pythagorean theorem problem solving questions

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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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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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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.

The next lessons are

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Pythagorean Theorem: Problems with Solutions

Pythagoras Theorem Questions (with Answers)

Pythagoras theorem.

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

• Length of the hypotenuse is c
• The hypotenuse is the longest side
• Lengths of the other sides are a, b

Right Triangle Questions – using the theorem

The Theorem helps us in:

• Finding Sides: If two sides are known, we can find the third side.
• Determining if a triangle is right-angled: If the sides of a triangle are known and satisfy the Pythagoras Formula, it is a right-angled triangle.

There is a proof of this theorem by a US president. Its simplicity makes it is easy enough for the grade 8 kids to understand.

Finding the missing sides (side lengths) of a Right Triangle

The theorem gives a relation among the three sides of a right-angled triangle. We can find one side if we know the other two sides. How?

Example: We are given (see figure) below the two sides of the right triangle. Find the third side.

Given: a = 3, c = 5

Which side is the hypotenuse?

✩ Always identify the hypotenuse first

Unknown side = BC = b ?

Putting the values in the Pythagoras Formula: a 2 + b 2 = c 2

3 2 + b 2 = 5 2

9 + b 2 = 25

b 2 = 25 − 9 = 16 = 4 2

Finding the Hypotenuse of a Triangle

Using the Pythagoras formula, finding hypotenuse is no different from any other side.

Example: Sides of a right triangle are 20 cm and 21 cm, find its hypotenuse.

Pythagoras Formula: a 2 + b 2 = c 2

• c is the length of the hypotenuse
• a, b are the lengths of the other two sides (you can assume any length as a or b ).

Let AB = a = 20, BC = b = 21

Putting values in the formula:

20 2 + 21 2 = c 2

400 + 441 = c 2

Finding Right Triangle

Given the sides, we can determine if a triangle is right-angled by applying the Pythagoras Formula. How?

• Assume the longest side to be hypotenuse Length = c . Find its square ( = c 2 )
• Find the sum of squares of the other two sides ( = a 2 + b 2 )
• If a 2 + b 2 ≠ c 2 it is a not right triangle
• If a 2 + b 2 = c 2 it is a right triangle

Example: A triangle has sides 8 cm, 11 cm, and 15 cm. Determine if it is a right triangle

Longest side = 15 cm. Let us assume it to be hypotenuse = c (as we know that it is always the longest)

c 2 = 15 2 = 225

Other sides a = 8 cm b = 11 cm. (You can assume any side length to be a or b ).

a 2 + b 2 = 8 2 + 11 2 = 64 + 121 = 185

a 2 + b 2 ≠ c 2

So this is not a right-angled triangle.

Example: The sides of a triangle are 8 cm, 17 cm, and 15 cm. Find if it is a right triangle.

Longest side = 17 cm. Let us assume it to be hypotenuse = c

c 2 = 17 2 = 289

Other sides a = 8 cm, b = 15 cm.

a 2 + b 2 = 8 2 + 15 2 = 64 + 225 = 289

So a 2 + b 2 = c 2

It is a right-angled triangle.

Pythagoras Questions Types

You will encounter the following types of questions related to this theorem:

• Find a side, given two sides These questions are the direct application of the theorem (formula) and are easiest to solve.
• Express the relation between the two sides in an equation
• Substitute one side by the other using the first equation in the Pythagoras Formula
• Given the Perimeter and one side, find other sides – Perimeter is the sum of the three sides. Since one side is known, we subtract it from the perimeter to get a relationship between the other two sides.
• Given Area and one side find other sides – Area = 2 1 ​ × ( ba se × a lt i t u d e ) . Base and altitude can be the sides with the right angle OR the hypotenuse and the altitude.

Pythagoras Questions

The questions chosen have minimal use of other concepts, yet, some of these are hard Pythagoras questions (See Ques 4 and Ques 10 ).

1 Question

ABC is a right triangle. AC is its hypotenuse. Length of side AB is 2√5 . Side BC is twice of side AB. Find the length of AC.

Can you express BC in terms of AB and apply the Pythagoras Theorem?

Ans.   AC = 10

2 Question

The hypotenuse of a right triangle is 6 cm . Its area is 9 cm 2 . Find its sides.

Can you form two equations – one using area and the other using the Pythagoras formula?

Ans.   Each side is 3√2 cm.

3 Question

One side of a right triangle is 4 10 ​ cm. Find the length of its other side if the hypotenuse is 13 cm.

Can you directly apply the Pythagoras Theorem?

Ans.   3 cm

4 Question

In a right triangle ABC, length of the medians to the sides AB and BC are 2 61 ​ and 601 ​ respectively. Find the length of its hypotenuse.

CE = 2 61 ​

Let AE = EB = x

BD = DC = y

Can you use the Pythagorean Theorem to form an equation between x, y, and AD?

Can you form another equation involving x, y, and EC?

Can you use this result to find A C 2 ? Use the triangle ABC.

Ans.   AC = 26

5 Question

In a right triangle, the longest side is 8 cm. One of the remaining sides is 4√3 cm long. Find the length of the other side.

Can you apply the Pythagoras Theorem directly?

Ans.   4 cm

6 Question

The first side of a right triangle is shorter than the second side by 1 cm. It is longer than the third side by 31 cm. Find the sides of the triangle.

Can you form equations between the first side and the other two sides? Which side is the hypotenuse?

Ans.   9 cm, 40 cm, 41 cm

7 Question

The perimeter of a right triangle is equal to 30 cm. The length of one of its sides is 10 cm. Find its hypotenuse.

Can you find the relation between the unknown side and the perimeter in terms of the hypotenuse?

Ans.   12.5 cm

8 Question

The sides of a triangle are 5 cm, 9 cm, and 12 cm. Is it a right-angled triangle?

Can you identify the possible hypotenuse? Also, test if the sides satisfy the Pythagoras Formula.

Ans.   Not a right triangle.

9 Question

In a right triangle, two sides are equal. The longest side is 7√2 cm, find the remaining sides.

Ans.   7 cm

10 Question

In the following right triangle altitude BD = 9 10 ​ cm and DC = 27 10 ​ cm. Find the sides of the triangle.

Can you apply the Pythagoras Theorem to the triangle BCD?

Can you form an equation between a and x using the triangle ABD?

Can you find the value of x using above equations and triangle ABC?

Ans.   AB = 30cm, BC = 90cm, A C = 30 10 ​ c m

Answers to Pythagoras Questions

1 answer.

Let AB = a , BC = b and AC = c .

AB = a = 2√5

BC is twice of AB, b = 2a = 4√5

AC = Hypotenuse = c

Applying the Pythagoras Theorem a 2 + b 2 = c 2 :

(2√5) 2 + (4√5) 2 = c 2

4(5) + 16(5) = c 2

c 2 = 20 + 80 = 100

2 Answer

Let AB = a , BC = b

In triangle ABC, base = b and altitude = a

Area of Triangle = 2 1 ​ × ( ba se × a lt i t u d e ) . So:

2 1 ​ × ( ab ) = 9

ab = 18 (Equation 1)

Using the Pythagoras Formula:

a 2 + b 2 = 6 2 = 36

We add and subtract 2ab to complete the square :

a 2 + b 2 − 2ab + 2ab = 36

(a − b) 2 + 2ab = 36

(a − b) 2 + 36 = 36 Using ab = 18 from Equation 1

(a − b) 2 = 0

Substituting a by b in Equation 1:

Side AB = BC = 3√2 cm

3 Answer

Let the length of sides be a, b and c , such that:

a = 4 10 ​ cm

b = unknown

From the Pythagoras formula a 2 + b 2 = c 2 , we get:

( 4 10 ​ ) 2 + b 2 = 1 3 2

(16 × 10) + b 2 = 169

160 + b 2 = 169

b 2 = 169 − 160 = 9

4 Answer

Using triangle ABD:

AB 2 + BD 2 = AD 2

( 2 x ) 2 + y 2 = ( 601 ​ ) 2

4 x 2 + y 2 = 601 (Equation 1)

Using triangle EBC :

EB 2 + BC 2 = EC 2

x 2 + ( 2 y ) 2 = ( 2 61 ​ ) 2

x 2 + 4 y 2 = ( 2 61 ​ ) 2

x 2 + 4 y 2 = 244 (Equation 2)

Adding Equation 1 and 2:

4x 2 + y 2 + x 2 + 4y 2 = 601 + 244

5x 2 + 5y 2 = 845

x 2 + y 2 = 169 (Equation 3)

Let us solve for the hypotenuse using the triangle ABC:

AB 2 + BC 2 = AC 2

(2x) 2 + (2y) 2 = AC 2

4x 2 + 4y 2 = AC 2

4(x 2 + y 2 ) = AC 2

Substituting the value of x 2 + y 2 from Equation 3:

4(169) = AC 2

A C = 4 ( 169 ) ​

AC = 2 × 13 = 26

5 Answer

Let the lengths of sides be a, b and c (hypotenuse).

Hypotenuse is the longest side. So c = 8 .

Let b = 4√3 .

From the Pythagoras Theorem:

a 2 + b 2 = c 2

a 2 + (4√3) 2 = 8 2

a 2 + 16(3) = 64

a 2 + 48 = 64

The third side is 4 cm.

6 Answer

The second side is the longest. It is the hypotenuse. Let its length be c .

Let the length of first side be b and third side a .

Applying the Pythagoras formula:

(b − 31) 2 + b 2 = (b + 1) 2

b 2 − 62b + 31 2 + b 2 = b 2 + 2b + 1

b 2 − 62b + 961 = 2b + 1

b 2 − 64b + 960 = 0

b 2 − 24b − 40b + 960 = 0

b(b − 24) − 40(b − 24) = 0

(b − 40)(b − 24) = 0

b = 40 Or b = 24

For b = 24 , we get a = 24 − 31 = − 7 . Length of a side cannot be negative, so we reject b = 24 .

For b = 40 , we get a = 40 − 31 = 9 and c = 40 + 1 = 41

The sides of triangle are 9 cm, 40 cm and 41 cm.

7 Answer

Side BC = b = 10 cm

Perimeter = Sum of the sides

= a + b + c = 30 (Given)

a + 10 + c = 30

a = 20 − c ( Equation 1 )

Applying the Pythagoras Theorem to find the hypotenuse:

Using Equation 1 to substitute the value of a

(20 − c) 2 + (10) 2 = c 2

400 − 40c + c 2 + 100 = c 2

500 − 40c = 0

The length of hypotenuse = 12.5 cm

8 Answer

Longest side = 12 cm. Let us assume it to be the hypotenuse = c

So c 2 = 12 2 = 144

The Pythagoras Formula: a 2 + b 2 = c 2

We can assume any side to be a or b.

Let a = 5 cm, b = 9 cm.

a 2 + b 2 = 5 2 + 9 2 = 25 + 81 = 106

So a 2 + b 2 ≠ c 2

This is a not a right angled triangle.

9 Answer

Let the length of the sides be a, b , and c (hypotenuse).

In a right triangle hypotenuse is the longest side. So c = 7√2

Other sides are equal. So a = b .

Applying the Pythagoras theorem:

b 2 + b 2 = (7√2) 2

2b 2 = 49(2)

Each side is 7 cm.

10 Answer

Let AB = a, BC = b, AC = c and AD = x

Given B D = 9 10 ​ D C = 27 10 ​

Applying Pythagoras Theorem to triangle BCD:

BD 2 + DC 2 = BC 2

( 9 10 ​ ) 2 + ( 27 10 ​ ) 2 = B C 2

(9 2 × 10) + (27 2 × 10) = b 2

810 + 7290 = b 2

Applying Pythagoras Theorem to triangle ABD:

BD 2 + AD 2 = AB 2

a 2 = ( 9 10 ​ ) 2 + x 2 (Equation 1)

From the figure:

AC = AD + BD

c = x + 27 10 ​ (Equation 2)

Applying Pythagoras Theorem to triangle ABC:

Using b = 90 and value of a 2 from Equation 1 and c from Equation 2:

( 9 10 ​ ) 2 + x 2 + 9 0 2 = ( x + 27 10 ​ ) 2

9 2 ( 10 ) + x 2 + 9 0 2 = x 2 + 54 10 ​ x + 2 7 2 ( 10 )

810 + 8100 = 54 10 ​ x + 7290

54 10 ​ x = 8910 − 7290 = 1620

x = 10 ​ 30 ​

Putting value of x in Equation 1:

a 2 = ( 9 10 ​ ) 2 + ( 3 10 ​ ) 2

a 2 = 810 + 90 = 900

Using value of a and b in Pythagoras Formula for triangle ABC:

30 2 + 90 2 = c 2

900 + 8100 = c 2

c = 30 10 ​

AB = 30cm, BC = 90cm, A C = 30 10 ​ c m

Difficult Pythagoras Questions (Year 10, Guided Answers) ➤

James Garfield Pythagorean Theorem (Illustration & Proof) ➤

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

• High School
• High school geometry
• Pythagorean Theorem

Expert Guidance for Your Pythagorean Theorem Questions

Two congruent circles with centres at (2,3) and (5,6), which intersect at right angles, have radius equal to?

State whether the given statement is true or false: 9, 40, 41 is a Pythagorean triplet. True or false

5, 12, 13 is a Pythagorean triplet. True or false

8, 15, 17 is a Pythagorean triplet

Based on Pythagorean identities, which equation is true? A. sin 2 ⁡ θ − 1 = cos 2 ⁡ θ B. sec 2 ⁡ θ − tan 2 ⁡ θ = − 1 C. - cos 2 ⁡ θ − 1 = sin 2 ⁡ θ D. cot 2 ⁡ θ − csc 2 ⁡ θ = − 1

The end rollers of bar AB(1.5R) are constrained to the slot. If roller A has a downward velocity of 1.2 m/s and this speed is constant over a small motion interval, determine the tangential acceleration of roller B as it passes the topmost position. The value of R is 0.5 m.

The area of the obtuse angle triangle shown below is: A17.5 sq. units B 25 sq. units C 14 sq. units D 15 sq. units

Replacing x with 1 2 in 2 x 2 will give you an answer of 1.

Why does the Pythagorean theorem only work for right triangles?

7, 24, 25 is a Pythagorean triplet. A.True B.False

A bicycle wheel with a 5 inch ray rotates 60 ∘ . What distance has the bicycle traveled?

The legs of a right triangle are 6 and 8 cm. Find the hypotenuse and the area of ​​the triangle.

Solve for X. Anlge A=8x+5 Angle B=4 x 2 -10 Angle C= x 2 +2x+10 I know that they equal 180 degrees. However I am drawinga blank on the factoring part of it

For each of the following, can the measures represent sides ofa right triangle? Explain your answers.a. 3 m, 4 m, 5 mb. 2 c m , 3 c m , 5 c m

A 10-m ladder is leaning against a building. The bottom of theladder is 5-m from the building. How high is the top of theladder?

Pythagorean theorem and its cause I'm in high school, and one of my problems with geometry is the Pythagorean theorem. I'm very curious, and everything I learn, I ask "but why?". I've reached a point where I understand what the Pythagorean theorem is, and I understand the equation, but I can't understand why it is that way. Like many things in math, I came to the conclusion that it is that way because it is; math is the laws of the universe, and it may reach a point where the "why" answers itself. So what I want to know is, is there an explication to why the addition of the squared lengths of the smaller sides is equal to the squared hypotenuse, or is it just a characteristic of the right triangle itself? And is math the answer to itself? Thank you.

pythagorean theorem extensions are there for a given integer N solutions to the equations ∑ n = 1 N x i 2 = z 2 for integers x i and zan easier equation given an integer number 'a' can be there solutions to the equation ∑ n = 1 N x i 2 = a 2 for N=2 this is pythagorean theorem

"Pythagorean theorem" for projection onto convex set I'm going through the book on online convex optimization by Hazan, and in the first chapter I saw this assertion (which Hazan calls the "pythagorean theorem"): Let K ⊂ R d be a convex set, y ∈ R d , and x = Π K ( y ) . Then for any z ∈ K we have: ‖ y − z ‖ ≥ ‖ x − z ‖ . It is presented without proof - what is a proof for this? Also, how does it relate to the pythagorean theorem?

Non-geometric Proof of Pythagorean Theorem Is there a purely algebraic proof for the Pythagorean theorem that doesn't rely on a geometric representation? Just algebra/calculus. I want to TRULY understand the WHY of how it is true. I know it works and I know the geometric proofs.

