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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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What is the Pythagorean Theorem?

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[FREE] Pythagorean Theorem Worksheet (Grades 8)

3D Pythagorean Theorem

Common Core State Standards

How to use Pythagorean Theorem

Pythagorean Theorem examples

Example 2: missing length c

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

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

Example 5: checking if a triangle has a right angle

Example 6: checking if a triangle has a right angle

Teaching tips for Pythagorean Theorem

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Practice Pythagorean Theorem questions

Pythagorean Theorem FAQs

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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 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

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.

Pythagoras Practice Questions

GCSE Revision Cards

5-a-day Workbooks

Primary Study Cards

Pythagoras Theorem Questions (with Answers)

Right Triangle Questions – using the theorem

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

Finding the Hypotenuse of a Triangle

Finding Right Triangle

Pythagoras Questions Types

Pythagoras Questions

1 Question

2 Question

3 Question

4 Question

5 Question

6 Question

7 Question

8 Question

9 Question

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

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Pythagorean theorem practice problems with answers.

There are eight (8) problems here about the Pythagorean Theorem for you to work on. When you do something a lot, you get better at it. Let’s get started!

Here’s the Pythagorean Theorem formula for your quick reference.

Pythagorean Theorem - a squared plus b squared is equal to c squared

Note: drawings not to scale

Problem 1: Find the value of [latex]x[/latex] in the right triangle.

the legs of a right triangle are 6 and 8

[latex]10[/latex]

Problem 2: Find the value of [latex]x[/latex] in the right triangle.

the leg of a right triangle is 12 and the hypotenuse is 16

[latex]4\sqrt 7 [/latex]

Problem 3: Find the value of [latex]x[/latex] in the right triangle.

right triangle with legs square root 3 and square root 6

[latex]3[/latex]

Problem 4: The legs of a right triangle are [latex]5[/latex] and [latex]12[/latex]. What is the length of the hypotenuse?

[latex]13[/latex]

Problem 5: The leg of a right triangle is [latex]8[/latex] and its hypotenuse is [latex]17[/latex]. What is the measure of its other leg?

[latex]15[/latex]

Problem 6: Suppose the shorter leg of a right triangle is [latex]\sqrt 2[/latex]. The longer leg is twice the shorter leg. Find the hypotenuse.

[latex]\sqrt {10} [/latex]

Problem 7: The hypotenuse of a right triangle is [latex]4\sqrt 2[/latex]. If the longest leg is half the hypotenuse, what is the length of the shortest leg?

[latex]2\sqrt 6[/latex]

Problem 8: Find the value of [latex]x[/latex] of the right isosceles triangle.

the hypotenuse of a right isosceles triangle is 2 times square root 3

[latex]\sqrt 6[/latex]

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.

= (Perpendicular) + (Base)

Click here to learn the proof of Pythagoras’ Theorem .

pythagorean theorem problem solving questions

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.

If three integers a, b and c are such that a + b = c , then (a, b, c) is called Pythagorean triples.

For any given integer m, (m – 1, 2m, m + 1) is the Pythagorean triplet.

Learn more about .

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.

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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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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Here are the instructions how to enable JavaScript in your web browser .

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.

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

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.

$Right triangle labelled$

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.

$Right triangle pythagoras$

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:

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

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

$Pythagoras theorem example 1$

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

$Pythagoras example 2$

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

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

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

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 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.

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.

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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.

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.

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.

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!

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

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.

In order to use Pythagorean Theorem:

Label the sides of the triangle.

Write down the formula and apply the numbers.

Record the answer.

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}

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}

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}

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}

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.

\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.

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.

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.

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 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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Terms and Conditions

Pythagoras theorem.

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

Right Triangle with Pythagoras Statement

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

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.

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

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

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.

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.

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

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

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

Right triangle ABC with hypotenuse AC = 6 cm

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

Ans. Each side is 3√2 cm.

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

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.

Right triangle ABC with medians to side AB and BC

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

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

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

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

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.

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

Ans. 7 cm

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

Right triangle ABC with altitude BD to the hypotenuse AC

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

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

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

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

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

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.

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.

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

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.

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.

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) ➤

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Problem Solver
College Math
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.

Let XY = x m

Therefore, YZ = (100 – x) m

In ∆ XYZ, ∠Z = 90°

Therefore, by Pythagoras theorem

XY = YZ + XZ

⇒ x = (100 – x) + 60

⇒ = 10000 – 200x + + 3600

$Pythagorean Theorem Word Problem$

⇒ 200x = 10000 + 3600

⇒ 200x = 13600

⇒ x = 13600/200

Therefore, distance between X and Y = 68 meters.

$Word problems on Pythagorean Theorem$

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.

$Word problem on Pythagorean Theorem$

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.

$Word problems using the Pythagorean Theorem$

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.

$Pythagorean Theorem: Word Problems$

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.

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Worksheet on unitary method | unitary method word problems | answers.

Aug 02, 24 04:40 PM

$Worksheet on Unitary Method$

Aug 02, 24 04:13 PM

$Unitary Method$

Aug 01, 24 02:50 PM

Aug 01, 24 02:47 PM

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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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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.

image showing the pythagorean theorem formula

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.

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.

figure for a word problem with the pythagorean theorem

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.

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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TemplateLab

48 pythagorean theorem worksheet with answers [word + pdf].

The simplicity of the Pythagorean Theorem worksheet is the best thing about it. What is the Pythagorean Theorem? Formulated in the 6th Century BC by Greek Philosopher and mathematician Pythagoras of Samos, Pythagorean Theorem is a mathematic equation used for a variety of purposes. Over the years, many engineers and architects have used Pythagorean Theorem worksheet to complete their projects .

Table of Contents

1 Pythagorean Theorem Worksheet
2 Knowing Pythagoras of Samos and how he came up with the Pythagorean equation
3 Understanding Pythagorean Theorem
4 Pythagorean Theorem Word Problems Worksheet
5 Using Pythagorean Theorem worksheet
6 Conclusion

A simple equation, Pythagorean Theorem states that the square of the hypotenuse (the side opposite to the right angle triangle) is equal to the sum of the other two sides . Following is how the Pythagorean equation is written:

In the aforementioned equation, c is the length of the hypotenuse while the length of the other two sides of the triangle are represented by b and a. Though the knowledge of the Pythagorean Theorem predates the Greek Philosopher, Pythagoras is generally credited for bringing the equation to the fore. This is the reason the Pythagorean equation is named after him. Before we discuss the Pythagorean Theorem and the Pythagorean Theorem worksheet in detail, let’s take a look at who Pythagoras of Samos was and how he came up with the Pythagorean equation.

A 6 th century BC Greek philosopher and mathematician, Pythagoras of Samos is widely credited for bringing the Pythagorean equation to the fore. Though others used the relationship long before his time, Pythagoras is the first one who made the relationship between the lengths of the sides on a right-angled triangle public. This is why he’s regarded as the inventor of the Pythagorean equation.

Apart from being a philosopher and mathematician, Pythagoras founded the Pythagoreanism movement. Born in Croton, Italy, Pythagoras travelled to many different countries including Greece, Egypt, and India. After moving back to Croton in 530 BC, Pythagoras established some kind of school. He returned to Samos in 520 BC. It was in late 6 th Century BC that Pythagoras started to make important contributions to philosophy and math. The Pythagorean equation was one of those contributions.

Though he revealed the Pythagorean equation to the world in the late 6 th Century BC while living in Samos, many historians believe that Pythagoras first thought about the equation during his time in Egypt. In fact, according to many historians, Pythagoras learned geometry, the Phoenicians arithmetic and other branches of mathematics from the Egyptians.

Though he has made many important contributions to philosophy, Pythagoras is widely known as the founder of the Pythagorean Theorem. As previously mentioned, the Pythagorean Theorem is a mathematical equation that states that the square of the hypotenuse (the side opposite to the right angle triangle) is equal to the sum of the other two sides .

Today the aforementioned equation bears Pythagoras’s name but it’s important to know that he wasn’t the first one to use the equation. Before Pythagoras’s time, the Indians and the Babylonians utilized the Pythagorean Theorem or equation. Since they constructed the first proof of the theorem, Pythagoras and his disciples are regarded as the inventors of the equation.

Many historians say that Pythagoras worked in a very secretive manner. This is the reason little evidence is available that the Greek Philosopher/ mathematician himself worked on and proved the Pythagorean Theorem. It is important to note that the first time Pythagoras was given credit for the Theorem was five centuries after his death. This makes Pythagoras’s contribution to the Theorem even more debatable. Nonetheless, since Pythagoras is the only one connected to the Pythagorean Theorem known today, we have to give him due credit. Now that we’ve discussed who Pythagoras of Samos was and how he came up with the Pythagorean equation, it’s time to take a detailed look at the Pythagorean Theorem and the Pythagorean Theorem worksheet.

According to Pythagorean Theorem, the sum of the squares on the right-angled triangle’s two smaller sides is equal to the side opposite to the right angle triangle (the square on hypotenuse). Using a Pythagorean Theorem worksheet is a good way to prove the aforementioned equation. An amazing discovery about triangles made over two thousand years ago, Pythagorean Theorem says that when a triangle has a 90° angle and squares are made on each of the triangle’s three sides, the size of the biggest square is equal to the size of the other two squares put together! A short equation, Pythagorean Theorem can be written in the following manner:

In Pythagorean Theorem, c is the triangle’s longest side while b and a make up the other two sides. The longest side of the triangle in the Pythagorean Theorem is referred to as the ‘hypotenuse’. Many people ask why Pythagorean Theorem is important. The answer to this is simple: you’ll be able to find the length of a right-angled triangle’s third side if you know the length of the other two sides. This equation works like magic and can be used to find any missing value. Following is an example that uses the Pythagorean Theorem to solve a triangle.

In this equation, the longest side of the triangle ‘c’ is missing. By finding out the sum of the squares of the two other sides, we were able to find the missing value. The most famous mathematical contribution of Pythagoras, the Pythagoras Theorem was one of the earliest documented theorems. Though Pythagoras is given most of the credit for the theorem, a major contribution to the theorem was made by his students.

When you look at a Pythagoras Theorem worksheet, you’ll notice that the theorem enables you to find the length of any right angle triangle side provided you know the length of the other two sides. Also, using the theorem, you can check whether a triangle is a right triangle. The Pythagoras Theorem is extremely useful in solving many math problems. Further, you can use it in many real life situations. This is illustrated by a Pythagoras Theorem worksheet.

A good way to review the Pythagoras Theorem and expand the mathematical equation is using a Pythagoras Theorem worksheet. By using the worksheet, you’ll be able to get a good understanding of geometry. Additionally, the worksheet will give you an opportunity to review the knowledge related to the different types of triangles. Finally and most importantly, you’ll be able to practice the ancient equation invented by the Greek mathematician and philosopher, Pythagoras. Before you start using the Pythagoras Theorem worksheet, just remember that ‘c’ is the hypotenuse while the shorter sides of the triangle are represented by ‘a’ and ‘b’.

A Pythagoras Theorem worksheet presents students with triangles of various orientations and asks them to identify the longest side of the triangle i.e. the hypotenuse. As you know by now, the formula used in Pythagoras Theorem is a²+b²=c². Regardless of what the worksheet asks the students to identify, the formula or equation of the theorem always remain the same. Though, the students could be presented with different challenges including solving triangles:

Labeled in different order
With a different set of letters
By using vertices to name the sides

The symbols used in the Pythagoras Theorem are something students will find on their calculators. Figuring out how to use these functions is what students need to establish. There is involvement of the Babylonians and the Egyptians in the invention of the Pythagoras Theorem but the earliest known proof of the theorem was produced by the school of Pythagoras.

Many Pythagorean triples were known to the Babylonians while the Egyptians knew and used the (3, 4, 5) triple. The Chinese and Indians also played a role in the invention of the Pythagoras Theorem. The first diagrammatic proof of the theorem was produced by the Chinese while the Indians discovered many triples. In 1995, the theorem became part of the Guinness Book of Records as the most proved theorem of all time.

The triples used in the Pythagoras Theorem include (3, 4, 5), (6, 8, 10), (5, 12, 13), (8,15,17), (7,24,25), (20,21,29), (12,35,37), (9,40,41), (28,45,53), (11,60,61), (16,63,65), (33,56,65) and (48,55,73). The aforementioned triples aren’t multiples of a smaller triple and the name given to them is ‘primitive’ triples. To solve a particular problem, the Pythagoras Theorem can be arranged. For example, if you’re asked to find b which is one of the two smaller sides of the right-angled triangle, you can rearrange the theorem to b²=c²-a². By doing this, you’ll be able to easily find the missing value.

The Pythagoras Theorem has many different proofs. However, when checking your answers, following are the two things that you must always remember:

The side opposite to the right angle or simply the hypotenuse is always the longest side of the triangle
Though it is the longest side of the triangle, the size of the hypotenuse can never exceed the sum of the other two squares

To understand this better, take a look at a Pythagoras Theorem worksheet. Today, you can get easy access to Pythagorean Theorem worksheet with answers. Nonetheless, we’re going to try and understand the Pythagoras Theorem as much as we can.

As mentioned earlier, if you know the size of the other two sides, you will be able to find out the length of the third side of the right angle triangle. Also, after being squared, the shorter length is subtracted from the square of the hypotenuse when the hypotenuse is one of the two known lengths. As seen earlier, the lengths of each side of the triangle in the Pythagoras Theorem are whole numbers. Such triangles are known as Pythagorean triangles.

Though there are many different proofs of the Pythagoras Theorem, only three of them can be constructed by students and other people on their own. The first proof starts off as rectangle and is then divided into three triangles that individually contain a right angle. To see first proof, you can use a computer or something as straight forward as an index card cut up into right triangles.

Beginning with a rectangle, the second proof of the Pythagoras Theorem starts off by constructing rectangle CADE with BA=DA. This is followed by the construction of the <BAD’s angle bisector. Once constructed, the bisector is allowed to intersect ED at point F. This makes <BAF and <DAF congruent, BA=DA, and AF=AF. This in turn makes the triangle DAF equal to triangle BAF which means that since ADF is a right angle, ABF will also be a right angle. The third and final proof of the Pythagorean Theorem that we’re going to discuss is the proof that starts off with a right angle. In this proof, triangle ABC is right angle and its right side is angle C.

The three proofs stated above are just few of the many Pythagoras Theorem. You’ll come across these proofs when you take a look at the Pythagorean Theorem worksheet with answers. Learning and understanding the Pythagorean concept is extremely important for students and other people who’ll use this theorem in their practical life.

It is important that you understand the algebraic representation of the Pythagoras Theorem as well as the geometric concepts behind it. You can accomplish this by using proofs, manipulatives, and computer technology. By using these methods to learn Pythagorean Theorem, you’ll be able to see the connections and benefit greatly.

Formulated in the 6th Century BC by Pythagoras of Samos, Pythagoras Theorem is widely used today. If you want to practice Pythagoras Theorem then you can do that easily. Pythagoras Theorem worksheets with answers are easily available and you can use these worksheets to get a good grip of the Theorem.

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Problem 1 :

A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?

the distance of his current position from the starting point = √18 2 + 24 2

= √(324 + 576)

= √900

= 30 m

So, the required distance is 30 m.

Problem 2 :

There are two paths that one can choose to go from Sarah’s house to James house. One way is to take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C street? (Using figure).

By choosing the C street, he has to cover the distance,

= √2 2 + 1.5 2

= √(4 + 2.25)

= √6.25

= 2.5 miles

By choosing the alternative way, he has to cover the distance = 2 + 1.5

= 3.5 miles

The difference between these two paths = 3.5 - 2.5

= 1 mile

So, by choosing the direct path, he may save 1 miles faster than other way.

Problem 3 :

To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?

By drawing the rough picture using the given information, we get

AC = √34 2 + 41 2

= √1156 + 1681

= √2837

= 53.26

Miles saved = (34 + 41) - 53.26

= 75 - 53.26

= 21.74 m

Problem 4 :

In the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm. Calculate the length and breadth of the rectangle?

XY + YZ = 17 cm

XZ + YW = 26 cm

To calculate : - Length and breadth of the rectangle.

We know that,

Diagonals of a rectangle are equal.

XZ = YW = 26/2 = 13 cm

In ∆XYZ, let YZ = P. Then

XY = 17 - P

Then, by Pythagoras theorem,

(P) 2 + (17 - P) 2 = (13) 2

P 2 + 289 - 34P + P 2 = 169

2P 2 - 34P = 169 - 289

2(P 2 - 17P) = - 120

P 2 - 17P = - 120/2

P 2 - 17P = - 60

P 2 - 17P + 60 = 0

P 2 - 12P - 5P + 60 = 0

P(P - 12) - 5(P - 12) = 0

(P - 12)(P - 5) = 0

P - 12 = 0 or P = 12

P = 12 cm or P = 5 cm

YZ = P = 12 cm [Because , YZ is the length of the rectangle ,so we will assign it the greatest value of P]

Again, XY = (17 - P) = (17 - 12) cm = 5 cm

[Because , XY is thee breadth]

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Ollie Lovell

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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.

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?

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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Pythagorean identities are fundamental trigonometric identities derived from the Pythagorean theorem. They establish essential relationships between the sine, cosine, and tangent functions. These identities are crucial in simplifying complex trigonometric expressions and solving various mathematical problems.

In this article, we will learn what are Pythagorean identities, derivations of other identities some more important trigonometry identities, and some practice questions on Pythagorean identities.

Pythagorean identities are composed of these six trigonometric functions present in mathematics: sine, cosine, tan, sec, cosec, and cot. The trigonometric functions sine (sin), cosine (cos), and tangent (tan) are involved in the three main Pythagorean identities , which are obtained from the Pythagorean theses. These identities are:

sin²θ + cos²θ = 1 1 + tan²θ = sec²θ 1 + cot²θ = cosec²θ

Problem 1: Verify the identity: sin 2 θ + cos 2 θ = 1 for θ = 45 o .

Sin45 o = √2/2 cos 45 o = √2/2 sin 2 45 o + cos 2 45 o = (√2/2) 2 + (√2/2) 2 = 2/4 + 2/4 = 4/4 = 1

Problem 2: If cosθ = 5/13 , then find sinθ.

Using the identity: sin 2 θ+ cos 2 θ = 1 sin 2 θ + (5/13) 2 = 1 sin 2 θ + 25/169 = 1 sin 2 θ = 1 – 25/169 sin 2 θ = 144/169 sinθ = ± 12/13

Problem 3: Solve for θ if sec 2 θ = 5.

Using the identity: sec 2 θ = 1 + tan 2 θ 5 = 1 + tan 2 θ tan 2 θ = 4 tanθ= ±2 θ = tan -1 (2)

Problem 4: Prove the identity: cos 2 θ = 1 – sin 2 θ.

Using the basic identity: sin 2 θ + cos 2 θ = 1 cos 2 θ = 1 – sin 2 θ

Problem 5: If sinθ = 3/5 , then find the value of cos θ using Pythagorean identity.

Using the identity: sin 2 θ+ cos 2 θ = 1 (3/5) 2 + cos 2 θ = 1 9/25 + cos 2 θ = 1 cos 2 θ = 1 – 9/25 cos 2 θ = 16/25 cosθ = ± 4/5

Problem 6: Find cosec θ if cot θ = 2.

Using the identity: cosec 2 θ – cot 2 θ = 1 1 + cot 2 θ = cosec 2 θ 1 + 2 2 = cosec 2 θ 1 + 4 = cosec 2 θ cosec 2 θ = 5 cosecθ = ±√5

Problem 7: Express cot 2 θ in terms of cosec 2 θ.

