# 13.6.4: Triangles, Rectangles, and the Pythagorean Theorem - Mathematics

Learning Objectives

By the end of this section, you will be able to:

• Solve applications using properties of triangles
• Use the Pythagorean Theorem
• Solve applications using rectangle properties

Before you get started, take this readiness quiz.

1. Simplify: (12(6h)).
If you missed this problem, review Exercise 1.10.1.
2. The length of a rectangle is three less than the width. Let w represent the width. Write an expression for the length of the rectangle.
If you missed this problem, review Exercise 1.3.43.
3. Solve: (A=frac{1}{2}bh) for b when A=260 and h=52.
If you missed this problem, review Exercise 2.6.10.
4. Simplify: (sqrt{144}).
If you missed this problem, review Exercise 1.9.10.

## Solve Applications Using Properties of Triangles

In this section we will use some common geometry formulas. The geometry formula will name the variables and give us the equation to solve. In addition, since these applications will all involve shapes of some sort, most people find it helpful to draw a figure and label it with the given information. We will include this in the first step of the problem solving strategy for geometry applications.

SOLVE GEOMETRY APPLICATIONS

1. Read the problem and make sure all the words and ideas are understood. Draw the figure and label it with the given information.
2. Identify what we are looking for.
3. Label what we are looking for by choosing a variable to represent it.
4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
5. Solve the equation using good algebra techniques.
6. Check the answer by substituting it back into the equation solved in step 5 and by making sure it makes sense in the context of the problem.
7. Answer the question with a complete sentence.

We will start geometry applications by looking at the properties of triangles. Let’s review some basic facts about triangles. Triangles have three sides and three interior angles. Usually each side is labeled with a lowercase letter to match the uppercase letter of the opposite vertex.

The plural of the word vertex is vertices. All triangles have three vertices. Triangles are named by their vertices: The triangle in Figure (PageIndex{1}) is called ( riangle{ABC}).

The three angles of a triangle are related in a special way. The sum of their measures is (180^{circ}). Note that we read (mangle{A}) as “the measure of angle A.” So in ( riangle{ABC}) in Figure (PageIndex{1}).

[m angle A+m angle B+m angle C=180^{circ} onumber]

Because the perimeter of a figure is the length of its boundary, the perimeter of ( riangle{ABC}) is the sum of the lengths of its three sides.

[P = a + b + c onumber]

To find the area of a triangle, we need to know its base and height. The height is a line that connects the base to the opposite vertex and makes a (90^circ) angle with the base. We will draw ( riangle{ABC}) again, and now show the height, (h). See Figure (PageIndex{2}).

TRIANGLE PROPERTIES

For ( riangle{ABC})

Angle measures:

[m angle A+m angle B+m angle C=180^{circ}]

• The sum of the measures of the angles of a triangle is 180°.

Perimeter:

[P = a + b + c]

• The perimeter is the sum of the lengths of the sides of the triangle.

Area:

(A = frac{1}{2}bh, b = ext{ base }, h = ext{ height })

• The area of a triangle is one-half the base times the height.

Example (PageIndex{1})

The measures of two angles of a triangle are 55 and 82 degrees. Find the measure of the third angle.

Solution

 Step 1. Read the problem. Draw the figure and label it with the given information. Step 2. Identify what you are looking for. the measure of the third angle in a triangle Step 3. Choose a variable to represent it. Let (x=) the measure of the angle. Step 4. Write the appropriate formula and substitute. (m angle A+m angle B+m angle C=180^{circ}) Step 5. Solve the equation. (egin{array} {rll} {55 + 82 + x} &{=} &{180} {137 + x} &{=} &{180} {x} &{=} &{43} end{array}) Step 6. (egin{array} {rll} {55 + 82 + 43} &{stackrel{?}{=}} &{180} {180} &{=} &{180checkmark} end{array}) Step 7. Answer the question. The measure of the third angle is 43 degrees.

Try It (PageIndex{1})

The measures of two angles of a triangle are 31 and 128 degrees. Find the measure of the third angle.

Answer

21 degrees

Try It (PageIndex{2})

The measures of two angles of a triangle are 49 and 75 degrees. Find the measure of the third angle.

Answer

56 degrees

Example (PageIndex{2})

The perimeter of a triangular garden is 24 feet. The lengths of two sides are four feet and nine feet. How long is the third side?

Solution

 Step 1. Draw the figure and label it with the given information. Step 2. Identify what you are looking for. length of the third side of a triangle Step 3. Choose a variable to represent it. Let (c=) the third side. Step 4. Write the appropriate formula and substitute. Substitute in the given information. Step 5. Solve the equation. Step 6. (egin{array} {rll} {P} &{=} &{a + b +c} {24} &{stackrel{?}{=}} &{4 + 9+11} {24} &{=} &{24checkmark} end{array}) Step 7. Answer the question. The third side is 11 feet long.

Try It (PageIndex{3})

The perimeter of a triangular garden is 48 feet. The lengths of two sides are 18 feet and 22 feet. How long is the third side?

Answer

8 feet

Try It (PageIndex{4})

The lengths of two sides of a triangular window are seven feet and five feet. The perimeter is 18 feet. How long is the third side?

Answer

6 feet

Example (PageIndex{3})

The area of a triangular church window is 90 square meters. The base of the window is 15 meters. What is the window’s height?

Solution

 Step 1. Draw the figure and label it with the given information. ( ext{ Area } = 90m^{2}) Step 2. Identify what you are looking for. height of a triangle Step 3. Choose a variable to represent it. Let (h=) the height. Step 4. Write the appropriate formula. Substitute in the given information. Step 5. Solve the equation. (90 = dfrac{15}{2}h)(12 = h) Step 6. (egin{array} {rll} {A} &{=} &{frac{1}{2}bh} {90} &{stackrel{?}{=}} &{frac{1}{2}cdot 15cdot 12} {90} &{=} &{90checkmark} end{array}) Step 7. Answer the question. The height of the triangle is 12 meters.

Try It (PageIndex{5})

The area of a triangular painting is 126 square inches. The base is 18 inches. What is the height?

Answer

14 inches

Try It (PageIndex{6})

A triangular tent door has area 15 square feet. The height is five feet. What is the base?

Answer

6 feet

The triangle properties we used so far apply to all triangles. Now we will look at one specific type of triangle—a right triangle. A right triangle has one 90° angle, which we usually mark with a small square in the corner.

Definition: RIGHT TRIANGLE

A right triangle has one 90° angle, which is often marked with a square at the vertex.

Example (PageIndex{4})

One angle of a right triangle measures 28°. What is the measure of the third angle?

Solution

 Step 1. Draw the figure and label it with the given information. Step 2. Identify what you are looking for. the measure of an angle Step 3. Choose a variable to represent it. Let (x=) the measure of an angle. Step 4. (mangle{A} + mangle{B} + mangle{C} = 180) Write the appropriate formula and substitute. (x+90+28=180) Step 5. Solve the equation. (x=62) Step 6. (egin{array} {rll} {180} &{stackrel{?}{=}} &{90+28+62} {180} &{=} &{180checkmark} end{array}) Step 7. Answer the question. The measure of the third angle is 62°.

Try It (PageIndex{7})

One angle of a right triangle measures 56°. What is the measure of the other small angle?

Answer

34°

Try It (PageIndex{8})

One angle of a right triangle measures 45°. What is the measure of the other small angle?

Answer

45°

In the examples we have seen so far, we could draw a figure and label it directly after reading the problem. In the next example, we will have to define one angle in terms of another. We will wait to draw the figure until we write expressions for all the angles we are looking for.

Example (PageIndex{5})

The measure of one angle of a right triangle is 20 degrees more than the measure of the smallest angle. Find the measures of all three angles.

Solution

 Step 1. Read the problem. Step 2. Identify what you are looking for. the measures of all three angles Step 3. Choose a variable to represent it. Let (a=1^{st}) angle.(a+20=2^{nd}) angle(90=3^{rd}) angle (the right angle) Draw the figure and label it with the given information Step 4. Translate Write the appropriate formula.Substitute into the formula. (a + (a + 20) + 90 = 180) Step 5. Solve the equation. (egin{align*} 2a + 110 &= 180 [3pt]2a &= 70 [3pt]a &= 35 ext{ first angle}[3pt]a &+ 20 ext{ second angle}[3pt]{color{red}{35}} &+ 20 = 55 end{align*})And the third angle is 90. Step 6. (egin{array} {rll} {35 + 55 + 90} &{stackrel{?}{=}} &{180} {180} &{=} &{180checkmark} end{array}) Step 7. Answer the question. The three angles measure 35°, 55°, and 90°.

Try It (PageIndex{9})

The measure of one angle of a right triangle is 50° more than the measure of the smallest angle. Find the measures of all three angles.

Answer

20°,70°,90°

Try It (PageIndex{10})

The measure of one angle of a right triangle is 30° more than the measure of the smallest angle. Find the measures of all three angles.

