use them is not important.

**Example 1**

*Determine if a triangle with sides 5, 12 and 13 is a right triangle.*

**Solution**

The triangle is right if its sides satisfy the Pythagorean Theorem.

If it is a right triangle, the longest side has to be the hypotenuse, so we let \begin{align*}c = 13\end{align*}.

We then designate the shorter sides as \begin{align*}a = 5\end{align*} and \begin{align*}b = 12\end{align*}.

We plug these values into the Pythagorean Theorem:

\begin{align*}a^2+b^2 = c^2 & \Rightarrow 5^2+12^2=c^2\\
25+144=169 = c^2 & \Rightarrow c=13\end{align*} The sides of the triangle satisfy the Pythagorean Theorem, thus **the triangle is a right triangle.**

**Example 2**

*Determine if a triangle with sides, \begin{align*}\sqrt{10}, \sqrt{15}\end{align*} and 5 is a right triangle.*

**Solution**

The longest side has to be the hypotenuse, so \begin{align*}c = 5\end{align*}.

We designate the shorter sides as \begin{align*}a = \sqrt{10}\end{align*} and \begin{align*}b = \sqrt{15}\end{align*}.

We plug these values into the Pythagorean Theorem:

\begin{align*}a^2+b^2 = c^2 & \Rightarrow \left(\sqrt{10}\right)^2+\left(\sqrt{15}\right)^2=c^2\\
10+15=25 = c^2 & \Rightarrow c=5\end{align*} The sides of the triangle satisfy the Pythagorean Theorem, thus **the triangle is a right triangle.**

The Pythagorean Theorem can also be used to find the missing hypotenuse of a right triangle if we know the legs of the triangle. (For a demonstration of this, see

.)

**Example 3**

*In a right triangle one leg has length 4 and the other has length 3. Find the length of the hypotenuse.*

**Solution**

\begin{align*}\text{Start with the Pythagorean Theorem:} && a^2+b^2& =c^2\\ \text{Plug in the known values of the legs:} && 3^2+4^2& =c^2\\ \text{Simplify:} && 9+16& =c^2\\ && 25& =c^2\\ \text{Take the square root of both sides:} && c& =5\end{align*}

## Use the Pythagorean Theorem with Variables

**Example 4**

*Determine the values of the missing sides. You may assume that each triangle is a right triangle.*

a)

b)

c)

**Solution**

Apply the Pythagorean Theorem.

a) \begin{align*}a^2+b^2 &= c^2\\ x^2+15^2 &= 21^2\\ x^2+225 &= 441\\ x^2 &= 216 \Rightarrow \\ x & =\sqrt{216}=6 \sqrt{6}\end{align*}

b) \begin{align*}a^2+b^2 &= c^2\\ y^2+3^2 &= 7^2\\ y^2+9 &= 49\\ y^2 &= 40 \Rightarrow\\ y & =\sqrt{40}=2 \sqrt{10}\end{align*}

c) \begin{align*}a^2+b^2 &= c^2\\
18^2+15^2 &= z^2\\
324+225 &= z\\
z^2 &= 549 \Rightarrow \\
z & =\sqrt{549}=3 \sqrt{61}\end{align*}**Example 5**

*One leg of a right triangle is 5 units longer than the other leg. The hypotenuse is one unit longer than twice the size of the short leg. Find the dimensions of the triangle.*

**Solution**

Let \begin{align*}x =\end{align*} length of the short leg.

Then \begin{align*}x + 5 =\end{align*} length of the long leg

And \begin{align*}2x + 1 =\end{align*} length of the hypotenuse.

The sides of the triangle must satisfy the Pythagorean Theorem.

\begin{align*}\text{Therefore:} && x^2+(x+5)^2& =(2x+1)^2\\ \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& =\underline{\underline{5.27}} \ \text{or} \ x=-2.27\end{align*}

The negative solution doesn’t make sense when we are looking for a physical distance, so we can discard it. Using the positive solution, we get: **\begin{align*}\text{short leg} = 5.27, \text{long leg} = 10.27\end{align*} and \begin{align*}\text{hypotenuse} = 11.54\end{align*}.**

## Solve Real-World Problems Using the Pythagorean Theorem and Its Converse

The Pythagorean Theorem and its converse have many applications for finding lengths and distances.

**Example 6**

*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.*

**Solution**

**Draw a sketch:**

**Define variables:** Let \begin{align*}c =\end{align*} length of the diagonal.

**Write a formula:** Use the Pythagorean Theorem: \begin{align*}a^2+b^2=c^2\end{align*}

**Solve the equation:**

\begin{align*}10^2+14^2 &= c^2\\ 100+196 &= c^2\\ c^2 = 296 & \Rightarrow c=\sqrt{296} \Rightarrow c=2 \sqrt{74} \ \text{or} \ c = 17.2 \ inches\end{align*}

**Check:** \begin{align*}10^2+14^2=100+196=296\end{align*} and \begin{align*}c^2=17.2^2=296\end{align*}. The solution checks out.