We know by identity that: 1 + cot 2 θ = cosec 2 θ Rearranging above equation we get: cot 2 θ = cosec 2 θ – 1

Problem 8: Find sinθ if secθ = 2.

secθ = 1/cosθ secθ = 2 cosθ = 1/2 sin 2 θ + cos 2 θ = 1 sin 2 θ + (1/2) 2 = 1 sin 2 θ= 1 – 1/4 sin 2 θ = 3/4 sinθ = ± √3/2

Problem 9: If sinθ = 1/2, find tanθ.

Using the identity: sin 2 θ + cos 2 θ = 1 (1/2) 2 + cos 2 θ = 1 cos 2 θ = 1 – 1/4 cos 2 θ = 3/4 cosθ = √3/2 tanθ = sinθ/cosθ tanθ = 1/2/√3/2 tanθ = 1/√3

Problem 10: Simplify the expression using Pythagorean identities: sin 2 θ + cos 2 θ + 1.

By Pythagorean identity, we know that: sin 2 θ + cos 2 θ = 1 So, the value of sin 2 θ + cos 2 θ + 1 = 1 + 1 = 2

Practice Questions on Pythagorean Identities

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You can download these Practice problems and their Solution from – Pythagorean Identities Practice Problems

Pythagorean identities, which in real life connect the basic trigonometric functions, are an essential part of trigonometry. It defines all three main formulas of trigonometry. After understanding these Pythagorean identities one has a deeper understanding of trigonometric identities and will be able to solve questions. The identities and practice questions given in this article will help you to understand it in a better way.

Pythagorean Identities Trigonometric Identities Trigonometric Identities Class 10

What are pythagorean identities.

Trigonometric identities also known as Pythagorean identities, which express relationships between the sine, cosine, and tangent functions, are derived using the famous Pythagorean theorem.

They are important for deriving solutions to trigonometric equations and serve as the basic understanding for different calculus and physics ideas.

They are derived by applying the Pythagorean theorem to a right triangle and then changing the fundamental identities(by exhanging LHS and RHS side of equation) to obtain other identities.

Yes, they are important for solving a variety of calculus problems, such as those involving the differentiation and integration of trigonometric functions.

The main relationships that are directly derived from the Pythagorean theorem are known as basic identities, and these basic relationships can be changed to create derived identities.

Rules, Formula and more

The sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse .

Usually, this theorem is expressed as $$ A^2 + B^2 = C^2 $$ .

A right triangle has one $$ 90^{\circ} $$ angle ($$ \angle $$ B in the picture on the left) and a variety of often-studied formulas such as:

The Pythagorean Theorem
Trigonometry Ratios (SOHCAHTOA)
Pythagorean Theorem vs Sohcahtoa (which to use)

SOHCAHTOA only applies to right triangles ( more here ) .

The hypotenuse is the largest side in a right triangle and is always opposite the right angle.

In the triangle above, the hypotenuse is the side AB which is opposite the right angle, $$ \angle C $$.

Online tool calculates the hypotenuse (or a leg) using the Pythagorean theorem.

Below are several practice problems involving the Pythagorean theorem, you can also get more detailed lesson on how to use the Pythagorean theorem here .

Find the length of side t in the triangle on the left.

Substitute the two known sides into the Pythagorean theorem's formula : A² + B² = C²

What is the value of x in the picture on the left?

Set up the Pythagorean Theorem : 14 2 + 48 2 = x 2 2,500 = X 2

$$ x = \sqrt{2500} = 50 $$

$$ x^2 = 21^2 + 72^2 \\ x^2= 5625 \\ x = \sqrt{5625} \\ x =75 $$

Find the length of side X in the triangle on on the left?

Substitue the two known sides into the pythagorean theorem's formula : $$ A^2 + B^2 = C^2 \\ 8^2 + 6^2 = x^2 \\ x = \sqrt{100}=10 $$

What is x in the triangle on the left?

x 2 + 4 2 = 5 2 x 2 + 16 = 25 x 2 = 25 - 16 = 9 x = 3

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