Answer

30°,60°,90°

## Use the Pythagorean Theorem

We have learned how the measures of the angles of a triangle relate to each other. Now, we will learn how the lengths of the sides relate to each other. An important property that describes the relationship among the lengths of the three sides of a right triangle is called the Pythagorean Theorem. This theorem has been used around the world since ancient times. It is named after the Greek philosopher and mathematician, Pythagoras, who lived around 500 BC.

Before we state the Pythagorean Theorem, we need to introduce some terms for the sides of a triangle. Remember that a right triangle has a 90° angle, marked with a small square in the corner. The side of the triangle opposite the 90°90° angle is called the hypotenuse and each of the other sides are called legs.

The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other. It states that in any right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse. In symbols we say: in any right triangle, (a^{2}+b^{2}=c^{2}), where a and b are the lengths of the legs and cc is the length of the hypotenuse.

Writing the formula in every exercise and saying it aloud as you write it, may help you remember the Pythagorean Theorem.

THE PYTHAGOREAN THEOREM

In any right triangle, where (a) and (b) are the lengths of the legs, (c) is the length of the hypotenuse.

Then

[a^{2}+b^{2}=c^{2} label{Ptheorem}]

To solve exercises that use the Pythagorean Theorem (Equation ef{Ptheorem}), we will need to find square roots. We have used the notation (sqrt{m}) and the definition:

If (m = n^{2}), then (sqrt{m} = n), for (ngeq 0).

For example, we found that (sqrt{25}) is 5 because (25=5^{2}).

Because the Pythagorean Theorem contains variables that are squared, to solve for the length of a side in a right triangle, we will have to use square roots.

Example (PageIndex{6})

Use the Pythagorean Theorem to find the length of the hypotenuse shown below.

Solution

 Step 1. Identify what you are looking for. the length of the hypotenuse of the triangle Step 3. Choose a variable to represent it.Label side c on the figure. Let c = the length of the hypotenuse. Step 4. Translate. Write the appropriate formula. (a^{2} + b^{2} = c^{2}) Substitute. (3^{2}+4^{2}=c^{2}) Step 5. Solve the equation. (9+16=c^{2}) Simplify. (25=c^{2}) Use the definition of square root. (sqrt{25} = c) Simplify. (5=c) Step 6. Step 7. Answer the question. The length of the hypotenuse is 5.

Try It (PageIndex{11})

Use the Pythagorean Theorem to find the length of the hypotenuse in the triangle shown below.

Answer

c=10

Try It (PageIndex{12})

Use the Pythagorean Theorem to find the length of the hypotenuse in the triangle shown below.

Answer

c=13

Example (PageIndex{7})

Use the Pythagorean Theorem to find the length of the leg shown below.

Solution

 Step 1. Identify what you are looking for. the length of the leg of the triangle Step 3. Name. Choose a variable to represent it. Let (b=) the leg of the triangle. Label side (b). Step 4. Translate Write the appropriate formula. (a^{2} + b^{2} = c^{2}) Substitute. (5^{2}+b^{2}=13^{2}) Step 5. Solve the equation. (25+b^{2}=169) Isolate the variable term. (b^{2}=144) Use the definition of square root. (b = sqrt{144}) Simplify. (b=12) Step 6. Step 7. Answer the question. The length of the leg is 12.

Try It (PageIndex{13})

Use the Pythagorean Theorem to find the length of the leg in the triangle shown below.

Answer

8

Try It (PageIndex{14})

Use the Pythagorean Theorem to find the length of the leg in the triangle shown below.

Answer

12

Example (PageIndex{8})

Kelvin is building a gazebo and wants to brace each corner by placing a 10″ piece of wood diagonally as shown above.

If he fastens the wood so that the ends of the brace are the same distance from the corner, what is the length of the legs of the right triangle formed? Approximate to the nearest tenth of an inch.

Solution

(egin{array} {ll} { extbf{Step 1. } ext{Read the problem.}} &{} { extbf{Step 2. } ext{Identify what we are looking for.}} &{ ext{the distance from the corner that the}} {} &{ ext{bracket should be attached}} { extbf{Step 3. } ext{Name. Choose a variable to represent it.}} &{ ext{Let x = distance from the corner.}} { extbf{Step 4.} ext{Translate}} &{} { ext{Write the appropriate formula and substitute.}} &{a^{2} + b^{2} = c^{2}} {} &{x^{2} + x^{2} = 10^{2}} { extbf{Step 5. Solve the equation.}} &{} {} &{2x^{2} = 100} { ext{Isolate the variable.}} &{x^{2} = 50} { ext{Simplify. Approximate to the nearest tenth.}} &{x approx 7.1} { extbf{Step 6. } ext{Check.}} &{} {a^{2} + b^{2} = c^{2}} &{} {(7.1)^{2} + (7.1)^{2} approx 10^{2} ext{ Yes.}} &{} { extbf{Step 7. Answer the question.}} &{ ext{Kelven should fasten each piece of}} {} &{ ext{wood approximately 7.1'' from the corner.}} end{array})

Try It (PageIndex{15})

John puts the base of a 13-foot ladder five feet from the wall of his house as shown below. How far up the wall does the ladder reach?

Answer

12 feet

Try It (PageIndex{16})

Randy wants to attach a 17 foot string of lights to the top of the 15 foot mast of his sailboat, as shown below. How far from the base of the mast should he attach the end of the light string?

Answer

8 feet

## Solve Applications Using Rectangle Properties

You may already be familiar with the properties of rectangles. Rectangles have four sides and four right (90°) angles. The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, (L), and its adjacent side as the width, (W).

The distance around this rectangle is (L+W+L+W), or (2L+2W). This is the perimeter, (P), of the rectangle.

[P=2L+2W]

What about the area of a rectangle? Imagine a rectangular rug that is 2-feet long by 3-feet wide. Its area is 6 square feet. There are six squares in the figure.

[egin{array} {l} {A=6} {A=2cdot3} {A=Lcdot W} end{array}]

The area is the length times the width. The formula for the area of a rectangle is

[A=LW.]

PROPERTIES OF RECTANGLES

Rectangles have four sides and four right (90°) angles.

The lengths of opposite sides are equal.

The perimeter of a rectangle is the sum of twice the length and twice the width.

[P=2L+2W]

The area of a rectangle is the product of the length and the width.

[A=L·W]

Example (PageIndex{9})

The length of a rectangle is 32 meters and the width is 20 meters. What is the perimeter?

Solution

 Step 1. Read the problem.Draw the figure and label it with the given information. Step 2. Identify what you are looking for. the perimeter of a rectangle Step 3. Name. Choose a variable to represent it. Let (P=) the perimeter. Step 4. Translate. Write the appropriate formula. Substitute. Step 5. Solve the equation. (P = 64 + 40)(P = 104) Step 6. (egin{array} {rcl} {P} &{stackrel{?}{=}} &{104} {20+32+20+32} &{stackrel{?}{=}} &{104} {104} &{=} &{104checkmark} end{array}) Step 7. Answer the question. The perimeter of the rectangle is 104 meters.

Try It (PageIndex{17})

The length of a rectangle is 120 yards and the width is 50 yards. What is the perimeter?

Answer

340 yards

Try It (PageIndex{18})

The length of a rectangle is 62 feet and the width is 48 feet. What is the perimeter?

Answer

220 feet

Example (PageIndex{10})

The area of a rectangular room is 168 square feet. The length is 14 feet. What is the width?

Solution

 Step 1. Read the problem.Draw the figure and label it with the given information. Step 2. Identify what you are looking for. the width of a rectangular room Step 3. Name. Choose a variable to represent it. Let (W=) the width. Step 4. Translate. Write the appropriate formula. (A=LW) Substitute. (168 = 14W) Step 5. Solve the equation. (frac{168}{14} = frac{14W}{14})(12 = W) Step 6. (egin{array} {rcl} {A} &{=} &{LW} {168} &{stackrel{?}{=}} &{14cdot 12} {168} &{=} &{168checkmark} end{array}) Step 7. Answer the question. The width of the room is 12 feet.

Try It (PageIndex{19})

The area of a rectangle is 598 square feet. The length is 23 feet. What is the width?

Answer

26 feet

Try It (PageIndex{20})

The width of a rectangle is 21 meters. The area is 609 square meters. What is the length?

Answer

29 meters

Example (PageIndex{11})

Find the length of a rectangle with perimeter 50 inches and width 10 inches.

Solution

 Step 1. Read the problem.Draw the figure and label it with the given information. Step 2. Identify what you are looking for. the length of the rectangle Step 3. Name. Choose a variable to represent it. Let (L=) the length. Step 4. Translate. Write the appropriate formula. (P = 2L + 2W) Substitute. (50 = 2L + 2(10)) Step 5. Solve the equation. Step 6. (egin{array} {rcl} {P} &{=} &{50} {15+10+15+10} &{stackrel{?}{=}} &{50} {50} &{=} &{50checkmark} end{array}) Step 7. Answer the question. The length is 15 inches.