**Example 7**

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

**Solution**

Draw the diagonal of the square in the figure:

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

**Define variables:** Let \begin{align*}c =\end{align*} diameter of the circle.

**Write the formula:** Use the Pythagorean Theorem: \begin{align*}a^2+b^2=c^2\end{align*}.

**Solve the equation:**

\begin{align*}2^2+2^2 &= c^2\\ 4+4 &= c^2\\ 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*}, therefore the radius \begin{align*}R=\sqrt{2}\end{align*}.

Area of a circle formula: \begin{align*}A=\pi \cdot R^2=\pi \left(\sqrt{2}\right)^2=2 \pi\end{align*}.

The area of the shaded region is therefore \begin{align*}2 \pi - 4 = 2.28\end{align*}.

**Example 8**

*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?*

**Solution**

**Make a sketch and define variables:**

Let: \begin{align*}a =\end{align*} length of the short leg

\begin{align*}2a =\end{align*} length of the long leg

\begin{align*}c =\end{align*} length of the hypotenuse

**Write formulas:**

The sides of the triangle are related in two different ways.

The perimeter is 28, so \begin{align*}a+2a+c=28 \Rightarrow 3a+c=28\end{align*}

The triangle is a right triangle, so the measures of the sides must satisfy the Pythagorean Theorem:

\begin{align*}&\qquad \qquad a^2+(2a)^2 = c^2 \Rightarrow a^2+4a^2=c^2 \Rightarrow 5a^2=c^2\\ &\text{or} \qquad \quad c = a\sqrt{5}=2.236a\end{align*}

**Solve the equation:**

Plug the value of \begin{align*}c\end{align*} we just obtained into the perimeter equation: \begin{align*}3a+c=28\end{align*}

\begin{align*}3a+2.236a=28 \Rightarrow 5.236a=28 \Rightarrow a=5.35\end{align*}

The short leg is: \begin{align*}a = 5.35\end{align*}

The long leg is: \begin{align*}2a = 10.70\end{align*}

The hypotenuse is: \begin{align*}c = 11.95\end{align*}

**Check:** The legs of the triangle should satisfy the Pythagorean Theorem:

\begin{align*}a^2+b^2=5.35^2+10.70^2=143.1, c^2=11.95^2=142.80\end{align*}. The results are approximately the same.

The perimeter of the triangle should be 28:

\begin{align*}a+b+c=5.35+10.70+11.95=28\end{align*} **The answer checks out.**

**Example 9**

*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?*

**Solution**

**Make a sketch:**

**Define variables:** Let \begin{align*}x =\end{align*} how far the ramp extends past the back of the van.

**Write a formula:** Use the Pythagorean Theorem: \begin{align*}x^2+2.5^2=10^2\end{align*}

**Solve the equation:**

\begin{align*}x^2+6.25 &= 100\\
x^2 &= 93.5\\
x &= \sqrt{93.5} = 9.7 \ ft\end{align*} **Check** by plugging the result in the Pythagorean Theorem:

\begin{align*}9.7^2+2.5^2=94.09+6.25=100.34 \approx 100\end{align*}. So the ramp is 10 feet long. **The answer checks out.**

## Review Questions

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*}
- \begin{align*}a =2 \sqrt{14}, b = 5, c = 9\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.
- Two sides of a right triangle are 5 units and 8 units respectively. Those sides could be the legs, or they could be one leg and the hypotenuse. What are the possible lengths of the third side?
- 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 \ \text{long} \ \times \ 10 \ cm \ \text{wide} \ \times \ 8 \ cm \ \text{deep}\end{align*}.
- What is the length of the diagonal across the bottom of the box?
- 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?
- If the edge of the roof is 10 feet off the ground and sticks out 1.5 feet beyond the wall, how far is it from the edge of the roof to the top of the ladder?

- Find the area of the triangle below if the area of a triangle is defined as \begin{align*}A=\frac{1}{2} \ base \times height\end{align*}:
- Instead of walking along the two sides of a rectangular field, Mario decided to cut across the diagonal. He thus 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.
- Find the length of the diagonal.

- Marcus sails due north and Sandra sails due east from the same starting point. In two hours Marcus’ 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?
- Sandra then sails 21 miles due north while Marcus stays put. How far is Sandra from the original starting point?
- How far is Sandra from Marcus now?

- Determine the area of the circle below. (Hint: the hypotenuse of the triangle is the diameter of the circle.)

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