pythagoras problem solving lesson

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

Module 11: Geometry

Using the pythagorean theorem to solve problems, learning outcomes.

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

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

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

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

The Pythagorean Theorem

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

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

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

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

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

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

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

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

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

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

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

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

Pythagorean Theorem

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

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

Common Pythagorean Triples

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

k can also be imaginary.

Introductory

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

Sample Problem

Solution 1 (Bash)

Solution 2 (Using 3-4-5)

Another Problem

Solution (Casework)

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

3 is a leg and 4 is the hypotenuse.

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

External links

• 122 proofs of the Pythagorean Theorem

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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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15 Pythagorean Theorem Practice Problems For 8th Grade

Beki Christian

Pythagorean Theorem practice problems involve using the relationship between the sides of a right triangle to calculate missing side lengths in triangles. The Pythagorean Theorem is introduced in 8th grade and is used to solve a variety of problems across high school.

Here, you’ll find a selection of Pythagorean Theorem questions that demonstrate the different types of questions students are likely to encounter in 8th grade.

What is the Pythagorean Theorem?

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

This can be written as a^2+b^2=c^2 for a triangle labeled like this:

15 Pythagoras Theorem Practice Problems

Find all of the Pythagoras Theorem questions included in this blog in this easy to download resource. Suitable for 8th grade students. Includes answer key.

How to answer Pythagorean Theorem questions

1 – Label the sides of the triangle a , b , and c . Note that the hypotenuse, the longest side of a right triangle, is opposite the right angle and will always be labeled .

2 – Write down the formula and substitute the values>

3 – Calculate the answer. You may be asked to give your answer in an exact form or round to a given degree of accuracy, such as a certain number of decimal places or significant figures.

Pythagorean Theorem in real life

Pythagorean Theorem has many real-life uses, including in architecture and construction, navigation and surveying.

Pythagorean Theorem in 8th grade

Pythagorean Theorem is usually introduced in middle school, as it is a part of the 8th grade Common Core Math Standards.

The emphasis in middle school is on students being able to:

• Explain the Pythagorean Theorem;
• Use the theorem to solve mathematical and real-world problems – with both 2D and 3D figures;
• Use the theorem to calculate the distance between two points on a coordinate grid.

The process for solving any Pythagoras Theorem problem always begins by identifying the relevant right-angled triangle and labeling the sides a , b , c. If there is not a diagram in the question, it can be helpful to draw one.

Where necessary, round your answers to 3 significant figures.

Pythagorean Theorem practice problems

1. A ship sails 6 \, km East and then 8 \, km North. Find the ship’s distance from its starting point.

The ship is 10 kilometers from its starting point.

2. A ladder is 5 \, m long. The base of the ladder is 3 \, m from the base of a vertical wall. How far up the wall does the ladder reach?

The ladder reaches 4 meters up the wall.

3. Alex and Sam start from the same point. Alex walks 400 meters west. Sam walks x meters south, until they are 600 \, m apart from each other. How far does Sam walk?

4. A television’s size is the measurement from the upper left hand corner of the television to the bottom right hand corner. Find the size of this television.

39.7 inches

55.1 inches

5. The pole of a sailing boat is supported by a rope from the top of the pole to an anchor point on the deck. The pole is 4 \, m long and the rope is 4.5 \, m long. Calculate the distance from the base of the pole to the anchor point of the rope on the deck.

6. Work out the length of the diagonal of a square with 8 \, cm sides.

The diagonal of the square has a length of 11.3 centimeters.

7. ABC is an isosceles triangle.

Work out the height of the triangle.

8. ABCD is an isosceles trapezoid.

Work out the length of AD.

9. Here is a cm square grid. Calculate the distance between the points A and B.

10. Which is a right angled triangle?

Not a right angled triangle because Pythagorean Theorem doesn’t work.

Right angled triangle because Pythagorean Theorem works.

11. PQRS is made from two right angled triangles.

Work out the length of QR.

Triangle \text{PQS:}

Triangle \text{QRS}

12. Here is a pattern made from right angled triangles. Work out the length x.

Triangle \text{ABC:}

Triangle \text{ACD:}

13. Here is a pyramid.

Work out the height of the pyramid.

14. Here is a cuboid.

Work out the length AG.

Give your answer in its exact form.

Length of \text{BG:}

Length of \text{AG:}

15. Here is a right angled triangle.

Form an equation and use it to work out the value of x.

x=4 \, or \, x=12

x cannot be 4 as you can’t have a negative side length so x=12

Pythagorean Theorem in middle school

In middle school, students…

• prove the Pythagorean Theorem;
• use the Pythagorean Theorem with trigonometric ratios to solve problems;
• use the Pythagorean Theorem in proofs.

Pythagoras Theorem may feature in questions alongside other topics, such as trigonometry, circle theorems or algebra.

The Pythagorean Theorem is used to calculate a missing length in a right triangle . If you have a right angled triangle and you know two of the lengths, label the sides of the triangle a,b and c (c must be the hypotenuse – the longest side). Pythagorean Theorem is a^2+b^2=c^2. Substitute the values you know into Pythagorean Theorem and solve to find the missing side.

The hypotenuse of a right triangle is the longest side. If you know the lengths of the other two sides, you can find the length of the hypotenuse by squaring the two shorter sides, adding those values together and then taking the square root. By doing this you are finding c in a^2+b^2=c^2

If your triangle is a right triangle and you know two of the sides, you can use Pythagorean Theorem to find the length of the third side. To do this, label the sides a , b and c (with c being the hypotenuse – the longest side). Substitute the values you know into a^2+b^2=c^2 and solve to find the missing side.

Looking for more Pythagorean theorem math questions?

• Ratio questions
• Algebra questions
• Probability questions
• Trigonometry questions
• Venn diagram questions
• Long division questions

RELATED RESOURCE :

• 6th Grade Math Problems
• 7th Grade Math Problems

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The content in this article was originally written by former UK Secondary teacher Beki Christian and has since been revised and adapted for US schools by elementary and middle school teacher Kathleen Epperson.

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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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Pythagorean Theorem Word Problems

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

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

How To Solve Word Problems Using The Pythagorean Theorem?

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

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

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

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

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

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

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

Examples Of Real Life Pythagorean Theorem Word Problems

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

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

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

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

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

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

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

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

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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.

Pythagoras' Theorem

Starts at the very beginning with using a calculator. May need editing depending on which calculators you use. Main activity differentiated and answers included. Includes 3D Pythagoras' theorem problems.

Problem solving with similarity and Pythagoras' theorem

I can use my knowledge of similarity and Pythagoras' theorem to solve problems.

Lesson details

Key learning points.

• Right-angled triangles can be seen in real-life (e.g. ladder against a vertical wall).
• A ratio table can help you find the scalar and functional multipliers in similar shapes.
• It can be initially difficult to identify whether Pythagoras' theorem can be used.

Common misconception

Every question that has a right-angled triangle must use Pythagoras' theorem to be solved.

It is easy to get into a habit of using Pythagoras' theorem when learning the topic, but it is likely you have seen several maths problems in the past with right-angled triangles, which ask to find areas and angles, without using Pythagoras' theorem.

Pythagoras’ theorem - Pythagoras’ theorem states that the sum of the squares of the two shorter sides of a right-angled triangle is equal to the square of the hypotenuse.

This content is © Oak National Academy Limited ( 2024 ), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

Starter quiz

6 questions.

Resourceaholic

Ideas and resources for teaching secondary school mathematics

• Blog Archive

2 September 2014

Pythagoras' theorem.

11 comments:

Great post, some lovely questions and resources here I've not seen before. On the subject of triples, you may find this interesting: http://wp.me/p2z9Lp-bP And my very first ever (albeit not great) blog post was about Pythagoras in 3d! http://wp.me/p2z9Lp-9

Thanks Cav. Aah, your first post from 2012! I bet that feels like a long time ago. Love your triples post, I'll definitely use that method for finding triples. Also the proof by induction for FP1, great stuff.

Here’s a lesson I did when I visited a teacher with a year 8 class in a middle school in the summer term. She asked me if I could do a lesson which tied in some assorted revision and practice while leading the way towards Pythagoras – without spoiling the fun for their teachers in secondary school. Phew! I asked for five minutes to think about it and this is what we did. 1…..I asked the children to construct a rectangle as perfect as they could make it. (Lots of revision of basic 2-D shapes and their properties, vocabulary, and of course accurate use of instruments.) 2…..I then asked them to draw a diagonal of their triangle and rub out one of the two triangles. 3…..I asked them to construct an equilateral triangle on each side of their triangle and to find the area of each of the three triangles. (So there was lots more practice in constructing triangles and using equipment, not to mention finding an assortment of methods to find the areas.) 4…..Finally, I asked them to post their three answers on the board and see what they could observe. I was pretty pleased that I'd managed to come up with the whole lesson quickly from scratch. The word Pythagoras was never mentioned, but I rather hoped the lesson would have to come mind when they met one of the standard diagrams in a few months’ time. Of course, constructing semi-circles on the sides would have worked just as well, but I thought there’d be more mileage in using equilateral triangles.

Thanks for sharing this lesson Alan, it's fantastic. I love the use of accurate constructions, a skill that's always worth practising. I'm certainly going to borrow these ideas.

Two animations that might be of interest to readers of this post: http://www.mathimation.co.uk/triangles/pythagoras-theorem-proof/ and http://burymathstutor.co.uk/Pythagoras.html

Interesting extensions of Pythagoras theorem. https://www.youtube.com/watch?v=YRdKI71tx-4 Bet you didn't know this about Pythagoras. https://www.youtube.com/watch?v=li8g0FMD3wc

Lovely, thank you!

Hi Jo, Just popping by for some inspiration! So many great ideas for my year 9s! A great activity I've done for Pythagoras is giving each half of the class a rope with 13 knots in at equal distances and then ask them to make me a right angled triangle. It works as a great little race for students to start problem solving. Dan

Thanks Dan, fantastic idea!

I'm not maths teacher (or a physics teacher) but a former engineer turned trade union official. I still love maths and physics though and during my attempt to get a decent understanding of special relativity I came across this very simple way of understanding special relativity and time dilatation using only Pythagoras. There is, of course, more complex maths involved in relativity, but this use of only Pythagoras is brilliant and could be used with older students or 6th formers to get a conceptual understanding of an important physics concept. I wish someone had shown this to me when I was at school as I may have studied physics instead of going to Uni to chemistry and dropping out. Is worth checking out http://www.emc2-explained.info/The-Light-Clock/#.VgBs6ZcYNOI

Great post and ideas, thanks. Been inspired to post a link to an interactive Pythagoras tool of our own: http://www.mathelize.co.uk/pt.html

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8th grade math - Pythagorean Theorem

Clip 1/5: lesson pythagorean theorem part 1.

Patty introduces her lesson by charging students to identify the “big ideas” they should be thinking about when they work with right triangles. Students pair-share their ideas, and Patty notes when they are making reference to available tools and supports, such as anchor charts, around the room. In her commentary, Patty notes that this lesson is intended to develop students’ capacity to engage in modeling mathematical situations. Students identify the Pythagorean Theorem, and Patty prompts them to attend to precision and communicate precisely. In a whole-group sharing, she engages all students to add on to, critique, extend, and clarify each other’s thinking. Students deepen their capacity to make sense of the problem or situation. Patty presents student work from a previous assessment and asks students to critique the person’s strategies and precision, giving advice to each exemplar learner about how to improve their approach.

Teacher Commentary

Some kids were really good with the geometry we had been doing. They were really good with all these procedures of writing equations, but they were still struggling with making sense of it. I like having it integrated, so students can shine in different ways. I was still making sense of a lot of it myself, too. The day before this lesson, students had a choice of two problems on comparing costs of gym memberships and rate plans. And they actually brainstormed very low-level questions like, "What is the constant rate of change?" But we got the high-end questions, like, "When will the cost be the same at the same time?" They chose which one they wanted to work on, and then they were going to be working on that. I wanted to address "modeling the situation," to help them solve that problem. I just think that that knowing what they know and don’t know makes every decision for me. I was excited to see what they could do, based on what their experiences were before. Then I really wanted them to write advice to their peers about an upcoming assessment. Like, "What do you do? What do you do if you get stuck? What kind of strategies can you do?" I found that many students understood the relationship and could verbalize and relate it to the areas of the land. When the length of the legs were given, students could use the Pythagorean Theorem. But when a hypotenuse and a leg were given, there was some difficulty finding out the length of the other leg. There was some confusion with the relationship. A lot of the language, you could still hear in their voice, they were struggling with being super precise all the time. I know they referred to the conceptual understanding, which I think supported them. They looked at the anchor poster, and they were able to make connections to the story of the area of land, and the relationship between the two areas and the one area.

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Pythagoras 2D Problem Solving PPT (Grade C-B).

Subject: Whole school

Age range: 14-16

Resource type: Worksheet/Activity

Last updated

21 March 2021

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This is a PowerPoint that I made for a Year7+8 Gifted and Talented Group on 2D Problem Solving involving Pythagoras. I made this ppt. to accompany a booklet that I had produced with aid from the CGP Higher Maths Revision Workbook that could be completed during one of their sessions. I have attached one version of the ppt. with fully annotated solutions as well as one that is completely blank with just the questions with no answers. Part 2 of Q15 is actually a trigonometry problem that I added in to add an extra challenge for the excelling students! This took an unbelievably long time to produce so I hope it serves its purpose well and feedback would be deeply appreciated.

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