## Word problems on Pythagorean Theorem

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

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

Three cases of word problems on Pythagorean Theorem :

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

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

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

Word problems using the Pythagorean Theorem:

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

⇒ 200x = 10000 + 3600

⇒ 200x = 13600

⇒ x = 13600/200

Therefore, distance between X and Y = 68 meters.

Therefore, length of each side is 8 cm.

Using the formula solve more word problems on Pythagorean Theorem.

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

In a rectangle, each angle measures 90°.

Therefore PSR is right angled at S

Using Pythagoras theorem, we get

⇒ PS = √6400

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

= 2 (150 + 80) m

= 2 (230) m

= 460 m

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

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

According to Pythagorean Theorem,

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

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

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

Draw DE ┴ AB

Then AE = AB – EB but EB = BC

Therefore AE = 34 m - 29 m = 5 m

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

⇒ AD = √169

Therefore the distance between their tops = 13 m.

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

Congruent Shapes

Congruent Line-segments

Congruent Angles

Congruent Triangles

Conditions for the Congruence of Triangles

Side Side Side Congruence

Side Angle Side Congruence

Angle Side Angle Congruence

Angle Angle Side Congruence

Right Angle Hypotenuse Side congruence

Pythagorean Theorem

Proof of Pythagorean Theorem

Converse of Pythagorean Theorem

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## Pythagorean Theorem 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 Word Problems Worksheets with Answers

The Pythagorean theorem is one of the most recognizable formulas in mathematics. Even many non-mathematics students recall that the sum of the squares of the two sides in a right triangle are equal to the square of the hypotenuse.

And while this formula may be considered basic after being introduced in the 8th grade, practicing applying it to solve math problems in the real-world is still very important! That’s why I am excited to shared this collection of Pythagorean theorem word problems worksheets with answers!

## What is Pythagorean Theorem Used For?

Pythagoras’ theorem is an example of a mathematical equation that is used to solve math problems involving right triangles. More specifically, we use the Pythagorean theorem to find the value of an unknown side length in a right triangle.

We can see the use of the Pythagorean theorem in many careers and real-world situations in everyday life . For example, carpenters, architects, and navigators apply this formula in order to find the length of the hypotenuse of a triangle found in their professions.

Whether it is a finding where to place the base of the ladder to reach a high window, or to find the total feet of the string needed to complete a triangular design, the Pythagorean theorem has a place in most sample questions you might face when working with real-world triangle problems.

## How to Use the Pythagorean Theorem

The first and most important thing I always start with when explaining Pythagoras’ theorem to my students is that it only holds true for right triangles. One of the most common mistakes that I see is students applying the formula when working with non-right triangles.

The Pythagorean theorem is an example of an algebraic formula . In order to understand how to use the Pythagorean Theorem, we first need to take a look at this famous formula:

$$a^2+b^2=c^2$$

In this formula:

- a represents the length of one of the straight lines that makes up a side length in a triangle
- b represents the length of the second side length
- c represents the length of the hypotenuse of a triangle

Note that a and b can be used interchangeably without impacting your calculation. However, it is important that the hypotenuse takes the place of c in the formula above.

Remember that the hypotenuse is the side length across from the right angle. Since the right angle is the biggest angle in a right triangle, it makes sense that the hypotenuse is also the longest side!

In words, this formula tells us that the sum of the squares of the two sides is equal to the square of the hypotenuse.

An alternate equation for the Pythagorean theorem has the length of the hypotenuse isolated:

$$c=\sqrt{a^2+b^2}$$

In words, this formula tells us that the length of the hypotenuse is equal to the square root of the sum of the squares of the two sides. This formula can be used as a quicker way of solving for the length of the hypotenuse of a triangle.

When working with math problems involving a missing side in a right triangle, we simply substitute the length of each side into either formula for either a, b, or c.

Let’s take a look at an example!