Try It (PageIndex{21})

Find the length of a rectangle with: perimeter 80 and width 25.

Answer

15

Try It (PageIndex{22})

Find the length of a rectangle with: perimeter 30 and width 6.

Answer

9

We have solved problems where either the length or width was given, along with the perimeter or area; now we will learn how to solve problems in which the width is defined in terms of the length. We will wait to draw the figure until we write an expression for the width so that we can label one side with that expression.

Example (PageIndex{12})

The width of a rectangle is two feet less than the length. The perimeter is 52 feet. Find the length and width.

Solution

 Step 1. Identify what you are looking for. the length and width of a rectangle Step 3. Name. Choose a variable to represent it.Since the width is defined in terms of the length, we let (L=) length. The width is two feet less than the length, so we let (L-2) width. (P=52) ft Step 4. Translate. Write the appropriate formula. The formula for the perimeter of a rectangle relates all the information. (P=2L+2W) Substitute in the given information. (52=2L+2(L−2)) Step 5. Solve the equation. (52=2L+2L−4) Combine like terms. (52=4L−4) Add 4 to each side. (56 = 4L) Divide by 4. (frac{56}{4} = frac{4L}{4})(14=L)The length is 14 feet. Now we need to find the width. The width is (L−2).The width is 12 feet. Step 6. Since (14+12+14+12=52), this works! Step 7. Answer the question. The length is 14 feet and the width is 12 feet.

Try It (PageIndex{23})

The width of a rectangle is seven meters less than the length. The perimeter is 58 meters. Find the length and width.

Answer

18 meters, 11 meters

Try It (PageIndex{24})

The length of a rectangle is eight feet more than the width. The perimeter is 60 feet. Find the length and width.

Answer

19 feet, 11 feet

Example (PageIndex{13})

The length of a rectangle is four centimeters more than twice the width. The perimeter is 32 centimeters. To the right of this, we have the length and width. To the right of this, we have let w equal the width. The length is four more than twice the width; hence, we let 2w plus 4 equal the length. Below this, we have a rectangle with length 2w plus 4 and width w. Below this, we have P equals 32 cm. Write the appropriate formula and substitute in the given information. Below this, we have 32 equals 2 times the quantity (2w plus 4) plus 2w. To the right of this, we have 32 equals 4w plus 8 plus 2w. Below this, we have 32 equals 6w plus 8. Below this, 24 equals 6w. Below this 4 equals w. Below this the length is determined: 2w plus 4, so 2 times 4 plus 4 equals 12. The length is 12 cm. We have a rectangle with length 12 cm and width 4 cm. Below this, we have P equals 2L plus 2W. Below this we have 32 equals with a question mark over it 2 times 12 plus 2 times 4. Below this, we have 32 equals 32. Answer the question: The length is 12 cm, and the width is 4 cm.">Step 1. Identify what you are looking for.the length and the widthStep 3. Name. Choose a variable to represent the width.The length is four more than twice the width.

Step 4. TranslateWrite the appropriate formula.(quad P=2L+2W)Substitute in the given information.Step 5. Solve the equation.

12
The length is 12 cm.

Step 6.

(egin{array} {rcl} {P} &{=} &{2L + 2W} {32} &{stackrel{?}{=}} &{2cdot 12 + 2cdot 4} {32} &{=} &{32checkmark} end{array})

Step 7. Answer the question.The length is 12 cm and the width is 4 cm.

Try It (PageIndex{25})

The length of a rectangle is eight more than twice the width. The perimeter is 64. Find the length and width.

Answer

24, 8

Try It (PageIndex{26})

The width of a rectangle is six less than twice the length. The perimeter is 18. Find the length and width.

Answer

5, 4

Try It (PageIndex{27})

The perimeter of a rectangular swimming pool is 200 feet. The length is 40 feet more than the width. Find the length and width.

Answer

70 feet, 30 feet

Try It (PageIndex{28})

The length of a rectangular garden is 30 yards more than the width. The perimeter is 300 yards. Find the length and width.

Answer

90 yards, 60 yards

## Key Concepts

• Problem-Solving Strategy for Geometry Applications
1. Read the problem and make all the words and ideas are understood. Draw the figure and label it with the given information.
2. Identify what we are looking for.
3. Name what we are looking for by choosing a variable to represent it.
4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.
• Triangle Properties For △ABC
Angle measures:
• (mangle{A}+mangle{B}+mangle{C}=180)
Perimeter:
• (P=a+b+c)
Area:
• (A=frac{1}{2}bh), b=base,h=height
A right triangle has one 90° angle.
• The Pythagorean Theorem In any right triangle, (a^{2} + b^{2} = c^{2}) where (c) is the length of the hypotenuse and (a) and (b) are the lengths of the legs.
• Properties of Rectangles
• Rectangles have four sides and four right (90°) angles.
• The lengths of opposite sides are equal.
• The perimeter of a rectangle is the sum of twice the length and twice the width: (P=2L+2W).
• The area of a rectangle is the length times the width: (A=LW).

Pythagorean Theorem
Investigate the areas of the squares on the sides of right angled triangles using this interactive figure.
www.interactive-maths.com/pythagoras-theorem-ggb.html

Proof Without Words: Pythagorean Theorem
Watch a dynamic, geometric "proof without words" of the Pythagorean Theorem. Can you explain the proof?
illuminations.nctm.org/Activity.aspx?id=4211

Pythagorean Explorer
Practice calculating the missing side length of right triangles. Includes three difficulty levels.
www.shodor.org/interactivate/activities/PythagoreanExplorer

The Pythagorean Theorem - a session from Geometry course by Learning Math
An online course that includes tutorials, problems, videos, interactive activities, homework problems and solutions. In this session, you will look at a few proofs and several applications of one of the most famous theorems in mathematics: the Pythagorean theorem.
www.learner.org/courses/learningmath/geometry/session6/index.html

Pythagorean Theorem Lesson
This lesson introduces and explores the Pythagorean Theorem using the Pythagorean Explorer. Three computer activities give students the opportunity to observe triangles, learn and use the Pythagorean Theorem and practice different ways of determining areas of triangles.
www.shodor.org/interactivate/lessons/pyth.html

Pythagorean Theorem for elementary school?
This is a lesson plan showing how you could introduce the Pythagorean Theorem in elementary school. I don't really recommend such because I feel it belongs to the middle school, but it seems some student-teachers are asked to write a lesson for such.
/teaching/pythagorean-theorem-elementary.php

## Geometry: Pythagorean Theorem

The following diagram gives the formula for the Pythagorean Theorem, scroll down the page for more examples and solutions that use the Pythagorean Theorem.

### What Is The Pythagorean Theorem?

A right triangle consists of two sides called the legs and one side called the hypotenuse. The hypotenuse is the longest side and is opposite the right angle.

The Pythagorean Theorem or Pythagoras' Theorem is a formula relating the lengths of the three sides of a right triangle.

If we take the length of the hypotenuse to be c and the length of the legs to be a and b then this theorem tells us that:

Pythagorean Theorem states that

In any right triangle, the sum of the squared lengths of the two legs is equal to the squared length of the hypotenuse.

Note: Pythagorean theorem only works for right triangles.

Proof Of The Pythagorean Theorem Using Similar Triangles

This proof is based on the proportionality of the sides of two similar triangles, that is, the ratio of any corresponding sides of similar triangles is the same regardless of the size of the triangles.

1. Determine the length of the missing side of the right triangle.
2. Mason wants to lay pavers in his families' backyard for the summer. It is important that he start the pavers at a right angle. If he want the dimensions of the patio to be 8 ft by 10 ft, what should the diagonal measure?

What Is The Converse Of The Pythagorean Theorem?

The converse of the Pythagorean Theorem is also true.

For any triangle with sides a, b, and c, if a 2 + b 2 = c 2 , then the angle between a and b measures 90° and the triangle is a right triangle.

How To Use The Converse Of The Pythagorean Theorem?

We can use the converse of the Pythagorean Theorem to check whether a given triangle is an acute triangle, a right triangle or an obtuse triangle.

For a triangle with sides a, b and c and c is the longest side then: If c 2 < a 2 + b 2 then it is an acute-angled triangle, i.e. the angle facing side c is an acute angle.

If c 2 = a 2 + b 2 then it is a right-angled triangle, i.e. the angle facing side c is a right angle.

If c 2 > a 2 + b 2 then it is an obtuse-angled triangle, i.e. the angle facing side c is an obtuse angle.

How to use the Converse of the Pythagorean Theorem?

This video shows how to use the Pythagorean Theorem and its Converse to determine if a triangle is acute, right, or obtuse.

According to the triangle inequality theorem, the sum of the two shorter sides of a triangle must be greater that the longest side.

Examples: Determine if the lengths represent the sides of an acute, right, or obtuse triangle if a triangle is possible.

How to use the Pythagorean Theorem?

The Pythagorean Theorem can be used when we know the length of two sides of a right triangle and we need to get the length of the third side.

Example 1:
Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 3 inches and 4 inches.

Solution:
Step 1: Write down the formula
c 2 = a 2 + b 2

Step 2: Plug in the values
c 2 = 3 2 + 4 2
c 2 = 9 + 16<
c 2 = 25
c = √25
c = 5

Answer: The length of the hypotenuse is 5 inches.

Example 2:
Find the length of one side of a right triangle if the length of the hypotenuse is 10 inches and the length of the other side is 9 inches.

Solution:
Step 1: Write down the formula
c 2 = a 2 + b 2

Step 2: Plug in the values
10 2 = 9 2 + b 2
100 = 81 + b 2
Step 3: Subtract 81 from both sides
19 = b 2
b = √19
b ≈ 4.36

Answer: The length of the side is 4.36 inches.

How to use the Pythagorean Theorem to solve real-world problems?

1. Claire wants to hang a banner from the sill of a second-story window in her house. She needs to find a ladder that, when rested against the outside wall of her house will be long enough to reach the second-story window. If the window is 16 feet above the ground and Claire places the foot of the ladder 12 feet from the wall, how long will the ladder need to be?
2. During a baseball game, the second baseman gets the ball and throws it to the catcher to stop a runner before he gets to home. If it is 90 feet between each base, how far did the second baseman throw the ball?
3. A water park wants to add a zipline into a pool. If the platform at the top of the zipline is 25 feet tall, and the pool is 40 feet ling, what is the maximum length needed for the zipline?
4. A wheelchair ramp is needed at the entrance to a building. There is only 10 feet of space available for the ramp. How long should the ramp be?
5. A television screen is advertised as 50 inches. If the television is 35 inches wide, how tall is it?
6. A kite at the end of a 40 feet line is 10 feet behind the runner. How high is the kite?
7. A roof is being placed on a frame that is 9 feet tall and 30 feet wide. How long are the diagonal pieces of the frame?

Proofs of the Pythagorean Theorem

There are many ways to proof the Pythagorean Theorem. We will look at three of them here.

How to proof the Pythagorean Theorem using Similar Triangles?

This proof is based on the fact that the ratio of any two corresponding sides of similar triangles is the same regardless of the size of the triangles.

Given Triangle ABC drawn above in the image and prove a 2 + b 2 = c 2 using Similar Triangles.

Triangle ABC ∼ Triangle ACD AA (Similarity Postulate)

Triangle ABC ∼ Triangle CBD AA (Similarity Postulate)

c/a = a/x (Converse of SSS Similarity Postulate)

c/b = b/y (Converse of SSS Similarity Postulate)

a 2 + b 2 = cx + cy (Adding the equations)

How to proof the Pythagorean Theorem using Algebra?

In this proof, we use four copies of the right triangle, rearrange them and use algebra to proof the theorem.

How to proof of the Pythagorean Theorem using Rearrangement of shapes?

The following video shows how a square with area c 2 can be cut up and rearranged such that it can fit into two other smaller squares with areas a 2 and b 2 .

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

## Pythagorean Theorem Worksheets

These printable worksheets have exercises on finding the leg and hypotenuse of a right triangle using the Pythagorean theorem. Pythagorean triple charts with exercises are provided here. Word problems on real time application are available. Moreover, descriptive charts on the application of the theorem in different shapes are included. These handouts are ideal for 7th grade, 8th grade, and high school students. Kick into gear with our free Pythagorean theorem worksheets!

Apply Pythagorean theorem to identify whether the given triangle is a right triangle. Each printable worksheet consists of six problems.

These descriptive charts explain the Pythagorean theorem with an illustration. These pdfs emphasize the relation of the theorem derived as an equation.

A set of three numbers is given in each problem. Grade 7 and grade 8 students need to apply the theorem and identify whether the set of numbers forms a Pythagorean triple.

This section comprises of Pythagorean triple sets up to 100. Besides, Pythagorean triple formulas with examples are provided in the charts.

Apply Pythagorean theorem to find the unknown side of the right triangle. Round the answer to the nearest tenth.

Apply the Pythagorean theorem to find the unknown length of each shape in these printable worksheets. Round the answer to the nearest tenth.

Brighten your math class with this bundle of real-life word problems based on the Pythagorean Theorem. Solve each word problem by finding the missing hypotenuse of the right triangle and rounding off the answer to the nearest tenth.

Presenting word problems with clear illustrations, these pdf worksheets require high school students to plug in the known values into the equation form of the Pythagorean Theorem and figure out the unknown side of the right triangle.

## 500 Lightstick-Toting Math Enthusiasts 'Pythagorized' NYC's Iconic Flatiron Building In Epic Fashion

"Pythagorizing the Flatiron" is the first in a planned series of "MathHappenings" to be run by MoMath.

The proof of the Flatiron building's right triangle nature is based on the Pythagorean theorem — the statement that for a right triangle with legs (shorter sides) of lengths a and b, and hypotenuse (long side) of length c, the sum of the squares of the two shorter lengths equals the square of the long length — a 2 + b 2 = c 2 .

The Museum of Math flipped this idea around — if the lengths of the sides of a triangle, like the Flatiron building, satisfy the Pythagorean theorem, then the triangle must be a right triangle.

MoMath measured the sides of the Flatiron building in a unique way. People lined up around the three sides of the building, and MoMath workers and volunteers handed out lightsticks that the math enthusiasts held end to end. By counting while handing out the glowing toys, MoMath was able to estimate the length of the building's sides in terms of lightsticks.

The shortest side of the building, along 22nd St and designated side A, measured 75 lightsticks. The longer leg of the building, going up 5th Ave and designated side B, measured 180 lightsticks. The longest side of the building, the hypotenuse C running along Broadway, had a length of 195 lightsticks.

Having found the lengths of the sides of the building in lightsticks, MoMath then projected on the side of the building the calculations that showed that the three sides do in fact match up with the Pythagorean Theorem, proving that the Flatiron is in fact a right triangle:

Right triangles whose sides all have whole number lengths are special. For most right triangles, at least one of the three sides will be an irrational number — a right triangle whose shorter sides both have length one will have a hypotenuse of length square root of 2 (since, again by the Pythagorean Theorem, 1 2 + 1 2 = 1 + 1 = 2, and 2 is, by definition, the square root of 2, squared).

Sets of three whole numbers that describe a right triangle, like 75, 180, and 195, are called Pythagorean triples, and they have been of interest to mathematicians since the time of the ancient Greeks.

This particular Pythagorean triple also has one other interesting property. All three of the numbers 75, 180, and 195 can be divided by 15 without leaving a remainder: 75 ÷ 15 = 5, 180 ÷ 15 = 12, and 195 ÷ 15 = 13. So, if we "rescale" the length measurements by performing this division on the three lengths, we get that the sides of the Flatiron building are 5, 12, and 13.

MoMath thus was very clever in choosing the date for the event — December 5, 2013, or 12/5/13, matching the sides of the building.

After making the measurements of the Flatiron building's sides and showing their Pythagorean relationship, MoMath projected a couple nice geometric proofs of the Pythagorean theorem onto the side of the building:

Events like this are a delightful way for all kinds of people to participate in the elegance and beauty of mathematics. The Pythagorean Theorem is a core concept in our understanding of geometry, and in many ways, it defines the shape of our world. It was exciting to see this fundamental mathematical principle brought to life in a fun and interactive way.

## 13.6.4: Triangles, Rectangles, and the Pythagorean Theorem - Mathematics

Let's build up squares on the sides of a right triangle. Pythagoras' Theorem then claims that the sum of (the areas of) two small squares equals (the area of) the large one.

In algebraic terms, a 2 + b 2 = c 2 where c is the hypotenuse while a and b are the sides of the triangle.

The theorem is of fundamental importance in the Euclidean Geometry where it serves as a basis for the definition of distance between two points. It's so basic and well known that, I believe, anyone who took geometry classes in high school couldn't fail to remember it long after other math notions got solidly forgotten.

I plan to present several geometric proofs of the Pythagorean Theorem. An impetus for this page was provided by a remarkable Java applet written by Jim Morey. This constitutes the first proof on this page. One of my first Java applets was written to illustrate another Euclidean proof. Presently, there are several Java illustrations of various proofs, but the majority have been rendered in plain HTML with simple graphic diagrams.

### Remark

The statement of the Theorem was discovered on a Babylonian tablet circa 1900-1600 B.C. Whether Pythagoras (c.560-c.480 B.C.) or someone else from his School was the first to discover its proof can't be claimed with any degree of credibility. Euclid's (c 300 B.C.) Elements furnish the first and, later, the standard reference in Geometry. Jim Morey's applet follows the Proposition I.47 (First Book, Proposition 47), mine VI.31. The Theorem is reversible which means that a triangle whose sides satisfy a 2 +b 2 =c 2 is right angled. Euclid was the first (I.48) to mention and prove this fact.

W. Dunham [Mathematical Universe] cites a book The Pythagorean Proposition by an early 20th century professor Elisha Scott Loomis. The book is a collection of 367 proofs of the Pythagorean Theorem and has been republished by NCTM in 1968.

Pythagorean Theorem generalizes to spaces of higher dimensions. Some of the generalizations are far from obvious.

Larry Hoehn came up with a plane generalization which is related to the law of cosines but is shorther and looks nicer.

The Theorem whose formulation leads to the notion of Euclidean distance and Euclidean and Hilbert spaces, plays an important role in Mathematics as a whole. I began collecting math facts whose proof may be based on the Pythagorean Theorem.

 (EWD) sign( a + b - g ) = sign(a 2 + b 2 - c 2 ),

where sign(t) is the signum function:

The theorem this page is devoted to is treated as "If then Dijkstra deservedly finds (EWD) more symmetric and more informative. Absence of transcendental quantities ( p ) is judged to be an additional advantage.

### Proof #2

We start with two squares with sides a and b, respectively, placed side by side. The total area of the two squares is a 2 +b 2 .

The construction did not start with a triangle but now we draw two of them, both with sides a and b and hypotenuse c. Note that the segment common to the two squares has been removed. At this point we therefore have two triangles and a strange looking shape.

As a last step, we rotate the triangles 90 o , each around its top vertex. The right one is rotated clockwise whereas the left triangle is rotated counterclockwise. Obviously the resulting shape is a square with the side c and area c 2 .

(A variant of this proof is found in an extant manuscript by Thâbit ibn Qurra located in the library of Aya Sofya Musium in Turkey, registered under the number 4832. [R. Shloming, Thâbit ibn Qurra and the Pythagorean Theorem , Mathematics Teacher 63 (Oct., 1970), 519-528]. ibn Qurra's diagram is similar to that in proof #27. The proof itself starts with noting the presence of four equal right triangles surrounding a strangenly looking shape as in the current proof #2. These four triangles correspond in pairs to the starting and ending positions of the rotated triangles in the current proof. This same configuration could be observed in a proof by tesselation.)

### Proof #3

Now we start with four copies of the same triangle. Three of these have been rotated 90 o , 180 o , and 270 o , respectively. Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c.

The square has a square hole with the side Summing up its area and 2ab, the area of the four triangles (4·ab/2), we get

### Proof #4

The fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side (a+b) and a hole with the side c. We can compute the area of the big square in two ways. Thus

 (a + b) 2 = 4·ab/2 + c 2

simplifying which we get the needed identity.

### Proof #5

This proof, discovered by President J.A. Garfield in 1876 [Pappas], is a variation on the previous one. But this time we draw no squares at all. The key now is the formula for the area of a trapezoid - half sum of the bases times the altitude - (a+b)/2·(a+b). Looking at the picture another way, this also can be computed as the sum of areas of the three triangles - ab/2 + ab/2 + c·c/2. As before, simplifications yield a 2 +b 2 =c 2 .

Two copies of the same trapezoid can be combined in two ways by attaching them along the slanted side of the trapezoid. One leads to the proof #4, the other to proof #52.

### Proof #6

We start with the original triangle, now denoted ABC, and need only one additional construct - the altitude AD. The triangles ABC, BDA and ADC are similar which leads to two ratios:

 AB/BC = BD/AB and AC/BC = DC/AC.

Written another way these become

 AB·AB = BD·BC and AC·AC = DC·BC

In a private correspondence, Dr. France Dacar, Ljubljana, Slovenia, has suggested that the diagram on the right may serve two purposes. First, it gives an additional graphical representation to the present proof #6. In addition, it highlights the relation of the latter to proof #1.

### Proof #7

The next proof is taken verbatim from Euclid VI.31 in translation by Sir Thomas L. Heath. The great G. Polya analyzes it in his Induction and Analogy in Mathematics (II.5) which is a recommended reading to students and teachers of Mathematics.

In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.

Let ABC be a right-angled triangle having the angle BAC right I say that the figure on BC is equal to the similar and similarly described figures on BA, AC.

Let AD be drawn perpendicular. Then since, in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC adjoining the perpendicular are similar both to the whole ABC and to one another [VI.8].

And, since ABC is similar to ABD, therefore, as CB is to BA so is AB to BD [VI.Def.1].

And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second [VI.19]. Therefore, as CB is to BD, so is the figure on CB to the similar and similarly described figure on BA.

For the same reason also, as BC is to CD, so is the figure on BC to that on CA so that, in addition, as BC is to BD, DC, so is the figure on BC to the similar and similarly described figures on BA, AC.

But BC is equal to BD, DC therefore the figure on BC is also equal to the similar and similarly described figures on BA, AC.

### Confession

I got a real appreciation of this proof only after reading the book by Polya I mentioned above. I hope that a Java applet will help you get to the bottom of this remarkable proof. Note that the statement actually proven is much more general than the theorem as it's generally known.

### Proof #8

Playing with the applet that demonstrates the Euclid's proof (#7), I have discovered another one which, although ugly, serves the purpose nonetheless.

Thus starting with the triangle 1 we add three more in the way suggested in proof #7: similar and similarly described triangles 2, 3, and 4. Deriving a couple of ratios as was done in proof #6 we arrive at the side lengths as depicted on the diagram. Now, it's possible to look at the final shape in two ways:

• as a union of the rectangle (1+3+4) and the triangle 2, or
• as a union of the rectangle (1+2) and two triangles 3 and 4.

 ab/c · (a 2 +b 2 )/c + ab/2 = ab + (ab/c · a 2 /c + ab/c · b 2 /c)/2

 ab/c · (a 2 +b 2 )/c/2 = ab/2, or (a 2 +b 2 )/c 2 = 1

### Remark

In hindsight, there is a simpler proof. Look at the rectangle (1+3+4). Its long side is, on one hand, plain c, while, on the other hand, it's a 2 /c+b 2 /c and we again have the same identity.

### Proof #9

Another proof stems from a rearrangement of rigid pieces, much like proof #2. It makes the algebraic part of proof #4 completely redundant. There is nothing much one can add to the two pictures.

(My sincere thanks go to Monty Phister for the kind permission to use the graphics.)

### Proof #10

This and the next 3 proofs came from [PWW].

The triangles in Proof #3 may be rearranged in yet another way that makes the Pythagorean identity obvious.

(A more elucidating diagram on the right was kindly sent to me by Monty Phister.)

### Proof #11

Draw a circle with radius c and a right triangle with sides a and b as shown. In this situation, one may apply any of a few well known facts. For example, in the diagram three points F, G, H located on the circle form another right triangle with the altitude FK of length a. Its hypotenuse GH is split in the ratio (c+b)/(c-b). So, as in Proof #6, we get a 2 = (c+b)(c-b) = c 2 - b 2 .

### Proof #12

This proof is a variation on #1, one of the original Euclid's proofs. In parts 1,2, and 3, the two small squares are sheared towards each other such that the total shaded area remains unchanged (and equal to a 2 +b 2 .) In part 3, the length of the vertical portion of the shaded area's border is exactly c because the two leftover triangles are copies of the original one. This means one may slide down the shaded area as in part 4. From here the Pythagorean Theorem follows easily.

(This proof can be found in H. Eves, In Mathematical Circles, MAA, 2002, pp. 74-75)

### Proof #13

In the diagram there is several similar triangles (abc, a'b'c', a'x, and b'y.) We successively have

y/b = b'/c, x/a = a'/c, cy + cx = aa' + bb'.

And, finally, cc' = aa' + bb'. This is very much like Proof #6 but the result is more general.

### Proof #14

This proof by H.E.Dudeney (1917) starts by cutting the square on the larger side into four parts that are then combined with the smaller one to form the square built on the hypotenuse.

Greg Frederickson from Purdue University, the author of a truly illuminating book, Dissections: Plane & Fancy (Cambridge University Press, 1997), pointed out the historical inaccuracy:

You attributed proof #14 to H.E. Dudeney (1917), but it was actually published earlier (1873) by Henry Perigal, a London stockbroker. A different dissection proof appeared much earlier, given by the Arabian mathematician/astronomer Thabit in the tenth century. I have included details about these and other dissections proofs (including proofs of the Law of Cosines) in my recent book "Dissections: Plane & Fancy", Cambridge University Press, 1997. You might enjoy the web page for the book:

Bill Casselman from the University of British Columbia seconds Greg's information. Mine came from Proofs Without Words by R.B.Nelsen (MAA, 1993).

### Proof #15

This remarkable proof by K. O. Friedrichs is a generalization of the previous one by Dudeney. It's indeed general. It's general in the sense that an infinite variety of specific geometric proofs may be derived from it. (Roger Nelsen ascribes [PWWII, p 3] this proof to Annairizi of Arabia (ca. 900 A.D.))

### Proof #16

This proof is ascribed to Leonardo da Vinci (1452-1519) [Eves]. Quadrilaterals ABHI, JHBC, ADGC, and EDGF are all equal. (This follows from the observation that the angle ABH is 45 o . This is so because ABC is right-angled, thus center O of the square ACJI lies on the circle circumscribing triangle ABC. Obviously, angle ABO is 45 o .) Now, area(ABHI)+area(JHBC)=area(ADGC)+area(EDGF). Each sum contains two areas of triangles equal to ABC (IJH or BEF) removing which one obtains the Pythagorean Theorem.

David King modifies the argument somewhat

The side lengths of the hexagons are identical. The angles at P (right angle + angle between a & c) are identical. The angles at Q (right angle + angle between b & c) are identical. Therefore all four hexagons are identical.

### Proof #17

This proof appears in the Book IV of Mathematical Collection by Pappus of Alexandria (ca A.D. 300) [Eves, Pappas]. It generalizes the Pythagorean Theorem in two ways: the triangle ABC is not required to be right-angled and the shapes built on its sides are arbitrary parallelograms instead of squares. Thus build parallelograms CADE and CBFG on sides AC and, respectively, BC. Let DE and FG meet in H and draw AL and BM parallel and equal to HC. Then area(ABML)=area(CADE)+area(CBFG). Indeed, with the sheering transformation already used in proofs #1 and #12, area(CADE)=area(CAUH)=area(SLAR) and also area(CBFG)=area(CBVH)=area(SMBR). Now, just add up what's equal.

### Proof #18

This is another generalization that does not require right angles. It's due to Thâbit ibn Qurra (836-901) [Eves]. If angles CAB, AC'B and AB'C are equal then Indeed, triangles ABC, AC'B and AB'C are similar. Thus we have and which immediately leads to the required identity. In case the angle A is right, the theorem reduces to the Pythagorean proposition and proof #6.

### Proof #19

This proof is a variation on #6. On the small side AB add a right-angled triangle ABD similar to ABC. Then, naturally, DBC is similar to the other two. From AD = AB 2 /AC and BD = AB·BC/AC we derive Dividing by AB/AC leads to

### Proof #20

This one is a cross between #7 and #19. Construct triangles ABC', BCA', and ACB' similar to ABC, as in the diagram. By construction, In addition, triangles ABB' and ABC' are also equal. Thus we conclude that From the similarity of triangles we get as before B'C = AC 2 /BC and BC' = AC·AB/BC. Putting it all together yields which is the same as

### Proof #21

The following is an excerpt from a letter by Dr. Scott Brodie from the Mount Sinai School of Medicine, NY who sent me a couple of proofs of the theorem proper and its generalization to the Law of Cosines:

 The first proof I merely pass on from the excellent discussion in the Project Mathematics series, based on Ptolemy's theorem on quadrilaterals inscribed in a circle: for such quadrilaterals, the sum of the products of the lengths of the opposite sides, taken in pairs equals the product of the lengths of the two diagonals. For the case of a rectangle, this reduces immediately to a 2 + b 2 = c 2 .

### Proof #22

Here is the second proof from Dr. Scott Brodie's letter.

We take as known a "power of the point" theorems: If a point is taken exterior to a circle, and from the point a segment is drawn tangent to the circle and another segment (a secant) is drawn which cuts the circle in two distinct points, then the square of the length of the tangent is equal to the product of the distance along the secant from the external point to the nearer point of intersection with the circle and the distance along the secant to the farther point of intersection with the circle.

Let ABC be a right triangle, with the right angle at C. Draw the altitude from C to the hypotenuse let P denote the foot of this altitude. Then since CPB is right, the point P lies on the circle with diameter BC and since CPA is right, the point P lies on the circle with diameter AC. Therefore the intersection of the two circles on the legs BC, CA of the original right triangle coincides with P, and in particular, lies on AB. Denote by x and y the lengths of segments BP and PA, respectively, and, as usual let a, b, c denote the lengths of the sides of ABC opposite the angles A, B, C respectively. Then, x + y = c.

Since angle C is right, BC is tangent to the circle with diameter CA, and the power theorem states that a 2 = xc similarly, AC is tangent to the circle with diameter BC, and b 2 = yc. Adding, we find a 2 + b 2 = xc + yc = c 2 , Q.E.D.

Dr. Brodie also created a Geometer's SketchPad file to illustrate this proof.

### Proof #23

Another proof is based on the Heron's formula which I already used in Proof #7 to display triangle areas. This is a rather convoluted way to prove the Pythagorean Theorem that, nonetheless reflects on the centrality of the Theorem in the geometry of the plane.

### Proof #24

[Swetz] ascribes this proof to abu' l'Hasan Thâbit ibn Qurra Marwân al'Harrani (826-901). It's the second of the proofs given by Thâbit ibn Qurra. The first one is essentially the #2 above.

The proof resembles part 3 from proof #12. ABC = FLC = FMC = BED = AGH = FGE. On one hand, the area of the shape ABDFH equals AC 2 + BC 2 + area(ABC + FMC + FLC). On the other hand, area(ABDFH) = AB 2 + area(BED + FGE + AGH).

This is an "unfolded" variant of the above proof. Two pentagonal regions - the red and the blue - are obviously equal and leave the same area upon removal of three equal triangles from each.

The proof is popularized by Monty Phister, author of the inimitable Gnarly Math CD-ROM.

### Proof #25

B.F.Yanney (1903, [Swetz]) gave a proof using the "sliding argument" also employed in the Proofs #1 and #12. Successively, areas of LMOA, LKCA, and ACDE (which is AC 2 ) are equal as are the areas of HMOB, HKCB, and HKDF (which BC 2 ). BC = DF. Thus AC 2 + BC 2 = area(LMOA) + area(HMOB) = area(ABHL) = AB 2 .

### Proof #26

This proof I discovered at the site maintained by Bill Casselman where it is presented by a Java applet.

With all the above proofs, this one must be simple. Similar triangles like in proofs #6 or #13.

### Proof #27

The same pieces as in proof #26 may be rearrangened in yet another manner.

This dissection is often attributed to the 17 th century Dutch mathematician Frans van Schooten. [Frederickson, p. 35] considers it as a hinged variant of one by ibn Qurra, see the note in parentheses following proof #2. Dr. France Dacar from Slovenia has pointed out that this same diagram is easily explained with a tesselation in proof #15. As a matter of fact, it may be better explained by a different tesselation. (I thank Douglas Rogers for setting this straight for me.)

### Proof #28

Melissa Running from MathForum has kindly sent me a link to A proof of the Pythagorean Theorem by Liu Hui (third century AD). The page is maintained by Donald B. Wagner, an expert on history of science and technology in China. The diagram is a reconstruction from a written description of an algorithm by Liu Hui (third century AD). For details you are referred to the original page.

### Proof #29

A mechanical proof of the theorem deserves a page of its own.

Pertinent to that proof is a page "Extra-geometric" proofs of the Pythagorean Theorem by Scott Brodie

### Proof #30

This proof I found in R. Nelsen's sequel Proofs Without Words II. (It's due to Poo-sung Park and was originally published in Mathematics Magazine, Dec 1999). Starting with one of the sides of a right triangle, construct 4 congruent right isosceles triangles with hypotenuses of any subsequent two perpendicular and apices away from the given triangle. The hypotenuse of the first of these triangles (in red in the diagram) should coincide with one of the sides.

The apices of the isosceles triangles form a square with the side equal to the hypotenuse of the given triangle. The hypotenuses of those triangles cut the sides of the square at their midpoints. So that there appear to be 4 pairs of equal triangles (one of the pairs is in green). One of the triangles in the pair is inside the square, the other is outside. Let the sides of the original triangle be a, b, c (hypotenuse). If the first isosceles triangle was built on side b, then each has area b 2 /4. We obtain

Here's a dynamic illustration and another diagram that shows how to dissect two smaller squares and rearrange them into the big one.

### Proof #31

Given right ABC, let, as usual, denote the lengths of sides BC, AC and that of the hypotenuse as a, b, and c, respectively. Erect squares on sides BC and AC as on the diagram. According to SAS, triangles ABC and PCQ are equal, so that Let M be the midpoint of the hypotenuse. Denote the intersection of MC and PQ as R. Let's show that

The median to the hypotenuse equals half of the latter. Therefore, CMB is isosceles and But we also have From here and it follows that angle CRP is right, or

With these preliminaries we turn to triangles MCP and MCQ. We evaluate their areas in two different ways:

One one hand, the altitude from M to PC equals AC/2 = b/2. But also Therefore, On the other hand, Similarly, and also

We may sum up the two identities: or

(My gratitude goes to Floor van Lamoen who brought this proof to my attention. It appeared in Pythagoras - a dutch math magazine for schoolkids - in the December 1998 issue, in an article by Bruno Ernst. The proof is attributed to an American High School student from 1938 by the name of Ann Condit.)

### Proof #32

Let ABC and DEF be two congruent right triangles such that B lies on DE and A, F, C, E are collinear. , , . Obviously, Compute the area of ADE in two different ways.

Area(ADE) = AB·DE/2 = c 2 /2 and also CE can be found from similar triangles BCE and DFE: Putting things together we obtain

(This proof is a simplification of one of the proofs by Michelle Watkins, a student at the University of North Florida, that appeared in Math Spectrum 1997/98, v30, n3, 53-54.)

Douglas Rogers observed that the same diagram can be treated differently:

Proof 32 can be tidied up a bit further, along the lines of the later proofs added more recently, and so avoiding similar triangles.

Of course, ADE is a triangle on base DE with height AB, so of area cc/2.

But it can be dissected into the triangle FEB and the quadrilateral ADBF. The former has base FE and height BC, so area aa/2. The latter in turn consists of two triangles back to back on base DF with combined heights AC, so area bb/2. An alternative dissection sees triangle ADE as consisting of triangle ADC and triangle CDE, which, in turn, consists of two triangles back to back on base BC, with combined heights EF.

The next two proofs have accompanied the following message from Shai Simonson, Professor at Stonehill College in Cambridge, MA:

I was enjoying looking through your site, and stumbled on the long list of Pyth Theorem Proofs.

In my course "The History of Mathematical Ingenuity" I use two proofs that use an inscribed circle in a right triangle. Each proof uses two diagrams, and each is a different geometric view of a single algebraic proof that I discovered many years ago and published in a letter to Mathematics Teacher.

The two geometric proofs require no words, but do require a little thought.

### Proof #35

Cracked Domino - a proof by Mario Pacek (aka Pakoslaw Gwizdalski) - also requires some thought.

The proof sent via email was accompanied by the following message:

This new, extraordinary and extremely elegant proof of quite probably the most fundamental theorem in mathematics (hands down winner with respect to the # of proofs 367?) is superior to all known to science including the Chinese and James A. Garfield's (20th US president), because it is direct, does not involve any formulas and even preschoolers can get it. Quite probably it is identical to the lost original one - but who can prove that? Not in the Guinness Book of Records yet!

The manner in which the pieces are combined may well be original. The dissection itself is well known (see Proofs 26 and 27) and is described in Frederickson's book, p. 29. It's remarked there that B. Brodie (1884) observed that the dissection like that also applies to similar rectangles. The dissection is also a particular instance of the superposition proof by K.O.Friedrichs.

### Proof #36

This proof is due to J. E. Böttcher and has been quoted by Nelsen (Proofs Without Words II, p. 6).

I think cracking this proof without words is a good exercise for middle or high school geometry class.

### Proof #37

An applet by David King that demonstrates this proof has been placed on a separate page.

### Proof #38

This proof was also communicated to me by David King. Squares and 2 triangles combine to produce two hexagon of equal area, which might have been established as in Proof #9. However, both hexagons tessellate the plane.

For every hexagon in the left tessellation there is a hexagon in the right tessellation. Both tessellations have the same lattice structure which is demonstrated by an applet. The Pythagorean theorem is proven after two triangles are removed from each of the hexagons.

### Proof #39

(By J. Barry Sutton, The Math Gazette , v 86, n 505, March 2002, p72.)

Let in ABC, angle C = 90 o . As usual, Define points D and E on AB so that

By construction, C lies on the circle with center A and radius b. Angle DCE subtends its diameter and thus is right: It follows that Since ACE is isosceles,

Triangles DBC and EBC share DBC. In addition, Therefore, triangles DBC and EBC are similar. We have or

 a 2 = c 2 - b 2 , a 2 + b 2 = c 2 .

The diagram reminds one of Thâbit ibn Qurra's proof. But the two are quite different.

### Proof #40

This one is by Michael Hardy from University of Toledo and was published in The Mathematical Intelligencer in 1988. It must be taken with a grain of salt.

Let ABC be a right triangle with hypotenuse BC. Denote and Then, as C moves along the line AC, x changes and so does y. Assume x changed by a small amount dx. Then y changed by a small amount dy. The triangle CDE may be approximately considered right. Assuming it is, it shares one angle (D) with triangle ABD, and is therefore similar to the latter. This leads to the proportion or a (separable) differential equation

which after integration gives y 2 - x 2 = const. The value of the constant is determined from the initial condition for Since for all x.

It is easy to take an issue with this proof. What does it mean for a triangle to be ? I can offer the following explanation. Triangles ABC and ABD are right by construction. We have, and also by the Pythagorean theorem. In terms of x and y, the theorem appears as

 x 2 + a 2 = y 2 (x + dx) 2 + a 2 = (y + dy) 2

which, after subtraction, gives

For small dx and dy, dx 2 and dy 2 are even smaller and might be neglected, leading to the approximate

The trick in Michael's vignette is in skipping the issue of approximation. But can one really justify the derivation without relying on the Pythagorean theorem in the first place? Regardless, I find it very much to my enjoyment to have the ubiquitous equation placed in that geometric context.

### Proof #41

This one was sent to me by Geoffrey Margrave from Lucent Technologies. It looks very much as #8, but is arrived at in a different way. Create 3 scaled copies of the triangle with sides a, b, c by multiplying it by a, b, and c in turn. Put together, the three similar triangles thus obtained form a rectangle whose upper side is , whereas the lower side is c 2 . (Which also shows that #8 might have been concluded in a shorter way.)

Also, picking just two triangles leads to a variant of Proofs #6 and #19:

In this form the proof appears in [Birkhoff, p. 92].

Yet another variant that could be related to #8 has been sent by James F.:

The latter has a twin with a and b swapping their roles.

### Proof #42

The proof is based on the same diagram as #33 [Pritchard, p. 226-227].

Area of a triangle is obviously rp, where r is the incircle and the semiperimeter of the triangle. From the diagram, the hypothenuse or The area of the triangle then is computed in two ways:

(The proof is due to Jack Oliver, and was originally published in Mathematical Gazette 81 (March 1997), p 117-118.)

### Proof #43

Apply the Power of a Point theorem to the diagram above where the side a serves as a tangent to a circle of radius b: The result follows immediately.

(The configuration here is essentially the same as in proof #39. The invocation of the Power of a Point theorem may be regarded as a shortcut to the argument in proof #39.)

### Proof #44

The following proof related to #39, have been submitted by Adam Rose (Sept. 23, 2004.)

Start with two identical right triangles: ABC and AFE, A the midpoint of BE and CF. Mark D on AB and G on extension of AF, such that

(For further notations refer to the above diagram.) BCD is isosceles. Therefore, Since angle C is right,

Since AFE is exterior to EFG, But EFG is also isosceles. Thus

We now have two lines, CD and EG, crossed by CG with two alternate interior angles, ACD and AGE, equal. Therefore, CD||EG. Triangles ACD and AGE are similar, and AD/AC = AE/AG:

and the Pythagorean theorem follows.

### Proof #45

This proof is due to Douglas Rogers who came upon it in the course of his investigation into the history of Chinese mathematics. The two have also online versions:

The proof is a variation on #33, #34, and #42. The proof proceeds in two steps. First, as it may be observed from

where d is the diameter of the circle inscribed into a right triangle with sides a and b and hypotenuse c. Based on that and rearranging the pieces in two ways supplies another proof without words of the Pythagorean theorem:

### Proof #46

This proof is due to Tao Tong ( Mathematics Teacher , Feb., 1994, Reader Reflections). I learned of it through the good services of Douglas Rogers who also brought to my attention Proofs #47, #48 and #49. In spirit, the proof resembles the proof #32.

Let ABC and BED be equal right triangles, with E on AB. We are going to evaluate the area of ABD in two ways:

Using the notations as indicated in the diagram we get can be found by noting the similarity of triangles BFC and ABC:

The two formulas easily combine into the Pythagorean identity.

### Proof #47

This proof which is due to a high school student John Kawamura was report by Chris Davis, his geometry teacher at Head-Rouce School, Oakland, CA (Mathematics Teacher , Apr., 2005, p. 518.)

The configuration is virtually identical to that of Proof #46, but this time we are interested in the area of the quadrilateral ABCD. Both of its perpendicular diagonals have length c, so that its area equals c 2 /2. On the other hand,

Multiplying by 2 yields the desired result.

### Proof #48

(W. J. Dobbs, The Mathematical Gazette , 8 (1915-1916), p. 268.)

In the diagram, two right triangles - ABC and ADE - are equal and E is located on AB. As in President Garfield's proof, we evaluate the area of a trapezoid ABCD in two ways:

where, as in the proof #47, c·c is the product of the two perpendicular diagonals of the quadrilateral AECD. On the other hand,

Combining the two we get c 2 /2 = a 2 /2 + b 2 /2, or, after multiplication by 2,

### Proof #49

In the previous proof we may proceed a little differently. Complete a square on sides AB and AD of the two triangles. Its area is, on one hand, b 2 and, on the other,

which amounts to the same identity as before.

Douglas Rogers who observed the relationship between the proofs 46-49 also remarked that a square could have been drawn on the smaller legs of the two triangles if the second triangle is drawn in the "bottom" position as in proofs 46 and 47. In this case, we will again evaluate the area of the quadrilateral ABCD in two ways. With a reference to the second of the diagrams above,

He also pointed out that it is possible to think of one of the right triangles as sliding from its position in proof #46 to its position in proof #48 so that its short leg glides along the long leg of the other triangle. At any intermediate position there is present a quadrilateral with equal and perpendicular diagonals, so that for all positions it is possible to construct proofs analogous to the above. The triangle always remains inside a square of side b - the length of the long leg of the two triangles. Now, we can also imagine the triangle ABC slide inside that square. Which leads to a proof that directly generalizes #49 and includes configurations of proofs 46-48. See below.

### Proof #50

The area of the big square KLMN is b 2 . The square is split into 4 triangles and one quadrilateral:

It's not an interesting derivation, but it shows that, when confronted with a task of simplifying algebraic expressions, multiplying through all terms as to remove all parentheses may not be the best strategy. In this case, however, there is even a better strategy that avoids lengthy computations altogether. On Douglas Rogers' suggestion, complete each of the four triangles to an appropriate rectangle:

The four rectangles always cut off a square of size a, so that their total area is b 2 - a 2 . Thus we can finish the proof as in the other proofs of this series:

### Proof #51

(W. J. Dobbs, The Mathematical Gazette , 7 (1913-1914), p. 168.)

This one comes courtesy of Douglas Rogers from his extensive collection. As in Proof #2, the triangle is rotated 90 o around one of its corners, such that the angle between the hypotenuses in two positions is right. The resulting shape of area b 2 is then dissected into two right triangles with side lengths and and areas c 2 /2 and

### Proof #52

This proof, discovered by a high school student, Jamie deLemos ( The Mathematics Teacher , 88 (1995), p. 79.), has been quoted by Larry Hoehn ( The Mathematics Teacher , 90 (1997), pp. 438-441.)

On one hand, the area of the trapezoid equals

Equating the two gives a 2 + b 2 = c 2 .

The proof is closely related to President Garfield's proof.

### Proof #53

Larry Hoehn also published the following proof ( The Mathematics Teacher , 88 (1995), p. 168.):

Extend the leg AC of the right triangle ABC to D so that as in the diagram. At D draw a perpendicular to CD. At A draw a bisector of the angle BAD. Let the two lines meet in E. Finally, let EF be perpendicular to CF.

By this construction, triangles ABE and ADE share side AE, have other two sides equal: as well as the angles formed by those sides: Therefore, triangles ABE and ADE are congruent by SAS. From here, angle ABE is right.

It then follows that in right triangles ABC and BEF angles ABC and EBF add up to 90 o . Thus

The two triangles are similar, so that

But, EF = CD, or x = b + c, which in combination with the above proportion gives

On the other hand, y = u + a, which leads to

which is easily simplified to c 2 = a 2 + b 2 .

### Proof #54k

Later ( The Mathematics Teacher , 90 (1997), pp. 438-441.) Larry Hoehn took a second look at his proof and produced a generic one, or rather a whole 1-parameter family of proofs, which, for various values of the parameter, included his older proof as well as #41. Below I offer a simplified variant inspired by Larry's work.

To reproduce the essential point of proof #53, i.e. having a right angled triangle ABE and another BEF, the latter being similar to ABC, we may simply place BEF with sides ka, kb, kc, for some k, as shown in the diagram. For the diagram to make sense we should restrict k so that (This insures that D does not go below A.)

Now, the area of the rectangle CDEF can be computed directly as the product of its sides ka and (kb + a), or as the sum of areas of triangles BEF, ABE, ABC, and ADE. Thus we get

which after simplification reduces to

which is just one step short of the Pythagorean proposition.

The proof works for any value of k satisfying kb/a. In particular, for we get proof #41. Further, leads to proof #53. Of course, we would get the same result by representing the area of the trapezoid AEFB in two ways. For this would lead to President Garfield's proof.

Obviously, dealing with a trapezoid is less restrictive and works for any positive value of k.

## 13.6.4: Triangles, Rectangles, and the Pythagorean Theorem - Mathematics

This is Proposition 47 in the first book of Euclid’s Elements. It is often illustrated by constructing three squares on the sides of a right triangle.

The theorem is illustrated above in the special case of a 5-12-13 right triangle, which is one Pythagorean triple with integer values:

In the above example, this means the total number of crackers in the two smaller squares (on the legs) equals the total number of crackers in the larger square (on the hypotenuse). Imagining mathematically ideal crackers, all of equal area, this is an illustration of the Pythagorean theorem: the area of the large square equals the sum of the areas of the two small squares.

The first thing to notice is that the triangle of edge length three contains nine crackers, the triangle of edge length four contains sixteen crackers, and the triangle of edge length five contains twenty five crackers. This squaring relationship between length and area is not specific to squares. We say nine is “three squared” but the picture shows that we could just as well say nine is “three triangled.” Or, in fact, we could say “three pentagoned.” It is true for any plane figure that if you scale its length by n then the area is scaled by n 2 . (I’m happy to stick with the traditional term “squared.”)

So in general one can choose any plane shape and make three similar copies of it scaled to fit on the sides of a right triangle. It is easy to prove that the area of the large one is equal to the sum of the areas of the smaller ones. In the example above, the blue triangle is a right triangle and the three squiggly shapes are similar, so the area of the red one is equal to the sum of the areas of the two green ones. They don’t make crackers in this shape, but if I baked my own, I’d know that the areas work out. This is because the small cracker has area proportional to A 2 , let’s call its area kA 2 (for some scaling factor, k ). The medium-size cracker has length B / A times the length of the small one, so its area is scaled by ( B / A ) 2 , and so its area is kB 2 . Similarly, the large one has area kC 2 . We easily prove that kA 2 + kB 2 = kC 2 just by multiplying the Pythagorean Theorem through by k .

For an introduction to the Pythagorean Theorem and two proofs of its correctness, see Steven Strogatz’s “Square Dancing” column from March 14, 2010, in the NY Times. For more hands-on math demonstrations, see our past Math Monday columns for Make: Online.

## Pythagorean Theorem Games and Worksheets

Our directory of Free Geometry Math Games available on the Internet - games that teach, build or strengthen your geometry math skills and concepts while having fun. We categorize and review the games listed here to help you find the math games you are looking for.

Check out the following Pythagorean Theorem Games, Worksheets and Simulations.

Practice using the Pythagorean' Theorem to calculate the sides of right triangles.

Pythagorean Theorem Game
In this Pythagorean Theorem game you will find the unknown side in a right triangle.
The Pythagorean Theorem takes place in a right triangle. The longest side in a right triangle is the hypotenuse and the other two sides are the legs. To find the unknown side, simply apply this formula: a 2 + b 2 = c 2 (where c is the hypotenuse, and a and b are the legs)

Pythagorean Theorem Jeopardy
In this Pythagorean Theorem Game, 8th grade students will practice calculating the hypotenuse and the unknown leg in a right triangle. The converse of the Pythagorean Theorem will also be utilized to verify if three numbers could be the sides of a right triangle.

Pythagoras' Crazy Theorem
"How much do you know about the theorem of Pythagoras? This quiz will involve questions about the theorem, determining side lengths, and will also have questions about Pythagorean Triples.

The Pythagorean Theorem and Baseball
You've just picked up a ground ball at first base, and you see the other team's player running towards third base. How far do you have to throw the ball to get it from first base to third base, and throw the runner out?

Pythagorean Theorem Jeopardy
Use the Pythagorean theorem to solve these problems.

Pythagorean Theorem Game
Use Pythagorean Theorem to find the distance.

Find the Hypotenuse Game
Practice using Pythagorean Theorem to find the hypotenuse of the right triangle.

Pythagorean Theorem Worksheets

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

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Write it down as an equation:

Then we use algebra to find any missing value, as in these examples:

### Example: Solve this triangle

You can also read about Squares and Square Roots to find out why √ 169 = 13

### Example: What is the diagonal distance across a square of size 1?

It works the other way around, too: when the three sides of a triangle make a 2 + b 2 = c 2 , then the triangle is right angled.

### Example: Does this triangle have a Right Angle?

Yes, it does have a Right Angle!

### Example: Does an 8, 15, 16 triangle have a Right Angle?

Does 8 2 + 15 2 = 16 2 ?

So, NO, it does not have a Right Angle

### Example: Does this triangle have a Right Angle?

So this is a right-angled triangle

## Pythagorean Theorem proof

There are many ways to prove the Pythagorean Theorem. One way to do so involves the use of the areas of squares and triangles.

• The green square is inscribed in the blue square above, creating four congruent right triangles with legs a and b, and hypotenuse c.
• The sides of the blue square are each (a + b), therefore the area of the blue square is (a + b) 2 .
• The area of the blue square also equals the area of the green square plus the area of the 4 right triangles around it. The area of the green square is c 2 and the area of each of the right triangles is . Therefore, the area of the blue square is equal to .
• Setting the formulas for the area of the blue square equal: