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Pythagorean Theorem: Problems with Solutions

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Pythagoras Theorem Questions

Welcome to our Pythagoras' Theorem Questions area. Here you will find help, support and questions to help you master Pythagoras' Theorem and apply it.

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Pythagoras' Theorem Questions

Here you will find our support page to help you learn to use and apply Pythagoras' theorem.

Please note: Pythagoras' Theorem is also called the Pythagorean Theorem

There are a range of sheets involving finding missing sides of right triangles, testing right triangles and solving word problems using Pythagoras' theorem.

Using these sheets will help your child to:

• learn Pythagoras' right triangle theorem;
• use and apply the theorem in a range of contexts to solve problems.

Pythagoras' Theorem

Pythagoras' Theorem - in more detail

Pythagoras' theorem states that in a right triangle (or right-angled triangle) the sum of the squares of the two smaller sides of the triangle is equal to the square of the hypotenuse.

In other words, \[ a^2 + b^2 = c^2 \]

where c is the hypotenuse (the longest side) and a and b are the other sides of the right triangle.

What does this mean?

This means that for any right triangle, the orange square (which is the square made using the longest side) has the same area as the other two blue squares added together.

Other formulas that can be deduced from the Pythagorean theorem

As a result of the formula \[ a^2 + b^2 = c^2 \] we can also deduce that:

• \[ b^2 = c^2 - a^2 \]
• \[ a^2 = c^2 - b^2 \]
• \[ c = \sqrt{a^2 + b^2} \]
• \[ b = \sqrt {c^2 - a^2} \]
• \[ a = \sqrt {c^2 - b^2} \]

Pythagarean Theorem Examples

Example 1) find the length of the missing side..

In this example, we need to find the hypotenuse (longest side of a right triangle).

So using pythagoras, the sum of the two smaller squares is equal to the square of the hypotenuse.

This gives us \[ 4^2 + 6^2 = ?^2 \]

So \[ ?^2 = 16 + 36 = 52 \]

This gives us \[ ? = \sqrt {52} = 7.21 \; cm \; to \; 2 \; decimal \; places \]

Example 2) Find the length of the missing side.

In this example, we need to find the length of the base of the triangle, given the other two sides.

This gives us \[ ?^2 + 5^2 = 8^2 \]

So \[ ?^2 = 8^2 - 5^2 = 64 - 25 = 39 \]

This gives us \[ ? = \sqrt {39} = 6.25 \; cm \; to \; 2 \; decimal \; places \]

Pythagoras' Theorem Question Worksheets

The following questions involve using Pythagoras' theorem to find the missing side of a right triangle.

The first sheet involves finding the hypotenuse only.

A range of different measurement units have been used in the triangles, which are not drawn to scale.

• Pythagoras Questions Sheet 1
• PDF version
• Pythagoras Questions Sheet 2
• Pythagoras Questions Sheet 3
• Pythagoras Questions Sheet 4

Pythagoras' Theorem Questions - Testing Right Triangles

The following questions involve using Pythagoras' theorem to find out whether or not a triangle is a right triangle, (whether the triangle has a right angle).

If Pythagoras' theorem is true for the triangle, and c 2 = a 2 + b 2 then the triangle is a right triangle.

If Pythagoras' theorem is false for the triangle, and c 2 = a 2 + b 2 then the triangle is not a right triangle.

• Pythagoras Triangle Test Sheet 1
• Pythagoras Triangle Test Sheet 2

Pythagoras' Theorem Questions - Word Problems

The following questions involve using Pythagoras' theorem to solve a range of word problems involving 'real-life' type questions.

On the first sheet, only the hypotenuse needs to be found, given the measurements of the other sides.

Illustrations have been provided to support students solving these word problems.

• Pythagoras Theorem Word Problems 1
• Pythagoras Theorem Word Problems 2

Geometry Formulas

• Geometry Formula Sheet

Here you will find a support page packed with a range of geometric formula.

Included in this page are formula for:

• areas and volumes of 2d and 3d shapes
• interior angles of polygons
• angles of 2d shapes
• triangle formulas and theorems

This page will provide a useful reference for anyone needing a geometric formula.

Triangle Formulas

Here you will find a support page to help you understand some of the special features that triangles have, particularly right triangles.

Using this support page will help you to:

• understand the different types and properties of triangles;
• understand how to find the area of a triangle;
• know and use Pythagoras' Theorem.

All the free printable geometry worksheets in this section support the Elementary Math Benchmarks.

• Geometry Formulas Triangles

Here you will find a range of geometry cheat sheets to help you answer a range of geometry questions.

The sheets contain information about angles, types and properties of 2d and 3d shapes, and also common formulas associated with 2d and 3d shapes.

Included in this page are:

• images of common 2d and 3d shapes;
• properties of 2d and 3d shapes;
• formulas involving 2d shapes, such as area and perimeter, pythagoras' theorem, trigonometry laws, etc;
• formulas involving 3d shapes about volume and surface area.

Using the sheets in this section will help you understand and answer a range of geometry questions.

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Word problems on Pythagorean Theorem

Learn how to solve different types of word problems on Pythagorean Theorem .

Pythagoras Theorem can be used to solve the problems step-by-step when we know the length of two sides of a right angled triangle and we need to get the length of the third side.

Three cases of word problems on Pythagorean Theorem :

Case 1: To find the hypotenuse where perpendicular and base are given.

Case 2: To find the base where perpendicular and hypotenuse are given.

Case 3: To find the perpendicular where base and hypotenuse are given.

Word problems using the Pythagorean Theorem:

1. A person has to walk 100 m to go from position X in the north of east direction to the position B and then to the west of Y to reach finally at position Z. The position Z is situated at the north of X and at a distance of 60 m from X. Find the distance between X and Y.

⇒ 200x = 10000 + 3600

⇒ 200x = 13600

⇒ x = 13600/200

Therefore, distance between X and Y = 68 meters.

Therefore, length of each side is 8 cm.

Using the formula solve more word problems on Pythagorean Theorem.

3. Find the perimeter of a rectangle whose length is 150 m and the diagonal is 170 m.

In a rectangle, each angle measures 90°.

Therefore PSR is right angled at S

Using Pythagoras theorem, we get

⇒ PS = √6400

Therefore perimeter of the rectangle PQRS = 2 (length + width)

                                                          = 2 (150 + 80) m

                                                          = 2 (230) m

                                                          = 460 m

4. A ladder 13 m long is placed on the ground in such a way that it touches the top of a vertical wall 12 m high. Find the distance of the foot of the ladder from the bottom of the wall.

Let the required distance be x meters. Here, the ladder, the wall and the ground from a right-angled triangle. The ladder is the hypotenuse of that triangle.

According to Pythagorean Theorem,

Therefore, distance of the foot of the ladder from the bottom of the wall = 5 meters.

5. The height of two building is 34 m and 29 m respectively. If the distance between the two building is 12 m, find the distance between their tops.

The vertical buildings AB and CD are 34 m and 29 m respectively.

Draw DE ┴ AB

Then AE = AB – EB but EB = BC

Therefore AE = 34 m - 29 m = 5 m

Now, AED is right angled triangle and right angled at E.

⇒ AD = √169

Therefore the distance between their tops = 13 m.

The examples will help us to solve various types of word problems on Pythagorean Theorem.

Congruent Shapes

Congruent Line-segments

Congruent Angles

Congruent Triangles

Conditions for the Congruence of Triangles

Side Side Side Congruence

Side Angle Side Congruence

Angle Side Angle Congruence

Angle Angle Side Congruence

Right Angle Hypotenuse Side congruence

Pythagorean Theorem

Proof of Pythagorean Theorem

Converse of Pythagorean Theorem

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Pythagorean Theorem

This is generalized by the Pythagorean Inequality and the Law of Cosines .

• 1.1 Proof 1
• 1.2 Proof 2
• 1.3 Proof 3
• 2 Common Pythagorean Triples
• 3.1 Introductory
• 3.2.1 Solution 1 (Bash)
• 3.2.2 Solution 2 (Using 3-4-5)
• 3.3.1 Solution (Casework)
• 4 External links

Common Pythagorean Triples

Also, if (a,b,c) are a pythagorean triplet it follows that (ka,kb,kc) will also form a pythagorean triplet for any constant k.

k can also be imaginary.

Introductory

• 2006 AIME I Problem 1
• 2007 AMC 12A Problem 10

Sample Problem

Solution 1 (Bash)

Solution 2 (Using 3-4-5)

Another Problem

Solution (Casework)

3 and 4 are the legs. Then 5 is the hypotenuse.

3 is a leg and 4 is the hypotenuse.

There are no more cases as the hypotenuse has to be greater than the leg.

External links

• 122 proofs of the Pythagorean Theorem

Something appears to not have loaded correctly.

Click to refresh .

• Maths Questions

Pythagoras Theorem Questions

Pythagoras theorem questions with detailed solutions are given for students to practice and understand the concept. Practising these questions will be a plus point in preparation for examinations. Let us discuss in brief about the Pythagoras theorem.

Pythagoras’ theorem is all about the relation between sides of a right-angled triangle. According to the theorem, the hypotenuse square equals the sum of squares of the perpendicular sides.

Click here to learn the proof of Pythagoras’ Theorem .

Video Lesson on Pythagoras Theorem

Pythagoras Theorem Questions with Solutions

Now that we have learnt about the Pythagoras Theorem, lets apply the same by solving the following questions.

Question 1: In a right-angled triangle, the measures of the perpendicular sides are 6 cm and 11 cm. Find the length of the third side.

Let ΔABC be the triangle, right-angled at B, such that AB and BC are the perpendicular sides. Let AB = 6 cm and BC = 11 cm

Then, by the Pythagoras theorem,

AC 2 = AB 2 + BC 2

\(\begin{array}{l}\Rightarrow AC=\sqrt{(AB^{2}+BC^{2})}=\sqrt{6^{2}+11^{2}}\end{array} \)

\(\begin{array}{l}=\sqrt{36+121}=\sqrt{157}\end{array} \)

∴ AC = √157 cm.

Question 2: A triangle is given whose sides are of length 21 cm, 20 cm and 29 cm. Check whether these are the sides of a right-angled triangle.

If these are the sides of a right-angled triangle, it must satisfy the Pythagoras theorem.

We have to check whether 21 2 + 20 2 = 29 2

Now, 21 2 + 20 2 = 441 + 400 = 841 = 29 2

Thus, the given triangle is a right-angled triangle.

Question 3: Find the Pythagorean triplet with whose one number is 6.

Now, m 2 + 1 = 9 + 1 = 10

and m 2 – 1 = 9 – 1 = 8

Therefore, the Pythagorean triplet is (6, 8, 10).

Question 4: The length of the diagonal of a square is 6 cm. Find the sides of the square.

Let ABCD be the square, and let AC be the diagonal of length 6 cm. Then triangle ABC is the right-angled triangle such that AB = BC (∵ all sides of a square are equal)

By Pythagoras theorem,

⇒ AC 2 = 2AB 2

⇒ AC = √2 AB

⇒ AB = (1/√2) AC = (1/√2)6 = 3√2 cm.

Question 5: A ladder is kept at a distance of 15 cm from the wall such that the top of the ladder is at the height of 8 cm from the bottom of the wall. Find the length of the wall.

Let AB be the ladder of length x.

AC 2 + BC 2 = AB 2

\(\begin{array}{l}\Rightarrow AB=\sqrt{AC^{2}+BC^{2}}\end{array} \)

\(\begin{array}{l}\Rightarrow x=\sqrt{8^{2}+15^{2}}=\sqrt{64+225}\end{array} \)

⇒ x = 17 cm

∴ Length of the ladder is 17 cm.

Question 6: Find the area of a rectangle whose length is 144 cm and the length of the diagonal 145 cm.

Let the rectangle be ABCD

\(\begin{array}{l}\Rightarrow AD=\sqrt{AC^{2}-CD^{2}}=\sqrt{145^{2}-144^{2)}\end{array} \)

⇒ AD = √(21025 – 20736) = √289

⇒ AD = 17 cm

Thus, area of the rectangle ABCD = 17 × 144 = 2448 cm 2 .

• Properties of Triangles
• Congruence of Triangles
• Similar Triangles
• Trigonometry

Question 7: A boy travels 24 km towards east from his house, then he turned his left and covers another 10 km. Find out his total displacement?

Let the boy’s house is at point O, then to find the total displacement, we have to find OB.

Clearly, ΔOAB is a right-angled triangle, by Pythagoras theorem,

\(\begin{array}{l} OB=\sqrt{OA^{2}+AB^{2}}=\sqrt{24^{2}-10^{2}}\end{array} \)

⇒ OB = √(576 + 100) = √676

⇒ OB = 26 km.

Question 8: Find the distance between a tower and a building of height 65 m and 34 m, respectively, such that the distance between their top is 29 m.

The figure below shows the situation. Let x be the distance between the tower and the building.

In right triangle DCE, by Pythagoras theorem,

CE = √(DE 2 – DC 2 ) = √(29 2 – 21 2 )

⇒ x = √(841 – 441) = √400

⇒ x = 20 m.

∴ the distance between the tower and the building is 20 m.

Question 9: Find the area of the triangle formed by the chord of length 10 cm of the circle whose radius is 13 cm.

Let AB be the chord of the circle with the centre at O such that AB = 10 and OA = OB = 13. Draw a perpendicular OM on AB.

By the property of circle, perpendicular dropped from the centre of the circle on a chord, bisects the chord.

Then, AM = MB = 5 cm.

Now, in right triangle OMB,

OB 2 = OM 2 + MB 2

⇒ OM = √(OB 2 – MB 2 )

⇒ OM = √(13 2 – 5 2 ) = √(169 – 25)

⇒ OM = √144 = 12 cm

Area of triangle OAB = ½ × AB × OM

= ½ × 10 × 12

= 60 cm 2 .

Question 10: Find the length of tangent PT where P is a point which is at a distance 10 cm from the centre O of the circle of radius 6 cm.

Given, OP = 10 cm and OT = 6m.

We have to find the value of PT.

By the property of tangents, the radius of the circle is perpendicular to the tangent at the point of contact.

Thus, triangle OTP is a right-angled triangle.

∴ by the Pythagoras theorem,

OP 2 = OT 2. + PT 2

⇒ PT = √(OP 2 – OT 2 ) = √(10 2 – 6 2 )

⇒ PT = √(100 – 36) = √64

⇒ PT = 8 cm.

Related Video on Pythagorean Triples

Practice Questions on Pythagoras Theorem

1. Find the area of a right-angled triangle whose hypotenuse is 13 cm and one of the perpendicular sides is 5 cm.

2. Find the Pythagorean triplet whose one member is 15.

3. Find the perimeter of a rectangle whose diagonal is 5 cm and one of its sides is 4 cm.

4 if a pole of length 65 cm is kept leaning against a wall such that the pole reaches up to a height of 63 cm on the wall from the ground. Find the distance between the pole and the wall.

5. Find the area of the triangle inscribed within a circle of radius 8.5 cm such that one of the sides of the triangle is the diameter of the circle and the length of the other side is 8 cm.

(Hint: The triangle is formed in semi-circular region and angle of a semi-circle is of 90 o )

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Pythagorean Theorem

Here you will learn about the Pythagorean Theorem, including how to find side measurements of a right triangle and using Pythagoras’ theorem to check and see if a triangle has a right angle or not.

Students will first learn about Pythagorean Theorem as part of geometry in 8 th grade and continue to use it in high school.

What is the Pythagorean Theorem?

The Pythagorean Theorem states that the square of the longest side of a right triangle (called the hypotenuse) is equal to the sum of the squares of the other two sides.

Pythagorean Theorem formula shown with triangle ABC is:

a^2+b^2=c^2

Side c is known as the hypotenuse . The hypotenuse is the longest side of a right triangle. Side a and side b are known as the adjacent sides. They are adjacent, or next to, the right angle.

You can only use the Pythagorean Theorem with right triangles.

For example,

Let’s look at this right triangle:

Above, three square grids have been drawn next to each of the sides of the triangle.

The area of the side of length 3=3 \times 3=3^2=9

The area of the side of length 4=4 \times 4=4^2=16

The area of the side of length 5=5 \times 5=5^2=25

The sum of the areas of the squares on the two shorter sides is equal to the area of the square on the longest side.

When you square the sides of the two shorter sides of a right triangle and add them together, you get the square of the longest side.

3^2+4^2=5^2

3, 4, 5 is known as a Pythagorean triple.

There are other Pythagorean triples such as 5, 12, 13 and 8, 15, 17.

If you know two lengths of a right triangle, you can use Pythagorean Theorem to work out the length of the third side.

Pythagorean Theorem Proof

Use the drawing of a square with a smaller square shown inside with the proof below.

The area of each triangle is \cfrac{1}{2} \, a b and the area of the smaller square is c^2.

There are two ways to find the area of the larger square.

• Combine the area of the four congruent triangles and the smaller square: \begin{aligned}& =\cfrac{1}{2} \, a b+\cfrac{1}{2} \, a b+\cfrac{1}{2} \, a b+\cfrac{1}{2} \, a b+c^2 \\\\ & =2 a b+c^2\end{aligned}
• Multiply the side lengths of the larger square together: \begin{aligned}& =(a+b)(a+b) \\\\ & =a^2+2 a b+b^2\end{aligned}

Now set the two expressions equal to each other to prove a^2+b^2=c^2 \text{:}

\begin{aligned}a^2 +2 a b+b^2 &=2 a b+c^2 \\\\ -2 a b \hspace{0.3cm} & \hspace{0.3cm} -2 a b \\\\ a^2+b^2&=c^2\end{aligned}

Since the triangles formed by the vertices of a square will also be right triangles, the proof above shows that a^2+b^2=c^2 will always be true for the sides of a right triangle.

[FREE] Pythagorean Theorem Worksheet (Grades 8)

Use this quiz to check your grade 8 students’ understanding of pythagorean theorem. 15+ questions with answers covering a range of 8th grade topics on pythagorean theorem to identify areas of strength and support!

3D Pythagorean Theorem

You can find the length AG in the cuboid ABCDEFGH using the Pythagorean Theorem.

You can make a right triangle ACG which you can use to calculate AG.

In order to use Pythagoras’ Theorem, you need to know two sides of the triangle. So in order to figure out the longest side AG, you first need to figure out one of the shorter sides AC.

Let’s call this side x and redraw this triangle.

You can see that the side labeled x forms the diagonal line of the base of the rectangular prism.

Triangle ABC is a right triangle, so we can use the Pythagorean Theorem to calculate x.

x=\sqrt{10^2 + 4^2} = 2\sqrt{29} = 10.7703…

AG=\sqrt{10.7703…^2 + 6^2} = \sqrt{152}=2\sqrt{38} = 12.328…

So the required length is 12.4 \, cm (rounded to the nearest tenth).

Step-by-step guide: 3D Pythagorean Theorem

Common Core State Standards

How does this relate to 8 th grade math?

• Grade 8 – Geometry (8.G.B.6) Explain a proof of the Pythagorean Theorem and its converse.
• Grade 8 – Geometry (8.G.B.7) Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions.

How to use Pythagorean Theorem

In order to use Pythagorean Theorem:

Label the sides of the triangle.

Write down the formula and apply the numbers.

Record the answer.

Pythagorean Theorem examples

Example 1: missing length of the hypotenuse c.

Find x and answer to the nearest hundredth.

Label the hypotenuse (the longest side) with c. The adjacent sides, next to the right angle can be labeled a and b (either way – they are interchangeable).

2 Write down the formula and apply the numbers.

\begin{aligned}& a^2+b^2=c^2 \\\\ & 3^2+8^2=x^2 \\\\ & 9+64=x^2 \\\\ & 73=x^2 \\\\ & \sqrt{73}=x\end{aligned}

An alternative method of rearranging the formula and to put one calculation into a calculator will also work.

\begin{aligned}& a^2+b^2=c^2 \\\\ & c^2=a^2+b^2 \\\\ & c=\sqrt{a^2+b^2} \\\\ & x=\sqrt{3^2+8^2}\end{aligned}

3 Record the answer.

Make sure you give your final answer in the correct form by calculating the square root value, including units where appropriate.

x=\sqrt{73}=8.5440037…

The final answer to the nearest hundredth is:

x=8.54 \mathrm{~cm}

Example 2: missing length c

Find x and answer to the nearest tenth.

\begin{aligned}& a^2 + b^2 = c^2 \\\\ & 7^2 + 9^2 = x^2 \\\\ & x^2 = 7^2 + 9^2 \\\\ & x^2 = 49+81 \\\\ & x^2 = 130 \\\\ & x = \sqrt{130}\end{aligned}

An alternative method is to rearrange the formula and put one calculation into a calculator.

\begin{aligned}& a^2+b^2=c^2 \\\\ & c^2=a^2+b^2 \\\\ & c=\sqrt{a^2+b^2} \\\\ & x=\sqrt{7^2+9^2}\end{aligned}

Make sure you give your final answer in the correct form; including units where appropriate.

x=\sqrt{130}=11.40175.…

The final answer to the nearest tenth is:

x=11.4 \mathrm{~cm}

Example 3: finding an adjacent side (a short side)

Find x and write your answer to the nearest hundredth.

\begin{aligned} & a^2 + b^2 = c^2 \\\\ & x^2 + 5^2 = 8^2 \\\\ & x^2+25 = 64 \\\\ & x^2 = 64 - 25 \\\\ & x^2 = 39 \\\\ & x =\sqrt{39}\ \end{aligned}

\begin{aligned}& a^2+b^2=c^2 \\\\ & a^2=c^2-b^2 \\\\ & a=\sqrt{c^2-b^2} \\\\ & x=\sqrt{8^2-5^2}\end{aligned}

Make sure you give your final answer in the correct form, including units where appropriate.

x=\sqrt{39}=6.244997.…

x=6.24 \mathrm{~cm}

Example 4: finding an adjacent side (a short side)

Find x and write your answer to the nearest tenth.

\begin{aligned} & a^2 + b^2 = c^2 \\\\ & x^2 + 11^2 = 20^2\\\\ & x^2+121 = 400 \\\\ & x^2= 400 - 121 \\\\ & x^2 = 279\\\\ & x =\sqrt{279}\end{aligned}

\begin{aligned}& a^2+b^2=c^2 \\\\ & a^2=c^2-b^2 \\\\ & a=\sqrt{c^2-b^2} \\\\ & x=\sqrt{20^2-11^2}\end{aligned}

x=\sqrt{279}=16.70329.…

The final answer rounded to the nearest tenth is:

x=16.7 \mathrm{~cm}

Example 5: checking if a triangle has a right angle

Is the triangle below a right triangle?

Label the longest side with c. The adjacent sides, next to the right angle can be labeled a and b (either way – they are interchangeable).

\begin{aligned} & a^2 + b^2 = c^2 \\\\ & 8^2 + 10^2 = 13^2 \\\\ & 64+100 = 169 \\\\ & 164 = 169\end{aligned}

But this is NOT correct. Pythagorean Theorem only works with right triangles.

Because 8^2 + 10^2 ≠ 13^2 , the sides of the triangles do not fit with Pythagorean Theorem. Therefore, the triangle is NOT a right triangle and c is not a hypotenuse.

Example 6: checking if a triangle has a right angle

\begin{aligned}& a^2 + b^2 = c^2 \\\\ & 6^2 + 8^2 = 10^2 \\\\ & 36+64 = 100 \\\\ & 100 = 100\end{aligned}

This is correct. Pythagorean Theorem only works with right triangles.

Because 6^2 + 8^2 = 10^2 , the sides of the triangles fit with Pythagorean Theorem. Therefore, the triangle is a right triangle and c is a hypotenuse.

Teaching tips for Pythagorean Theorem

• In the beginning, give examples that are on grids and have the sides of the right triangle as positive integers. This allows students to draw the corresponding square for each side of the triangle and test to see that a^2+b^2=c^2.
• Give students a chance to try and prove the Pythagorean Theorem on their own. Then give students examples of a few different theorems and challenge them to find the one that makes the most sense for them.
• The Pythagorean Theorem has connections outside of math classes. If time allows, students can explore the history of this theorem.

Easy mistakes to make

• Not correctly identifying the hypotenuse It is very important to make sure that the hypotenuse, the long side, is correctly identified and labeled c.

• Thinking the lengths of sides can only be whole numbers Lengths can be decimals, fractions or even irrational numbers such as, \sqrt{2}.

• Rounding too early If you need to use Pythagorean Theorem in a question with multiple steps, do not round until the very end of the question or you will lose accuracy. For example, you may need to find the height of a triangle, and then use that height to find its area.

Practice Pythagorean Theorem questions

1. Find side x. Give your answer to the nearest hundredth:

\begin{aligned}& a^2 + b^2 = c^2 \\\\ & 7^2 + 5^2 = x^2 \\\\ & x^2 = 7^2 + 5^2\\\\ & x^2 = 49+25 \\\\ & x^2 = 74 \\\\& x = \sqrt{74}\\\\ &x = 8.602325…\end{aligned}

x=8.60 \mathrm{~cm}

2. Find side x. Give your answer to the nearest hundredth:

\begin{aligned}&a^2 + b^2 = c^2 \\\\ & 14^2 + 10^2 = x^2 \\\\ & x^2 = 14^2 + 10^2 \\\\ & x^2 = 196+10 \\\\ & x^2 = 296 \\\\ & x = \sqrt{296} \\\\ & x = 17.20465… \end{aligned}

x=17.20 \mathrm{~cm}

3. Find side x. Give your answer to the nearest hundredth:

\begin{aligned}& a^2 + b^2 = c^2 \\\\ & x^2 + 10^2 = 18^2\\\\ & x^2+100 = 324 \\\\ & x^2 = 324 – 100 \\\\ & x^2 = 224 \\\\ & x =\sqrt{224} \\\\ & x = 14.96662…\end{aligned}

x=14.97 \mathrm{~cm}

4. Find side x. Give your answer to the nearest hundredth:

\begin{aligned}& a^2 + b^2 = c^2 \\\\ & x^2 + 2.5^2 = 7.2^2 \\\\ & x^2+6.25 = 51.84 \\\\ & x^2 = 51.84-6.25 \\\\ & x^2 = 45.59 \\\\ & x =\sqrt{45.59} \\\\ & x = 6.75203…\end{aligned}

x=6.75 \mathrm{~cm}

5. Is this a right triangle? Justify your answer with the Pythagorean Theorem.

No, because 12^2+5^2 ≠ 13^2

Yes, because I measured the angle and it was 90^{\circ}

Yes, because 12^2+5^2=13^2

No, because I measured the angle and it was not 90^{\circ}

\begin{aligned}& a^2 + b^2 = c^2 \\\\ & 12^2 + 5^2 = 13^2 \\\\ & 144+25 = 169 \\\\ & 169 = 169 \end{aligned}

Therefore the triangle is a right triangle.

6. Is this a right triangle? Justify your answer with the Pythagorean Theorem.

Yes, because 6^2+13^2=14^2

No, because 6^2+13^2 ≠ 14^2

\begin{aligned}& a^2 + b^2 = c^2 \\\\ & 6^2 + 13^2 = 14^2 \\\\ & 36+169 = 196 \\\\ & 205 = 196\end{aligned}

This is NOT correct. Pythagorean Theorem only works with right triangles.

Therefore the triangle is NOT a right-angled triangle.

Pythagorean Theorem FAQs

Pythagorean Theorem is named after a Greek mathematician who lived about 2,500 years ago, however, the ancient Babylonians used this rule about 4 thousand years ago! At the same time, the Egyptians were using the theorem to help them with right angles when building structures.

No, this is only true for the sides of a right triangle. The sum of the squares of the lengths for a and b will only be equal to the square of side c if the triangle is right. However, you can use this relationship to decide if a triangle is acute or obtuse. For an acute triangle, the square of the hypotenuse will be less than the sum of the squares of a and b. For an obtuse triangle, the square of the hypotenuse will be more than the sum of the squares of a and b.

Yes, there are many algebraic proofs and geometric proofs that address the Pythagorean Theorem. The proof shown at the top of this page is one of the simplest ways to prove the Pythagorean Theorem.

No, though they all follow the Pythagorean Theorem this does not mean they are similar. They are only similar if there is a multiplicative relationship between each corresponding side of the triangle.

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• Circle math
• Sectors, arcs and segments

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The pythagorean theorem with examples.

The Pythagorean theorem is a way of relating the leg lengths of a right triangle to the length of the hypotenuse, which is the side opposite the right angle. Even though it is written in these terms, it can be used to find any of the side as long as you know the lengths of the other two sides. In this lesson, we will look at several different types of examples of applying this theorem.

Table of Contents

• Examples of using the Pythagorean theorem
• Solving applied problems (word problems)
• Solving algebraic problems

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Applying the Pythagorean theorem (examples)

In the examples below, we will see how to apply this rule to find any side of a right triangle triangle. As in the formula below, we will let a and b be the lengths of the legs and c be the length of the hypotenuse. Remember though, that you could use any variables to represent these lengths.

In each example, pay close attention to the information given and what we are trying to find. This helps you determine the correct values to use in the different parts of the formula.

Find the value of \(x\).

The side opposite the right angle is the side labelled \(x\). This is the hypotenuse. When applying the Pythagorean theorem, this squared is equal to the sum of the other two sides squared. Mathematically, this means:

\(6^2 + 8^2 = x^2\)

Which is the same as:

\(100 = x^2\)

Therefore, we can write:

\(\begin{align}x &= \sqrt{100}\\ &= \bbox[border: 1px solid black; padding: 2px]{10}\end{align}\)

Maybe you remember that in an equation like this, \(x\) could also be –10, since –10 squared is also 100. But, the length of any side of a triangle can never be negative and therefore we only consider the positive square root.

In other situations, you will be trying to find the length of one of the legs of a right triangle. You can still use the Pythagorean theorem in these types of problems, but you will need to be careful about the order you use the values in the formula.

Find the value of \(y\).

The side opposite the right angle has a length of 12. Therefore, we will write:

\(8^2 + y^2 = 12^2\)

This is the same as:

\(64 + y^2 = 144\)

Subtracting 64 from both sides:

\(y^2 = 80\)

\(\begin{align}y &= \sqrt{80} \\ &= \sqrt{16 \times 5} \\ &= \bbox[border: 1px solid black; padding: 2px]{4\sqrt{5}}\end{align}\)

In this last example, we left the answer in exact form instead of finding a decimal approximation. This is common unless you are working on an applied problem.

Applications (word problems) with the Pythagorean theorem

There are many different kinds of real-life problems that can be solved using the Pythagorean theorem. The easiest way to see that you should be applying this theorem is by drawing a picture of whatever situation is described.

Two hikers leave a cabin at the same time, one heading due south and the other headed due west. After one hour, the hiker walking south has covered 2.8 miles and the hiker walking west has covered 3.1 miles. At that moment, what is the shortest distance between the two hikers?

First, sketch a picture of the information given. Label any unknown value with a variable name, like x.

Due south and due west form a right angle, and the shortest distance between any two points is a straight line. Therefore, we can apply the Pythagorean theorem and write:

\(3.1^2 + 2.8^2 = x^2\)

Here, you will need to use a calculator to simplify the left-hand side:

\(17.45 = x^2\)

Now use your calculator to take the square root. You will likely need to round your answer.

\(\begin{align}x &= \sqrt{17.45} \\ &\approx 4.18 \text{ miles}\end{align}\)

As you can see, it will be up to you to determine that a right angle is part of the situation given in the word problem. If it isn’t, then you can’t use the Pythagorean theorem.

Algebra style problems with the Pythagorean theorem

There is one last type of problem you might run into where you use the Pythagorean theorem to write some type of algebraic expression. This is something that you will not need to do in every course, but it does come up.

A right triangle has a hypotenuse of length \(2x\), a leg of length \(x\), and a leg of length y. Write an expression that shows the value of \(y\) in terms of \(x\).

Since no figure was given, your first step should be to draw one. The order of the legs isn’t important, but remember that the hypotenuse is opposite the right angle.

Now you can apply the Pythagorean theorem to write:

\(x^2 + y^2 = (2x)^2\)

Squaring the right-hand side:

\(x^2 + y^2 = 4x^2\)

When the problem says “the value of \(y\)”, it means you must solve for \(y\). Therefore, we will write:

\(y^2 = 4x^2 – x^2\)

Combining like terms:

\(y^2 = 3x^2\)

Now, use the square root to write:

\(y = \sqrt{3x^2}\)

Finally, this simplifies to give us the expression we are looking for:

\(y = \bbox[border: 1px solid black; padding: 2px]{x\sqrt{3x}}\)

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The Pythagorean theorem allows you to find the length of any of the three sides of a right triangle. It is one of those things that you should memorize, as it comes up in all areas of math, and therefore in many different math courses you will probably take. Remember to avoid the common mistake of mixing up where the legs go in the formula vs. the hypotenuse and to always draw a picture when one isn’t given.

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Pythagorean Theorem – Definition, Formula, Problems

In mathematic, the Pythagorean theorem states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of its other two sides . Another way of stating the theorem is that the sum of the areas of the squares formed by the sides of a right triangle equals the area of the square whose side is the hypotenuse. The theorem is a key relation in Euclidean geometry. It is named for the Greek philosopher Pythagorus.

Remember: The Pythagorean theorem only applies to right triangles!

Pythagorean Theorem Formula

The formula for the Pythagorean theorem describes the relationship between the sides a and b of a right triangle to its hypotenuse, c . A right triangle is one containing a 90° or right angle. The hypotenuse is the side of the triangle opposite from the right angle (which is the largest angle in a right triangle).

a 2 + b 2 = c 2

Solving for a, b, and c

Rearranging the equation gives the formulas solving for a, b, and c:

• a = (c 2 – b 2 ) ½
• b = (c 2 – a 2 ) ½
• c = (a 2 + b 2 ) ½

How to Solve the Pythagorean Theorem – Example Problems

For example, find the hypotenuse of a right triangle with side that have lengths of 5 and 12.

Start with the formula for the Pythagorean theorem and plug in the numbers for the sides a and b to solve for c .

a 2 + b 2 = c 2 5 2 + 12 2 = c 2 c 2 = 5 2 + 12 2 = 25 + 144 = 169 c2 = 169 c = √169 or 169 ½ = 13

For example, solve for side b of a triangle where a is 9 and the hypotenuse c is 15.

a 2 + b 2 = c 2 9 2 + b 2 = 15 2 b 2 = 15 2 – 9 2 = 225 – 81 = 144 b = √144 = 12

Now, let’s combine a bit of algebra with the geometry. Solve for x where the sides of a right triangle are 5x and 4x +5 and the hypotenuse has a length of 8x -3.

a 2 + b 2 = c 2 (5x) 2 + (4x +5) 2 = (8x-3) 2

The (4x + 5) 2 and (8x -3)2 terms are the squares of binomial expressions. So, expanding the equation gives the following:

25x 2 + (4x +5)(4x +5) = (8x -3)(8x -3) 25x 2 = 16×2 + 20x +20x + 25 = 64x – 24x – 24x + 9

Combine like terms:

41x 2 + 40x + 25 = 64x 2 – 48x + 9

Rewrite the equation and solve for x.

0 = 23x 2 – 88x – 16

Apply the quadratic equation and solve for x:

x = [-b ± √(b 2 -4ac)]/2a x = [-(-88) ± √[-88 2 – 4(23)(-16)] / 2(23) = [88 ± √(7744 + 1472)] / 46 = (88 ± 96) / 46

So, there are two answers:

x = (88 + 96)/46 = 4 and (88 – 96).46 = -4/23

A triangle does not have a negative length for its side, so x is 4.

Plugging in”4″ in place of x, the sides of the right triangle are 20, 21, and 29.

Pythagorean Triples

Pythagorean triples are integers a, b, and c, that represent the sides of a right triangle and satisfy the Pythagorean theorem. Here is the list of Pythagorean triples for integers with values less than 100:

(3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41), (11, 60, 61), (12, 35, 37), (13, 84, 85), (16, 63, 65), (20, 21, 29), (28, 45, 53), (33, 56, 65), (36, 77, 85), (39, 80, 89), (48, 55, 73), (65, 72, 97)

Proof of the Pythagorean Theorem

There are more proofs for the Pythagorean theorem than for any other theorem in geometry! At least 370 proofs are known. Some of these proofs use the parallel postulate. Some rely on the complementarity of acute angles in a right triangle. Proofs using shearing use the properties of parallelograms.

History – Did Pythagoras Discover the Pythagorean Theorem?

While the Pythagorean theorem takes its name from Pythagorus, he did not discover it. Exactly who gets the credit or whether many different places made the discovery independently is a matter of debate. The Mesopotamians made calculations using the formula as early as 2000 BC, which was over a thousand years before Pythagorus. A papyrus from the Egyptian Middle Kingdom, dating between 2000 and 1786 BC, references a math problem describing Pythagorean triples. The Baudhayana Shulba Sutra from India (dating between the 8th and 5th century BC) lists both Pythagorean triples and the Pythagorean theorem. The “Gougu theorem” from China offers a proof for the Pythagorean theorem, which came into use long before its oldest surviving description from the 1st century BC.

Pythagorus of Samos lived between 570 and 495 BC. While he was not the original person who formulated the Pythagorean theorem, he (or his students) may have introduced its proof to ancient Greece. In any case, his philosophical treatment of math left a lasting impression on the world.

• Bell, John L. (1999). The Art of the Intelligible: An Elementary Survey of Mathematics in its Conceptual Development . Kluwer. ISBN 0-7923-5972-0.
• Heath, Sir Thomas (1921). “ The ‘Theorem of Pythagoras ‘”. A History of Greek Mathematics (2 Vols.) (Dover Publications, Inc. (1981) ed.). Oxford: Clarendon Press. ISBN 0-486-24073-8.
• Maor, Eli (2007). The Pythagorean Theorem: A 4,000-Year History . Princeton, New Jersey: Princeton University Press. ISBN 978-0-691-12526-8.
• Swetz, Frank; Kao, T. I. (1977). Was Pythagoras Chinese?: An Examination of Right Triangle Theory in Ancient China . Pennsylvania State University Press. ISBN 0-271-01238-2.

Related Posts

Pythagorean Theorem

How to Use The Pythagorean Theorem

The Formula

The picture below shows the formula for the Pythagorean theorem. For the purposes of the formula, side $$ \overline{c}$$ is always the hypotenuse . Remember that this formula only applies to right triangles .

Examples of the Pythagorean Theorem

When you use the Pythagorean theorem, just remember that the hypotenuse is always 'C' in the formula above. Look at the following examples to see pictures of the formula.

Conceptual Animation of Pythagorean Theorem

Demonstration #1.

More on the Pythagorean theorem

Demonstration #2

Video tutorial on how to use the pythagorean theorem.

Step By Step Examples of Using the Pythagorean Theorem

Example 1 (solving for the hypotenuse).

Use the Pythagorean theorem to determine the length of X.

Identify the legs and the hypotenuse of the right triangle .

The legs have length 6 and 8 . $$X $$ is the hypotenuse because it is opposite the right angle.

Substitute values into the formula (remember 'C' is the hypotenuse).

$ A^2+ B^2= \red C^2 \\ 6^2+ 8^2= \red X^2 $

$A^2+ B^2= \red X^2 \\ 100= \red X^2 \\ \sqrt {100} = \red X \\ 10= \red X $

Example 2 (solving for a Leg)

The legs have length 24 and $$X$$ are the legs. The hypotenuse is 26.

$ \red A^2+ B^2= C^2 \\ \red x^2 + 24^2= {26}^2 $

$ \red x^2 + 24^2= 26^2 \\ \red x^2 + 576= 676 \\ \red x^2 = 676 - 576 \\ \red x^2 = 100 \\ \red x = \sqrt { 100} \\ \red x = 10 $

Practice Problems

Find the length of X.

Remember our steps for how to use this theorem. This problems is like example 1 because we are solving for the hypotenuse .

The legs have length 14 and 48 . The hypotenuse is X.

$ A^2 + B^2 = C^2 \\ 14^2 + 48^2 = x^2 $

Solve for the unknown.

$ 14^2 + 48^2 = x^2 \\ 196 + 2304 = x^2 \\ \sqrt{2500} = x \\ \boxed{ 50 = x} $

Use the Pythagorean theorem to calculate the value of X. Round your answer to the nearest tenth.

Remember our steps for how to use this theorem. This problems is like example 2 because we are solving for one of the legs .

The legs have length 9 and X . The hypotenuse is 10.

$ A^2 + B^2 = C^2 \\ 9^2 + x^2 = 10^2 $

$ 9^2 + x^2 = 10^2 \\ 81 + x^2 = 100 \\ x^2 = 100 - 81 \\ x^2 = 19 \\ x = \sqrt{19} \approx 4.4 $

Use the Pythagorean theorem to calculate the value of X. Round your answer to the nearest hundredth.

The legs have length '10' and 'X'. The hypotenuse is 20.

$ A^2 + B^2 = C^2 \\ 10^2 + \red x^2 = 20^2 $

$ 10^2 + \red x^2 = 20^2 \\ 100 + \red x^2 = 400 \\ \red x^2 = 400 -100 \\ \red x^2 = 300 \\ \red x = \sqrt{300} \approx 17.32 $

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Module 11: Geometry

Using the pythagorean theorem to solve problems, learning outcomes.

• Use the pythagorean theorem to find the unknown length of a right triangle given the two other lengths

The Pythagorean Theorem is a special property of right triangles that has been used since ancient times. It is named after the Greek philosopher and mathematician Pythagoras who lived around [latex]500[/latex] BCE.

Remember that a right triangle has a [latex]90^\circ [/latex] angle, which we usually mark with a small square in the corner. The side of the triangle opposite the [latex]90^\circ [/latex] angle is called the hypotenuse, and the other two sides are called the legs. See the triangles below.

In a right triangle, the side opposite the [latex]90^\circ [/latex] angle is called the hypotenuse and each of the other sides is called a leg.

The Pythagorean Theorem

In any right triangle [latex]\Delta ABC[/latex],

[latex]{a}^{2}+{b}^{2}={c}^{2}[/latex]

where [latex]c[/latex] is the length of the hypotenuse [latex]a[/latex] and [latex]b[/latex] are the lengths of the legs.

To solve problems that use the Pythagorean Theorem, we will need to find square roots. In Simplify and Use Square Roots we introduced the notation [latex]\sqrt{m}[/latex] and defined it in this way:

[latex]\text{If }m={n}^{2},\text{ then }\sqrt{m}=n\text{ for }n\ge 0[/latex]

For example, we found that [latex]\sqrt{25}[/latex] is [latex]5[/latex] because [latex]{5}^{2}=25[/latex].

We will use this definition of square roots to solve for the length of a side in a right triangle.

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

Use the Pythagorean Theorem to find the length of the longer leg.

Kelvin is building a gazebo and wants to brace each corner by placing a [latex]\text{10-inch}[/latex] wooden bracket diagonally as shown. How far below the corner should he fasten the bracket if he wants the distances from the corner to each end of the bracket to be equal? Approximate to the nearest tenth of an inch.

In the following video we show two more examples of how to use the Pythagorean Theorem to solve application problems.

• Question ID 146918, 146916, 146914, 146913. Authored by : Lumen Learning. License : CC BY: Attribution
• Solve Applications Using the Pythagorean Theorem (c only). Authored by : James Sousa (mathispower4u.com). Located at : https://youtu.be/2P0dJxpwFMY . License : CC BY: Attribution
• Prealgebra. Provided by : OpenStax. License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

Pythagorean Theorem Word Problems

In these lessons, we will be looking at how to solve different types of word problems using the Pythagorean Theorem.

Related Pages Pythagorean Theorem Converse Of Pythagorean Theorem Applications Of Pythagorean Theorem More Geometry Lessons

How To Solve Word Problems Using The Pythagorean Theorem?

• Determine whether the word problem can be modeled by a right triangle.
• Use the Pythagorean Theorem to find the missing side if you are given two sides.

Example: Shane marched 3 m east and 6 m north. How far is he from his starting point?

Solution: First, sketch the scenario. The path taken by Shane forms a right-angled triangle. The distance from the starting point forms the hypotenuse.

Example: The rectangle PQRS represents the floor of a room.

Ivan stands at point A. Calculate the distance of Ivan from a) the corner R of the room b) the corner S of the room

Example: In the following diagram of a circle, O is the centre and the radius is 12 cm. AB and EF are straight lines.

Find the length of EF if the length of OP is 6 cm.

Examples Of Real Life Pythagorean Theorem Word Problems

Problem 1: A 35-foot ladder is leaning against the side of a building and is positioned such that the base of the ladder is 21 feet from the base of the building. How far above the ground is the point where the ladder touches the building?

Problem 2: The main mast of a fishing boat is supported by a sturdy rope that extends from the top of the mast to the deck. If the mast is 20 feet tall and the rope attached to the deck 15 feet away from the base of the mast, how long is the rope?

Problem 3: If an equilateral triangle has a height of 8, find the length of each side.

Problem 4: Two cyclist start from the same location. One cyclist travels due north and the other due east, at the same speed. Find the speed of each in miles per hour if after two hours they are 17sqrt(2) miles apart.

Problem 5: Two cars start from the same intersection with one traveling southbound while the other travels eastbound going 10 mph faster. If after two hours they are 10sqrt(34) apart, how fast was each car traveling?

Problem 6: A carpet measures 7 feet long and has a diagonal measurement of sqrt(74) feet. Find the width of the carpet.

Problem 7: Jim and Eileen decided to take a short cut through the woods to go to their friend’s house. When they went home they decided to take the long way around the woods to avoid getting muddy shoes. What total distance did they walk to and from their friend’s house? Dimensions are in meters.

Problem 8: Shari went to a level field to fly a kite. She let out all 650 feet of the string and tied it to a stake. Then, she walked out on the field until she was directly under the kite, which was 600 feet from the stake. How high was the kite from the ground?

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How to Solve Pythagorean Theorem Problems? (+FREE Worksheet!)

In mathematics, the Pythagorean Theorem is the relationship between three sides of a right triangle.

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Step by step guide to solve Pythagorean Theorem problems

• We can use the Pythagorean Theorem to find a missing side in a right triangle.

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The pythagorean theorem – example 1:.

Right triangle ABC has two legs of lengths \(9\) cm (AB) and \(12\) cm (AC). What is the length of the third side (BC)?

Use Pythagorean Theorem: \(\color{blue}{a^2+b^2= c^2}\) Then: \(a^2+b^2= c^2 →9^2+12^2= c^2 →81+144=c^2\) \(c^2=225 →\) \(c=\sqrt{225}=15\) \(cm\) → \(c=15 cm\)

The Pythagorean Theorem – Example 2:

Use Pythagorean Theorem: \(\color{blue}{a^2+b^2= c^2}\) Then: \(a^2+b^2= c^2 →8^2+6^2= c^2 →64+36=c^2\) \(c^2=100 →\) \(c=\sqrt{100}=10\) → \(c=10\)

The Pythagorean Theorem – Example 3:

Find the hypotenuse of the following right triangle.

Use Pythagorean Theorem: \(\color{blue}{a^2+b^2= c^2}\) Then: \(a^2+b^2= c^2 →3^2+4^2= c^2 →9+16=c^2\) \(c^2=25 →\) \(c=\sqrt{25}=5\) → \(c=5\)

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The pythagorean theorem – example 4:.

Right triangle ABC has two legs of lengths \(6\) cm (AB) and \(8\) cm (AC). What is the length of the third side (BC)?

Use Pythagorean Theorem: \(\color{blue}{a^2+b^2= c^2}\) Then: \(a^2+b^2= c^2 →6^2+8^2= c^2 →36+64=c^2\) \( c^2=100 →\) \(c=\sqrt{100}=10\) \(cm\) → \(c= 10 cm\)

Exercises for Solveing the Pythagorean Theorem

Find the missing side in each right triangle., download pythagorean relationship worksheet.

• \(\color{blue}{13}\)
• \(\color{blue}{5}\)
• \(\color{blue}{15}\)
• \(\color{blue}{8}\)

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Pythagorean theorem.

The Pythagorean Theorem, also known as Pythagoras theorem is a mathematical relation between the 3 sides of a right triangle, a triangle in which one of 3 angles is 90°. It was discovered and named after the Greek philosopher and mathematician of Samos, Pythagoras.

Does Pythagorean Theorem Work on All Triangles

No, the Pythagorean Theorem works only for right triangles. Thus, it helps to test whether a triangle is right triangle or not.  The theorem is also used to find the length of one side of a right triangle when the other two sides are known.

The theorem was first evident on a 4000-year-old Babylonian tablet (beginning about 1900 B.C.) now known as Plimpton 322. Much later in 570–500/490 BCE, the relationship was popularized when Pythagoras stated it explicitly.

When a triangle has a right angle (90°) and squares are made on each of the 3 sides, then the biggest square has an area equal to the sum of the areas of the other 2 squares.

Thus, the Pythagorean Theorem states that the area of the square formed by the longest side of the right triangle (the hypotenuse) is equal to the sum of the area of the squares formed by the other two sides.

In the above figure,

Area of □ A = a 2

Area of □ B = b 2

Area of □ C = c 2

Now, according to the Pythagorean Theorem,

Area of □ A + Area of □ B = Area of □ C

(a × a) + (b × b) = (c × c)

a 2 + b 2 = c 2

The above relation is useful to find an unknown side of a right triangle when the lengths of the other 2 sides are known.

The equation that represents the Pythagorean Theorem in mathematical form is given below:

Let us find out how it works using an example.

Consider a right triangle with side lengths 3, 4, 5.

In the above figure, let us use the Pythagorean Theorem

3 2 + 4 2 = 5 2

=> 9 + 16 =25

9, 16, & 25 are the areas of the three squares

This proves how the Pythagorean Theorem works

Let us solve an example to understand the concept better

Find the hypotenuse of a right-angle triangle with the other 2 sides 15 cm and 20 cm.

As we know, In a right angle triangle, a 2 + b 2 = c 2 , here c = length of the hypotenuse, a = 15 cm, c = 20 cm => 15 2 + 20 2 = c 2 => 225 + 400 = c 2 => c 2 = 625 => c = 25 cm Thus, the hypotenuse of a right triangle with sides 15 cm and 20 cm is 25 cm

As we know, In a right angle triangle, a 2 + b 2 = c 2 , here c = 5, a = 3 cm => 3 2 + b 2 = 5 2 => b 2 = 5 2 – 3 2 => b 2 = 25 – 9 => b 2 = 16 => b = 4 cm

As we know, In a right angle triangle, a 2 + b 2 = c 2 , here a = 9, b = 12, c = 15 cm => 9 2 + 12 2 => 81 + 144 => 225 As, 15 2 = 225 Thus, the above relation holds true for the Pythagorean Theorem, thus it is a right angle triangle.

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

As we know, In a right angle triangle, a 2 + b 2 = c 2 Here, => 8 2 + 15 2 => 64 + 225 => 289 But, 16 2 = 256 Thus, the above relation does not hold true for the Pythagorean Theorem, thus it is not a right-angle triangle.

The Pythagorean Theorem can be proved in many ways. The 2 most common ways of proving the theorem are described below:

Algebraic Method

This method helps us to prove the Pythagorean Theorem by using the side lengths.

Let us consider 4 right triangles with side lengths a, b, & c, where c is the length of the hypotenuse and ‘a’ and ‘b’ are the lengths of the other 2 sides

If we arrange the 4 right triangles in a square of length (a + b), we can derive the equation of the Pythagorean Theorem as shown below:

Area of □ EFGH = c × c = c 2

Area of □ ABCD = Area of □ EFGH + Area of 4 right △s

=> (a + b) 2 = c 2 + 4(1/2 × b × a)

=> (a + b) 2 = c 2 + 2ab

=> a 2 + b 2 + 2ab = c 2 + 2ab

=> a 2 + b 2 = c 2

This proves the Pythagorean Theorem

Using Similar Triangles

As we know,

△ADB ~ △ABC

∴ CD/CB = CB/CA (Corresponding Sides of Similar Triangles are Equal)

=> CB 2 = CD × CA …….. (1)

△BDA ~ △CBA

∴ AD/BA = BA/CA (Corresponding Sides of Similar Triangles are Equal)

=> BA 2 = AD × CA …….. (2)

Adding equations (1) and (2), we get

CB 2 + BA 2 = (CD × CA) + (AD × CA)

=> CB 2 + BA 2 = CA (CD + AD)

Since, CD + AD = CA

∴ CA 2 = CB 2 + BA 2

Applications: What is It Used For

Through learning the basic concepts of the Pythagorean Theorem is important to determine whether a triangle is a right triangle or not. But, we are even more curious in understanding the applications of the Pythagorean Theorem.

Some common real-life applications of the Pythagorean Theorem are given below:

• Two-dimensional navigation where it is used for calculate the distance between the ship and the point of navigation. The same principles are also used for air navigation. For example, a plane can use its height above the ground and its distance from the source to its destination.
• Cartographers use to calculate and survey the numerical distances and heights between different points before a map is drawn. It is also used to calculate the steepness of slopes of hills or mountains.
• Given 2 straight lines, it is used for calculating the length of the diagonal line connecting them. This application is frequently used in architecture , woodworks , or other physical construction projects. It is also used for laying out square angles for making design in buildings.
• In oceanography, it is used for calculating speed of sound waves in water.
• Making of tv screens, computer screens , and solar panels
• Works to determine the height of ladder required to painta wall of certain height
• Converse of Pythagoras Theorem
• Pythagorean Triples
• Pythagorean Identities

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Pythagorean Theorem

The pythagorean theorem.

If we have a right triangle, and we construct squares using the edges or sides of the right triangle (gray triangle in the middle), the area of the largest square built on the hypotenuse (the longest side) is equal to the sum of the areas of the squares built on the other two sides. This is the Pythagorean Theorem in a nutshell. By the way, this is also known as the Pythagoras’ Theorem .

Notice that we square (raised to the second power) the variables [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] to indicate areas. The sum of the smaller squares (orange and yellow) is equal to the largest square (blue).

The Pythagorean Theorem relates the three sides in a right triangle. To be specific, relating the two legs and the hypotenuse, the longest side.

The Pythagorean Theorem can be summarized in a short and compact equation as shown below.

Definition of Pythagorean Theorem

For a given right triangle, it states that the square of the hypotenuse, [latex]c[/latex], is equal to the sum of the squares of the legs, [latex]a[/latex] and [latex]b[/latex]. That is, [latex]{a^2} + {b^2} = {c^2}[/latex].

For a more general definition, we have:

In right a triangle, the square of longest side known as the hypotenuse is equal to the sum of the squares of the other two sides.

The Pythagorean Theorem guarantees that if we know the lengths of two sides of a right triangle, we can always determine the length of the third side.

Here are the three variations of the Pythagorean Theorem formulas:

Let’s go over some examples!

Examples of Applying the Pythagorean Theorem

Example 1: Find the length of the hypotenuse.

Our goal is to solve for the length of the hypotenuse. We are given the lengths of the two legs. We know two sides out of the three! This is enough information for the formula to work.

For the legs, it doesn’t matter which one we assign for [latex]a[/latex] or [latex]b[/latex]. The result will be the same. So if we let [latex]a=5[/latex], then [latex]b=7[/latex]. Substituting these values into the Pythagorean Formula equation, we get

To isolate the variable [latex]c[/latex], we take the square roots of both sides of the equation. That eliminates the square (power of 2) on the right side. And on the left, we simply have a square root of a number which is no big deal.

However, we need to be mindful here when we take the square root of a number. We want to consider only the principal square root or the positive square root since we are dealing with length. It doesn’t make any sense to have a negative length, thus we disregard the negative length!

Therefore, the length of the hypotenuse is [latex]\sqrt {74}[/latex] inches. If we wish to approximate it to the nearest tenth, we have [latex]8.6[/latex] inches.

Example 2: Find the length of the leg.

Just by looking at the figure above, we know that we have enough information to solve for the missing side. The reason is the measure of the two sides are given and the other leg is left as unknown. That’s two sides given out of the possible three.

Here, we can let [latex]a[/latex] or [latex]b[/latex] equal [latex]7[/latex]. It really doesn’t matter. So, for this, we let [latex]a=7[/latex]. That means we are solving for the leg [latex]b[/latex]. But for the hypotenuse, there’s no room for error. We have to be certain that we are assigning [latex]c[/latex] for the length, that is, for the longest side. In this case, the longest side has a measure of [latex]9[/latex] cm and that is the value we will assign for [latex]c[/latex], therefore [latex]c=9[/latex].

Let’s calculate the length of leg [latex]b[/latex]. We have [latex]a=7[/latex] and [latex]c=9[/latex].

Therefore, the length of the missing leg is [latex]4\sqrt 2[/latex] cm. Rounding it to two decimal places, we have [latex]5.66[/latex] cm.

Example 3: Do the sides [latex]17[/latex], [latex]15[/latex] and [latex]8[/latex] form a right triangle? If so, which sides are the legs and the hypotenuse?

If these are the sides of a right triangle then it must satisfy the Pythagorean Theorem. The sum of the squares of the shorter sides must be equal to the square to the longest side. Obviously, the sides [latex]8[/latex] and [latex]15[/latex] are shorter than [latex]17[/latex] so we will assume that they are the legs and [latex]17[/latex] is the hypotenuse. So we let [latex]a=8[/latex], [latex]b=15[/latex], and [latex]c=17[/latex].

Let’s plug these values into the Pythagorean equation and check if the equation is true.

Since we have a true statement, then we have a case of a right triangle! We can now say for sure that the shorter sides [latex]8[/latex] and [latex]15[/latex] are the legs of the right triangle while the longest side [latex]17[/latex] is the hypotenuse.

Example 4: A rectangle has a length of [latex]8[/latex] meters and a width of [latex]6[/latex] meters. What is the length of the diagonal of the rectangle?

The diagonal of a rectangle is just the line segment that connects two non-adjacent vertices. In the figure below, it is obvious that the diagonal is the hypotenuse of the right triangle while the two other sides are the legs which are [latex]8[/latex] and [latex]6[/latex].

If we let [latex]a=6[/latex] and [latex]b=8[/latex], we can solve for [latex]c[/latex] in the Pythagorean equation which is just the diagonal.

Therefore, the measure of the diagonal is [latex]10[/latex] meters.

Example 5: A ladder is leaning against a wall. The distance from the top of the ladder to the ground is [latex]20[/latex] feet. If the base of the ladder is [latex]4[/latex] feet away from the wall, how long is the ladder?

If you study the illustration, the length of the ladder is just the hypotenuse of the right triangle with legs [latex]20[/latex] feet and [latex]4[/latex] feet.

Again, we just need to perform direct substitution into the Pythagorean Theorem formula using the known values then solve for [latex]c[/latex] or the hypotenuse.

Therefore, the length of the ladder is [latex]4\sqrt {26}[/latex] feet or approximately [latex]20.4[/latex] feet.

Example 6: In a right isosceles triangle, the hypotenuse measures [latex]12[/latex] feet. What is the length of each leg?

Remember that a right isosceles triangle is a triangle that contains a 90-degree angle and two of its sides are congruent.

In the figure below, the hypotenuse is [latex]12[/latex] feet. The two legs are both labeled as [latex]x[/latex] since they are congruent.

Let’s substitute these values into the formula then solve for the value of [latex]x[/latex]. We know that [latex]x[/latex] is just the leg of the right isosceles triangle which is the unknown that we are trying to solve for.

Therefore, the leg of the right isosceles triangle is [latex]6\sqrt 2[/latex] feet. If we want an approximate value, it is [latex]8.49[/latex] feet, rounded to the nearest hundredth.

Example 7: The diagonal of the square below is [latex]2\sqrt 2[/latex]. Find its area.

We know the area of the square is given by the formula [latex]A=s^2[/latex] where [latex]s[/latex] is the side of the square. So that means we need to find the side of the square given its diagonal. If we look closely, the diagonal is simply the hypotenuse of a right triangle. More importantly, the legs of the right triangle are also congruent.

Since the legs are congruent, we can let it equal to [latex]x[/latex].

Substitute these values into the Pythagorean Theorem formula then solve for [latex]x[/latex].

We calculated the length of the leg to be [latex]2[/latex] units. It is also the side of the square. So to find the area of the square, we use the formula

[latex]A = {s^2}[/latex]

That means, the area is

[latex]A = {s^2} = {\left( 2 \right)^2} = 4[/latex]

Therefore, the area of the square is [latex]4[/latex] square units.

You might also like these tutorials:

• Pythagorean Theorem Practice Problems with Answers
• Pythagorean Triples
• Generating Pythagorean Triples

Pythagorean Theorem and Problems with Solutions

Explore some simple proofs of the Pythagorean theorem and its converse and use them to solve problems. Detailed solutions to the problems are also presented.

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How The Pythagorean Theorem Helps Solve a Right Triangle

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Key Takeaways

• The Pythagorean theorem is crucial in various fields, including construction, manufacturing and navigation, enabling precise measurements and the creation of right angles for large structures.
• It underpins our entire system of measurement, allowing for accurate navigation by pilots and ships, and making GPS measurements possible through the calculation of distances and angles.
• Beyond navigation, the theorem is essential in geometry, physics, geology, engineering and even practical applications by carpenters and machinists.

The Pythagorean theorem is an ancient mathematical theorem which is one of the most fundamental and important concepts in two-dimensional Euclidean geometry going back thousands of years. It can help students find the sides of a right triangle on a piece of paper, but it also has greater implications in the fields of engineering, physics and architecture.

Since triangles always follow concrete rules we can use concepts like the Pythagorean theorem formula — and later, trigonometry — to find all the parameters of a triangle (angle values, lengths) if at least one of them is known.

The Pythagorean theorem is the simplest of these concepts and lets us easily solve the length of a third side of a right triangle if two sides are currently known.

Basics of The Pythagorean Theorem

Solving a right angled triangle using pythagorean formula, common forms of pythagorean triples, who was pythagoras, pythagorean theorem calculator, how is the pythagorean theorem useful today.

In its simplest form , the Pythagorean theorem states that in a hypothetical right triangle abc : a² + b² = c² .

The value of c² is equal to the sum of the squares, where hypotenuse c is the longest side of a right triangle. It's also always the side opposite the right angle.

Using this formula, we can always find the length of the hypotenuse if the other two sides are known values. After adding the numbers, we will need to apply a square root operation to arrive at the value of c .

Going back to triangle abc , what do we do if one of the known sides is the hypotenuse? We can reverse the Pythagorean theorem formula and turn it into a subtraction problem, then apply a square root just as before.

If a triangle contains two unknown sides, then more complex trigonometric formulas and algebraic proofs will have to be applied in order to find them. This same mathematical theorem can also be applied to physics problems like triangular force vectors.

What Is a Right Triangle?

A right angled triangle has exactly one of its angle values equal to 90 degrees, which is where the Pythagorean theorem formula can be applied. The side opposite the right angle is known as the hypotenuse and will always be the longest side of the right angled triangle.

Triangles without a right angle, like a scalene or isosceles triangle, cannot be solved using the Pythagorean theorem. They must be broken up into smaller shapes or have more complex formulas applied.

Like all triangles, the angle values of a right angled triangle add up to a sum of 180 degrees. This also means that the two non-right angles of the triangle must add up to 90 degrees.

Now that we know a bit about solving right angled triangle abc , let's replace our variables with real numbers and run through the formula again. The side lengths we know are 16 and 20, and our hypotenuse is the unknown side.

Based on these calculations, we now know that the hypotenuse of the triangle equals 25.61.

Another useful idea related to Pythagoras theorem proof is the concept of a Pythagorean triple. These are essentially forms of right triangles which have sides that are all equal to whole numbers.

The most common form of a Pythagorean triple you are likely to see math education is known as the (3, 4, 5) triangle. If two sides of a right triangle equal 3 and 4, then the hypotenuse will always be 5.

Shown using the Pythagorean theorem formula:

Learning to spot Pythagorean triples by eye can help you easily solve them without resorting to the Pythagorean theorem formula every time. There are theoretically infinite Pythagorean triples out there, but some other common ones include:

Similar triangles to a Pythagorean triple will themselves be triples. So we can multiply all the values of our previous example by 2 to get a triangle of (6, 8, 10).

Multiplying any Pythagorean triple by any positive integer (as in, applying the same multiplier to all sides of the same triangle) will give you similar results.

The Pythagorean equation is most often attributed to Pythagoras of Somos , but we now know that many ancient civilizations like those in Egypt, India and China had discovered the mathematical relationship independently.

That said, the man whom this math trick is named for is nearly as fascinating. Pythagoras, an ancient Greek thinker who was born on the island of Samos and lived from 570 to 490 B.C.E, was kind of a trippy character — equal parts philosopher, mathematician and mystical cult leader.

In his lifetime, Pythagoras wasn't known as much for solving for the length of the hypotenuse as he was for his belief in reincarnation and adherence to an ascetic lifestyle that emphasized a strict vegetarian diet, adherence to religious rituals and plenty of self-discipline that he taught to his followers.

Pythagoras biographer Christoph Riedweg describes him as a tall, handsome and charismatic figure, whose aura was enhanced by his eccentric attire — a white robe, trousers and a golden wreath on his head. Odd rumors swirled around him — that he could perform miracles, that he had a golden artificial leg concealed beneath his clothes and that he possessed the power to be in two places at one time.

Pythagoras founded a school near what is now the port city of Crotone in southern Italy, which was named the Semicircle of Pythagoras. Followers, who were sworn to a code of secrecy, learned to contemplate numbers in a fashion similar to the Jewish mysticism of Kaballah. In Pythagoras' philosophy, each number had a divine meaning, and their combination revealed a greater truth.

With a hyperbolic reputation like that, it's little wonder that Pythagoras was credited with devising one of the most famous theorems of all time, even though he wasn't actually the first to come up with the concept. Chinese and Babylonian mathematicians beat him to it by a millennium.

"What we have is evidence they knew the Pythagorean relationship through specific examples," writes G. Donald Allen , a math professor and director of the Center for Technology-Mediated Instruction in Mathematics at Texas A&M University, in an email. "An entire Babylonian tablet was found that shows various triples of numbers that meet the condition: a 2 + b 2 = c 2 ."

The earliest known example of the Pythagorean theorem formula is on a clay tablet unearthed in modern-day Iraq and now resides in a museum in Istanbul. This tablet of Babylonian origin displays various trigonometric functions, including what we now know as the Pythagorean theorem, but it predates Pythagoras by more than 1,000 years. Historians estimate the tablet was drawn as early as 1,900 B.C.E.

The Pythagorean theorem isn't just an intriguing mathematical exercise. It's utilized in a wide range of fields, from construction and manufacturing to navigation.

As Allen explains, one of the classic uses of the Pythagorean theorem is in laying the foundations of buildings. "You see, to make a rectangular foundation for, say, a temple, you need to make right angles. But how can you do that? By eyeballing it? This wouldn't work for a large structure. But, when you have the length and width, you can use the Pythagorean theorem to make a precise right angle to any precision."

Beyond that, "This theorem and those related to it have given us our entire system of measurement," Allen says. "It allows pilots to navigate in windy skies, and ships to set their course. All GPS measurements are possible because of this theorem."

In navigation, the Pythagorean theorem provides a ship's navigator with a way of calculating the distance to a point in the ocean that's, say, 300 miles north and 400 miles west (480 kilometers north and 640 kilometers west). It's also useful to cartographers, who use it to calculate the steepness of hills and mountains.

"This theorem is important in all of geometry, including solid geometry," Allen continues. "It is also foundational in other branches of mathematics, much of physics, geology, all of mechanical and aeronautical engineering. Carpenters use it and so do machinists. When you have angles, and you need measurements, you need this theorem."

One of the formative experiences in the life of Albert Einstein was writing his own mathematical proof of the Pythagorean theorem at age 12. Einstein's fascination with geometry eventually played a role in his development of the theories of special and general relativity.

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Problem solving using Pythagoras' Theorem

Subject: Mathematics

Age range: 14-16

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Last updated

8 March 2020

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5 Problems that require careful thinking using Pythagoras’ Theorem. Suitable for anyone who has already covered the basics Answers included on P2

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Mathematics > Optimization and Control

Title: numerical exploration of the pythagorean theorem using hobo algorithm.

Abstract: This paper introduces a novel method for finding integer sets that satisfy the Pythagorean theorem by leveraging the Higher-Order Binary Optimization (HOBO) formulation. Unlike the Quadratic Unconstrained Binary Optimization (QUBO) formulation, which struggles to express complex mathematical equations, HOBO's ability to model higher-order interactions between binary variables makes it well-suited for addressing more complex and expressive problem settings.

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