## Pythagorean Theorem Word Problem Example

A ship’s mast stands 40 metres tall in the middle of a ship. To ensure the mast’s stability, the captain decides to reinforce it with a support cable that will be attached from the top of the mast to a point on the ground that is 25 metres from the base of the mast. How long should the cable be?

Since the cable connecting the top of the mast to the point from the base of the mast forms a right triangle, we know that we can apply the Pythagorean theorem to solve for the length of the cable. Your starting point should be to draw a diagram in order to help you identify and label the sides of the right triangle.

Next, we substitute the information from the problem into the formula:

$$25^2+40^2=c^2$$

Simplifying the left side and taking the square root of both sides will allow us to solve for the missing side:

$$\begin{split}25^2+40^2&=c^2 \\ \\ 2225&=c^2 \\ \\ c&=47.17 \end{split} $$

We could also solve this problem using the alternate formula shown above:

$$\begin{split} c&=\sqrt{25^2+40^2} \\ \\ c&=47.17 \end{split} $$

Therefore, the length of the cable will be approximately 47.17 feet.

This first worksheet will focus on basic math problems involving right triangles found in everyday life. To solve these problems, keep the equation for the Pythagorean theorem handy. Your starting point should always be drawing a sketch of the right triangle to help you identify the side of the right triangle. Remember to check the answer key to ensure you fully understand!

Download the PDF worksheet by clicking below!

This second Pythagorean theorem word problems worksheet with answers will focus on more challenging math problems involving right triangles. The sample questions found on this worksheet involve problems where the hypotenuse is given, and the missing side is one of the other two side lengths.

Since you are working with 2-step equations , you will have to apply algebra to rearrange the equation in order to solve for the measurement of the unknown side length. You can watch this short video example of a problem like this for more help! Be sure to check the answer key if you are feeling stuck!

## Using the Pythagorean Theorem to Solve Word Problems

Like any other real-life word problems , the first step you always want to take is drawing a diagram. This is a key step that helps you make decisions about the side lengths you are given and the missing side length in the problem. After you have done this, it becomes much easier to determine what value to substitute in each position of the Pythagorean theorem equation!

As is the case with any math concept, the best thing you can do to learn how to apply the Pythagorean theorem is practice! Use the Pythagorean theorem word problems worksheets with answers I have provided here to help you practice solving the various math problems you can encounter involving right triangles!

Did you find these Pythagorean theorem word problems worksheets with answers helpful? Share this post and subscribe to Math By The Pixel on YouTube for more helpful mathematics content!

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## {FREE} Pythagorean Theorem Word Problems Task Cards

Probably one of the most important formulas for students to know, remember and feel confident using is the Pythagorean theorem. I always like to introduce it in a fun, hands on way to make sure kids understand the meaning before I present a formal definition. But then, kids need lots of practice. And since mindlessly solving problems for a or b or c with no context gets tedious and boring, why not practice with these Pythagorean Theorem Word Problems ?

*Please Note: This post contains affiliate links which support the work of this site. Read our full disclosure here .*

This set of problems can be used in a lot of different ways and will give your kids a context for using and applying the formula.

If Pythagorean theorem is still new for your kids, you may want to start with these introductory lessons .

These real life examples provide further context and give space for kids to draw pictures and show their thinking.

Once they feel more confident, this set of Pythagorean theorem word problems task cards can help reinforce their understanding. Plus, it provides an opportunity to practice critical thinking and problem solving !

–>Pssst! Do your students need help breaking down and solving word problems? You might like this set of editable word problem templates ! Use with any grade and any type of word problem:

## Pythagorean Theorem Word Problems Set:

This download includes a set of 6 task cards . Each card includes a word problem that requires kids to use the Pythagorean theorem to solve.

Some of the problems ask for the length of one of the sides of the given right triangle, while a couple of the problems are actually two-step problems . This means using the Pythagorean theorem is a necessary step, but doesn’t provide the final answer.

Also included is a recording page . On this page, kids have space to draw a picture (ideal for these types of problems) and show their thinking. Then there is space for them to write the final answer.

Finally, there is a teaching tips page with examples and ideas for how to use the task cards, plus an answer key , making this super easy on you, the teacher!

## To Prepare the Task Cards:

First, print the page of task cards . I suggest printing on card stock and laminating them for durability.

Cut out the cards, then print a recording page for each student.

Now you’re ready to decide how to use them!

## How to Use the Pythagorean Theorem Task Cards:

Task cards are a fun way to practice because they can be used in whole class or small group games, or as individual review.

To play as a whole class or small group, use them in a game of scoot . This means each card is at a different “station” around the room.

Kids then (probably in small groups of 2-3) make their way around the room, solving each problem. You set a timer for each station and then call out, “Scoot!” to let kids know it’s time to move to the next problem.

You could also include the entire set as a single math station . Then when kids get to the station, they work through as many task card problems as they can, recording their answers.

This could also be used with small groups who need more review or intervention with you, or as individual practice at home as homework .

However you work the problems into your math routine, I hope it helps your kids see the benefits of math in real life !

Want to grab this free set of Pythagorean theorem word problems? Just click the link below to grab them in my shop!

## {Click HERE to go to my shop and grab this set of Pythagorean Theorem Word Problem Task Cards!}

More pythagorean theorem resources:.

- Proving the Pythagorean Theorem with LEGO Bricks
- Real Life Pythagorean Theorem Practice

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## Pythagorean Theorem

How to Use The Pythagorean Theorem

## The Formula

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

## Examples of the Pythagorean Theorem

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

## Conceptual Animation of Pythagorean Theorem

Demonstration #1.

More on the Pythagorean theorem

## Demonstration #2

Video tutorial on how to use the pythagorean theorem.

## Step By Step Examples of Using the Pythagorean Theorem

Example 1 (solving for the hypotenuse).

Use the Pythagorean theorem to determine the length of X.

Identify the legs and the hypotenuse of the right triangle .

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

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

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

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

## Example 2 (solving for a Leg)

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

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

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

## Practice Problems

Find the length of X.

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

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

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

Solve for the unknown.

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

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

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

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

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

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

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

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

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

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

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## Word Problems on Pythagorean Theorem | Pythagorean Theorem Problems with Answers

Check Word Problems On Pythagorean Theorem in the below sections. Refer Pythagorean Theorem Study Material along with the solutions and steps to solve them. Get various model questions on Pythagoreans theorem and improve your mathematical knowledge along with other math skills. Know the shortcuts, tips, and tricks to solve Pythagorean Theorem Problems. Assess your preparation standard taking the help of the Pythagorean Theorem Questions available and cross-check your solutions here.

## How to Solve Problems on Pythagorean Theorem?

Follow the simple steps listed here to solve problems related to the Pythagorean Theorem. They are along the lines

Step 1: Look at all the terms in the final equation

Step 2: Find out which right triangles contain those terms

Step 3: Start with those right triangles and apply the Pythagorean Theorem

## Pythagorean Theorem Word Problems

A 35-foot ladder is leaning against the side of a building and its positioned such that the base of the ladder is 21 feet from the base of the building. Find the distance above the ground where the point of the ladder touches the building?

Let the point above the ground = x

As given in the question,

Length of the foot ladder = 35

Base of the ladder = 21 feet

Let the equation be a² + b² = c²

(21)² + (x)² = (35)²

441 + x = 1225

Subtracting by 441 on both sides

Squaring and both sides

28 feet far above the ground is the point where the ladder touches the building.

Lainey runs a string of lights from the ground straight up a door frame that is 2.5 meters tall. Then they run the rest of the string in a straight line to a point on the ground that is 6 meters from the base of the door frame. There are 10 lights per meter of a string. How many total lights are on the string?

The length of the lights = 2.5 meters

The base of the door frame = 6 meters

No of lights per meter = 10

To find the total number of lights on a string, we write the equation

h² = (2.5)² + (6)²

h²= 6.25 + 36

String Length = 2.5 + √42.25

Therefore, no of lights = 9m * 10 lights = 90 lights

Thus, the total lights on the string = 90 lights

Fencing at a hardware store costs $16.97 per yard. How much would it cost to fence in a triangular-shaped yard that has a leg of 11 feet and another leg of 60 feet?

Note: The yard is in the shape of a right triangle.

Amount of fencing at hardware store = $16.97 per yard

Length of one leg = 11 feet

Length of another leg = 60 feet

As given the yard is in the shape of right triangle

We know the equation,

a² + b² = c²

(11)² + (60)² = c²

121 + 3600 = c²

Squaring on both sides

√3721 = √c²

The amount for a fence in a triangular-shaped yard = $61

Thus, the final solution is $61

If a 34-foot ladder is placed against the top of a 30-foot building. How many feet will be at the bottom of the ladder from the bottom of the building?

Length of the ladder = 34 foot

Length of the building = 30 foot

Let the length of bottom of the ladder = x

(30)² + x² = (34)²

900 + x² = 1156

Add square root on both sides

x = 16 feets

Therefore, the bottom of the ladder from the bottom of the building will be 16 feets

Thus, the final solution is 16 feet

A certain television is advertised as a 34-inch TV. If the width of the TV is 30 inches, how many inches tall is the TV?

Size of the TV = 34 inch

Width of the TV = 30 inch

Length of TV = x

To know the length of TV, we have to find the equation

√x² = √1156

TV is 34 inches tall.

Thus, the final solution is 34 inches

Ramu starts driving north for 9 miles, then takes a right turn, and then he drives east for another 40 miles. At the end of driving, what is the distance of a straight line from the starting point?

Driving for north = 9 miles

Driving for east =40 miles

Let the straight line distance = x

The equation will be

(9)² + (40)² = c²

81 + 1600 = c²

Add square roots on both sides

√1681 = √c²

Therefore, the straight line distance from the starting point is 41 miles

Thus, the final solution is 41 miles

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

Given: A right-angle triangle in which one of the sides is 90°

To prove: AC² = AB² + BC²

Construction: Draw BD ⊥ AC

In Δ’s ADB and ABC

∠ADB = ∠ABC (Each 90°)

∠A = ∠A (Common)

ΔADB ∼ ΔABC (By A-A Criteria)

AD/AB = AB/AC

AB² = AD * AC is the first equation

In Δ’s BDC and ABC

∠CDB = ∠ABC (Each 90°)

∠C = ∠C (Common)

ΔBDC ∼ ΔABC (By A-A Criteria)

DC/BC = BC/AC

BC² = AC * DC is the (2) equation

AB² = AD * AC is the (3) equation

Adding the equations (1) and (2)

AB² + BC² = AD * AC + AC * AD

AB² + BC² = AC (AD + DC)

AB² + BC² = AC (AC)

AB² + BC² = AC²

∴ Hence proved

In the triangle, if the square of one side of the triangle is equal to the sum of the squares of another two sides of the triangle, then the angle that is opposite to the first side is a right angle triangle?

Given: ΔABC such that AB² + BC² = AC²

To prove: ∠B = 90°

Construction: Draw ΔDEF such that DE = AB, EF = BC and ∠E = 90°

Since ΔDEF is right-angled,

By Pythagoras theorem

DE² + EF² = DF²

AB² + BC² = DF² (By const DE = AB and EF = BC)

AC² = DF² (Given AB² + BC² = AC²)

In Δs ABC and DEF

AB = DE (By const)

BC = EF (By const)

AC = DF (Proved)

ΔABC ≅ ΔDEF (SSS)

∠B = ∠E = 90°

ΔABC is a right triangle

Mary wants to cut across a rectangular lot rather than walk around it. Of the lot is 120 feet long and 50 feet wide. Mary walks diagonally across the lot, how many feet is the short cut?

Length of the lot = 120 feet

Width of the lot = 50 feet

Let the shortcut distance be x

As per the Pythagorean theorem,

(50)² + (120)² = x²

2500 + 14,400 = x²

16,900 = x²

Therefore, Mary walks 130 feet shortcut

Thus, the final solution is 130 feet

Problem 10:

The length of a living room is 2 feet less than twice its width. If the diagonal is 2 feet more than twice the width, find the dimensions of the room?