The Pythagorean theorem and Hilbert axioms Can one state and prove the Pythagorean theorem using Hilbert's axioms of geometry, without any reference to arithmetic? Edit: Here is a possible motivation for this question (and in particular for the "state" part of this question). It is known that the theory of Euclidean geometry is complete. Every true statement in this theory is provable. On the other hand, it is known that the axioms of (Peano) arithmetic cannot be proven to be consistent. So, basically, I ask if there is a reasonable theory which is known to be consistent and complete, and in which the Pythagorean theorem can be stated and proved. In summary, I guess I am asking - can we be sure that the Pythagorean theorem is true? :)

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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.

MathBootCamps

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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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]

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

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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Pythagoras’ Theorem: My attempt to write challenging questions and engage students

On my last placement I taught a unit on Pythagoras' theorem to my year 9 class. I found it deceptively difficult. Why? The challenge was twofold: First, students had seen the theorem in previous years so it was a bit old hat to them. Second, I wondered how to stretch students and provide them with variety and challenge beyond the old trick of calling the hypotenuse a ladder?

I was thinking to myself ‘ WHY does Pythagoras’ theorem actually matter, and how can I make that tangible and relevant to students? I came up with the following slideshow.

I had this first slide on the screen as students walked into the classroom, it created a bit of a buzz in the room…

I got students to vote (hands up) on which corner they thought was square (which do you think it is?), then revealed the answer.

I've presented the remainder of the slideshow in the following video with examples of the kind of teacher questions and the script that I used. Seemed easier and clearer than writing it all out…

(If you want to use any of this you can download the slideshow file here:  Pythagoras Slides_Oliver Lovell_www.ollielovell.com )

Clarifying points…

• I used the qualifier of ‘on the macro scale' when talking about ‘no squares in nature' because I thought it's probably possible for there to be perfect squares in crystalline atomic structures or something like that.
• Perhaps an oversight in the video is a discussion of the meaning of the word ‘breadth' (I've become more aware of the vocab that I'm using in class since my resent teaching placement in Myanmar where the students had very limited english. Post to come).

The general structure of lessons in this unit from then on was a bit of book work (using the ‘points' system that I'll write about in another post) with a challenge question for those students who got all of their points done. Lessons were wrapped up with students sharing on the board their approach to the challenge question and what they had learned from attempting it. I also book-ended most lessons with a micro-revision, 3 questions on the basics as a mini-test at the start, and an exit card at the end to help me to keep track of how students were progressing.

Below are the questions that I used as challenge questions. I wrote all of the following questions, excepting the final question, which I found here . I'd also like to acknowledge Dr Max Stephens who suggested that I create questions that encourage students to solve for exact values rather than untidy decimals. I've left out answers on purpose in the knowledge that future students will possibly visit this post.

Pythagoras Extension Questions:

You’ve always got to include a ladder question don’t you? (this one is a little bit different).

A Cat is stuck in a tall tree and needs to be saved by the Firefighters. The fire truck has a 35m ladder. The closest that it can get to the tree (because there are parked cars in the way) is 12m. The cat is in the tree, 39m up. Comment on the likelihood of the cat getting saved and justify your suggestion mathematically.

Q:  A cube has side length of 10cm. what is the length of the diagonal through its centre? (could accompany this with the picture below).

Alternatively you could ask, Q:  ‘What is the length of d in the following if the side lengths are 10cm?' (referring to the above image)

Prior to giving students the above question we’d had a discussion about the formula of the diagonal of a box and students had worked out that it’s just the square root of the sum of the squares of the box’s dimensions. This helped them to solve this question. One of the students actually proposed this ‘pythagoras in 3 dimensions' formula, so I encouraged them all to check it and see if the proposal seemed plausible.

An additional question that I also posed to students was,  Q:  “Show that the formula for the diagonal of a box is √a²+b²+c².” A discussion of this also helped to scaffold the students.

Q:  A rectangular prism has sides of length a, 4a/3, and 4a. It has an internal diagonal of 26cm. What are the dimensions of the rectangular prism?

Or, you could word it in a more tricky way:   Q:  A rectangular prism has a width one quarter of its length and a height one third of its width. It has an internal diagonal of 26cm. What are the dimensions of the rectangular prism?

Some other interesting shapes…

Another question:

In the 6 sided pyramid pictured below, what is the height of the apex ‘V’ above the point  ‘O’ ? (You can assume that O is in the centre of the base, the base is a regular hexagon, and V is directly above O).

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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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