Let the width of the living room = x

Length of the living room = 2 feet less than twice its width

Diagonal = 2 feet more than twice the width

From the given equations,

Width represents x² + (2x – 2)² = (2x + 2)²

Length represents x² = (2x + 2)² – (2x-2)²

From both the equations,

x² = (4x + 8x + 4) – (4x – 8x + 4)

(x²-16x) = 0

x(x-16) = 0

x = 0, x=16

Therefore, the width = 16

Length = (2x – 2) = 2(16) – 2 = 30

Thus, the length of the living room = 30 feet

Width = 16 feet

Hence, the final solution is l = 30 feet and w = 16 feet

Problem 11:

A man goes 12m east and 9m north. Find the distance from the initial point?

The distance man goes to east = 12m

i.e., BC = 12m

The distance man goes to north = 9m

i.e., AB = 9m

To find the initial point (AC), apply the Pythagorean theorem

AC² = AB² + BC²

AC² = (9)² + (12)²

AC² = 81 + 144

Therefore, he is 15m far from his initial point

Thus, the final solution is 15m

Problem 12:

Hari wants to hang a 7m long banner from the roof of her shop. The hooks for the strings are 10m apart. Hari wants the top of the banner to hang 1m below the roof. How long should each of the strings be?

As per the question,

Length of the long banner = 7m

Hooks for strings = 10m apart

Length where the banner to hang below the roof = 1m

Let the length of the string = x

Hence, the equation will be

7 + 2x = 10

Now, apply the Pythagorean theorem,

C² = (1)² + (1.5)²

C² = 1 + 2.25

Hence, each string should be 1.80m

Thus, the final solution is 1.80m

Problem 13:

Two kids are flying a kite with a string of 50 meters long. If the kids are 35 meters apart, how high is the kite off the ground?

Length of the string = 50 meters

The distance of kids apart = 35 meters

The height of the kite off the ground = b

Applying the Pythagorean theorem, we get

(35)² + b² = (50)²

1225 + b² = 2500

Therefore, 35.70m is the kite off the ground

Thus, the final solution is 35.70 m

Problem 14:

A carpenter needs to add 2 braces to a barn door. If the door measures 12 * 16 feet, how much wood will he need for both braces?

No of braces = 2

Length of the door = 12 feet

The breadth of the door = 16 feet

To find the amount of wood for braces, we apply Pythagorean theorem

(12)² + (16)² = c²

144 + 256 = c²

Amount of wood for both the braces = 2 * 20 = 40ft

Therefore, the carpenter requires 40 ft of wood for both the braces.

Thus, the final solution is 40 ft

Problem 15:

M and N are points on the sides RP and RQ respectively of triangle PQR right-angled at R. Prove that PN² + QM² = PQ² + MN²?

PN² = PR² + RN² is the (1) equation

QM² = MR² + RQ² is the (2) equation

Adding the equation (1) and (2)

PN² + QM² = MR² + RQ² + PR² + RN²

PN² + QM² = (MR² + RN²) + (RQ² + PR²)

PN² + QM² = MN² + PQ²

∴ Hence, it is proved

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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?

How to answer pythagorean theorem questions, pythagorean theorem in real life, pythagorean theorem in 8th grade, pythagorean theorem practice problems.

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

Wish you could have the 15 Pythagoras Theorem questions from this blog in a ready-to-go worksheet? We've done just that!

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

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.

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

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

Do you have students who need extra support in math? Give your students more opportunities to consolidate learning and practice skills through personalized math tutoring with their own dedicated online math tutor. Each student receives differentiated instruction designed to close their individual learning gaps, and scaffolded learning ensures every student learns at the right pace. Lessons are aligned with your state’s standards and assessments, plus you’ll receive regular reports every step of the way. Personalized one-on-one math tutoring programs are available for: – 2nd grade tutoring – 3rd grade tutoring – 4th grade tutoring – 5th grade tutoring – 6th grade tutoring – 7th grade tutoring – 8th grade tutoring Why not learn more about how it works ?

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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- How do you solve word problems?
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- How do you identify word problems in math?
- Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.
- Is there a calculator that can solve word problems?
- Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems.
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- An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age.

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## How to Solve the Converse of Pythagoras’ Theorem Problems?

The converse of Pythagoras' Theorem is the inverse of Pythagoras's theorem and helps distinguish acute, right, or obtuse triangles. In this guide, you can learn the converse of Pythagoras’ theorem.

The converse of the Pythagoras theorem helps determine whether a triangle is a right triangle or not. While a Pythagorean theorem helps determine the length of the missing side of a right triangle.

## Related Topics

- How to Solve Pythagorean Theorem Problems?

## A step-by-step guide to the converse of Pythagoras’s theorem

The converse of Pythagoras’ theorem states that if the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, that triangle is right-angled. The converse is the complete reverse of the Pythagorean theorem.

The primary use of the converse of the Pythagorean theorem is that measurements help determine the type of triangle – right, acute, or obtuse. Once the triangle is identified, it becomes easy to make that triangle. Three things happen:

- If the sum of the squares of the two sides of the triangle is equal to the square of the hypotenuse, the triangle is right-angled.
- Whether the sum of the squares of the two sides of a triangle is less than the square of the hypotenuse, the triangle is obtuse.
- If the sum of the squares of the two sides of a triangle is greater than the square of the hypotenuse, the triangle is acute.

## Pythagoras Theorem

Pythagoras’s theorem states that if a triangle is right-angled (\(90\) degrees), the square of the hypotenuse is equal to the sum of the squares of the other two sides. In the given triangle \(ABC\), we have \(BC^2 = AB^2 + AC^2\). Here, \(AB\) is the base, \(AC\) is the altitude or the height, and \(BC\) is the hypotenuse. In other words, we can say, in a right triangle, \(\color{blue}{\left(Opposite\right)^2+\:\left(Adjacent\right)^2=\:\left(Hypotenuse\right)^2}\).

The converse of Pythagoras’ theorem formula

The converse of Pythagoras theorem formula is \(\color{blue}{c^2=a^2 + b^2}\), where \(a, b,\) and \(c\) are the sides of the triangle.

## The Converse of Pythagoras’s Theorem – Example 1:

The side of the triangle is of lengths \(8\) units, \(10\) units, and \(6\) units. Is this triangle a right triangle?

Using the converse of Pythagoras’ theorem, we obtain,

\((10)^2 = (8)^2 + (6)^2\)

\(100 = 64 + 36\)

\(100=100\)

Since both sides are equal, the triangle is right-angled.

## Exercises for the Converse of Pythagoras’s Theorem

- The sides of a triangle are \(7, 11\), and \(13\). Check whether the given triangle is a right triangle or not.
- Determine whether a triangle with sides \(3 cm\), \(5 cm\), and \(7 cm\) is an acute, right or obtuse triangle.
- Classify a triangle whose side lengths are given as;\(11 in, 13 in\), and \(17 in\).
- \(\color{blue}{Not}\)
- \(\color{blue}{Obtuse\:triangle}\)
- \(\color{blue}{Acute\:triangle}\)

by: Effortless Math Team about 2 years ago (category: Articles )

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Course: 8th grade > Unit 5 Pythagorean theorem word problems Google Classroom You might need: Calculator Steve is turning half of his backyard into a chicken pen. His backyard is a 24 meter by 45 meter rectangle. He wants to put a chicken wire fence that stretches diagonally from one corner to the opposite corner.

Step 1: Analyzing the Problem The first step in solving any word problem is understanding what the problem is asking. Translate the problem into an understandable format and identify the elements related to the Pythagorean Theorem. Step 2: Sketching the Problem Visualizing the problem can be extremely helpful.

How to Solve Pythagorean Theorem Word Problems: Example 1 Find the length of the diagonal of the following rectangle. From the figure, we see that the length of the rectangle is 12 inches,...

1.1K Share Save 307K views 15 years ago Geometry Video Playlist from MathHelp.com For a complete lesson on Pythagorean Theorem word problems, go to https://www.MathHelp.com - 1000+ online math...

0:00 / 1:25 Pythagorean Theorem Word Problems | MathHelp.com MathHelp.com 342K subscribers Subscribe 173 28K views 7 years ago Need a custom math course? Visit https://www.MathHelp.com . This...

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.

Barry Desborough 10 years ago Good question! Draw a right-angled triangle, with a square drawn on each of the sides. The areas of the two smaller squares, added together, is equal to the area of the largest square. Now draw two more triangles.

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.

No cable box. No problems. Get your free lessons: https://vividmath.comUse Pythagoras' Theorem to find the Hypotenuse of a Right Angled Triangle (Word Problem).See all Pythagoras' Theo...

The side opposite the right angle, or the 90 degrees, is a hypotenuse, or the longest side. It is the square root of 74. And the shorter sides are w and 7. And the Pythagorean Theorem tells us that the sum of the squares of the shorter side will be equal to the square of the hypotenuse, so the square of the longer side.

How to Solve a Word Problem Involving the Pythagorean Theorem Step 1: Use the information given in the problem to create a right triangle representing the scenario described. Include the...

c = a2 + b2− −−−−−√ In words, this formula tells us that the length of the hypotenuse is equal to the square root of the sum of the squares of the two sides. This formula can be used as a quicker way of solving for the length of the hypotenuse of a triangle.

Here are eight (8) Pythagorean Theorem problems for you to solve. You might need to find either the leg or the hypotenuse of the right triangle. These problems vary in type and difficulty, providing you an opportunity to level up your skills. ... There are eight (8) problems here about the Pythagorean Theorem for you to work on. When you do ...

Pythagorean Theorem Word Problems Set: This download includes a set of 6 task cards. Each card includes a word problem that requires kids to use the Pythagorean theorem to solve. Some of the problems ask for the length of one of the sides of the given right triangle, while a couple of the problems are actually two-step problems.

Step 1 Identify the legs and the hypotenuse of the right triangle . The legs have length 6 and 8. X X is the hypotenuse because it is opposite the right angle. Step 2 Substitute values into the formula (remember 'C' is the hypotenuse). A2 + B2 = C2 62 + 82 = X2 A 2 + B 2 = C 2 6 2 + 8 2 = X 2 Step 3 Solve for the unknown.

Problem 1: A 35-foot ladder is leaning against the side of a building and its positioned such that the base of the ladder is 21 feet from the base of the building. Find the distance above the ground where the point of the ladder touches the building? Solution: Let the point above the ground = x As given in the question,

The Pythagorean theorem describes a special relationship between the sides of a right triangle. Even the ancients knew of this relationship. ... Pythagorean theorem word problem: fishing boat (Opens a modal) Pythagorean theorem in 3D (Opens a modal) Practice. Use Pythagorean theorem to find perimeter. 4 questions.

This video shows how to solve 2 different word problems using the Pythagorean Theorem.

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>. a^2+b^2=c^2 a2 +b2 = c2. 3 - Calculate the answer.

How do you apply the pythagorean theorem in word problems? Algebra Radicals and Geometry Connections Pythagorean Theorem and its Converse 1 Answer Sally Jan 15, 2015 So when you have word problems, the most important thing to understand how to convert the words to mathematical models.

An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age. Show more

Find the height of a parallelogram using the Pythagorean theoremOn MooMooMath we provide helpful Math and Geometry videos to help you figure out how to solve...

The converse is the complete reverse of the Pythagorean theorem. The primary use of the converse of the Pythagorean theorem is that measurements help determine the type of triangle - right, acute, or obtuse. Once the triangle is identified, it becomes easy to make that triangle. Three things happen: