pythagoras and trigonometry problem solving questions

Pythagorean Theorem Exercises

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.

Note: drawings not to scale

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

[latex]10[/latex]

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

[latex]4\sqrt 7 [/latex]

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

[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.

[latex]\sqrt 6[/latex]

You might also like these tutorials:

• Pythagorean Theorem
• Pythagorean Triples
• Generating Pythagorean Triples

• Maths Questions

Pythagoras Theorem Questions

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

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

Click here to learn the proof of Pythagoras’ Theorem .

Video Lesson on Pythagoras Theorem

Pythagoras Theorem Questions with Solutions

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

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

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

Then, by the Pythagoras theorem,

AC 2 = AB 2 + BC 2

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

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

∴ AC = √157 cm.

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

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

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

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

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

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

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

and m 2 – 1 = 9 – 1 = 8

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

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

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

By Pythagoras theorem,

⇒ AC 2 = 2AB 2

⇒ AC = √2 AB

⇒ AB = (1/√2) AC = (1/√2)6 = 3√2 cm.

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

Let AB be the ladder of length x.

AC 2 + BC 2 = AB 2

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

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

⇒ x = 17 cm

∴ Length of the ladder is 17 cm.

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

Let the rectangle be ABCD

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

⇒ AD = √(21025 – 20736) = √289

⇒ AD = 17 cm

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

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

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

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

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

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

⇒ OB = √(576 + 100) = √676

⇒ OB = 26 km.

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

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

In right triangle DCE, by Pythagoras theorem,

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

⇒ x = √(841 – 441) = √400

⇒ x = 20 m.

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

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

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

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

Then, AM = MB = 5 cm.

Now, in right triangle OMB,

OB 2 = OM 2 + MB 2

⇒ OM = √(OB 2 – MB 2 )

⇒ OM = √(13 2 – 5 2 ) = √(169 – 25)

⇒ OM = √144 = 12 cm

Area of triangle OAB = ½ × AB × OM

= ½ × 10 × 12

= 60 cm 2 .

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

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

We have to find the value of PT.

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

Thus, triangle OTP is a right-angled triangle.

∴ by the Pythagoras theorem,

OP 2 = OT 2. + PT 2

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

⇒ PT = √(100 – 36) = √64

⇒ PT = 8 cm.

Related Video on Pythagorean Triples

Practice Questions on Pythagoras Theorem

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

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

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

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

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

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

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Pythagoras' Theorem and Trigonometry - Short Problems

A collection of short problems on Pythagoras's Theorem and Trigonometry.

Pythagoras Theorem Questions (with Answers)

Pythagoras theorem.

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

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

Right Triangle Questions – using the theorem

The Theorem helps us in:

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

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

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

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

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

Given: a = 3, c = 5

Which side is the hypotenuse?

✩ Always identify the hypotenuse first

Unknown side = BC = b ?

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

3 2 + b 2 = 5 2

9 + b 2 = 25

b 2 = 25 − 9 = 16 = 4 2

Finding the Hypotenuse of a Triangle

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

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

Pythagoras Formula: a 2 + b 2 = c 2

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

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

Putting values in the formula:

20 2 + 21 2 = c 2

400 + 441 = c 2

Finding Right Triangle

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

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

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

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

c 2 = 15 2 = 225

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

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

a 2 + b 2 ≠ c 2

So this is not a right-angled triangle.

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

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

c 2 = 17 2 = 289

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

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

So a 2 + b 2 = c 2

It is a right-angled triangle.

Pythagoras Questions Types

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

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

Pythagoras Questions

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

1 Question

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

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

Ans.   AC = 10

2 Question

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

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

Ans.   Each side is 3√2 cm.

3 Question

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

Can you directly apply the Pythagoras Theorem?

Ans.   3 cm

4 Question

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

CE = 2 61 ​

Let AE = EB = x

BD = DC = y

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

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

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

Ans.   AC = 26

5 Question

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

Can you apply the Pythagoras Theorem directly?

Ans.   4 cm

6 Question

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

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

Ans.   9 cm, 40 cm, 41 cm

7 Question

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

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

Ans.   12.5 cm

8 Question

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

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

Ans.   Not a right triangle.

9 Question

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

Ans.   7 cm

10 Question

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

Can you apply the Pythagoras Theorem to the triangle BCD?

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

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

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

Answers to Pythagoras Questions

1 answer.

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

AB = a = 2√5

BC is twice of AB, b = 2a = 4√5

AC = Hypotenuse = c

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

(2√5) 2 + (4√5) 2 = c 2

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

c 2 = 20 + 80 = 100

2 Answer

Let AB = a , BC = b

In triangle ABC, base = b and altitude = a

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

2 1 ​ × ( ab ) = 9

ab = 18 (Equation 1)

Using the Pythagoras Formula:

a 2 + b 2 = 6 2 = 36

We add and subtract 2ab to complete the square :

a 2 + b 2 − 2ab + 2ab = 36

(a − b) 2 + 2ab = 36

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

(a − b) 2 = 0

Substituting a by b in Equation 1:

Side AB = BC = 3√2 cm

3 Answer

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

a = 4 10 ​ cm

b = unknown

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

( 4 10 ​ ) 2 + b 2 = 1 3 2

(16 × 10) + b 2 = 169

160 + b 2 = 169

b 2 = 169 − 160 = 9

4 Answer

Using triangle ABD:

AB 2 + BD 2 = AD 2

( 2 x ) 2 + y 2 = ( 601 ​ ) 2

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

Using triangle EBC :

EB 2 + BC 2 = EC 2

x 2 + ( 2 y ) 2 = ( 2 61 ​ ) 2

x 2 + 4 y 2 = ( 2 61 ​ ) 2

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

Adding Equation 1 and 2:

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

5x 2 + 5y 2 = 845

x 2 + y 2 = 169 (Equation 3)

Let us solve for the hypotenuse using the triangle ABC:

AB 2 + BC 2 = AC 2

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

4x 2 + 4y 2 = AC 2

4(x 2 + y 2 ) = AC 2

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

4(169) = AC 2

A C = 4 ( 169 ) ​

AC = 2 × 13 = 26

5 Answer

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

Hypotenuse is the longest side. So c = 8 .

Let b = 4√3 .

From the Pythagoras Theorem:

a 2 + b 2 = c 2

a 2 + (4√3) 2 = 8 2

a 2 + 16(3) = 64

a 2 + 48 = 64

The third side is 4 cm.

6 Answer

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

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

Applying the Pythagoras formula:

(b − 31) 2 + b 2 = (b + 1) 2

b 2 − 62b + 31 2 + b 2 = b 2 + 2b + 1

b 2 − 62b + 961 = 2b + 1

b 2 − 64b + 960 = 0

b 2 − 24b − 40b + 960 = 0

b(b − 24) − 40(b − 24) = 0

(b − 40)(b − 24) = 0

b = 40 Or b = 24

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

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

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

7 Answer

Side BC = b = 10 cm

Perimeter = Sum of the sides

= a + b + c = 30 (Given)

a + 10 + c = 30

a = 20 − c ( Equation 1 )

Applying the Pythagoras Theorem to find the hypotenuse:

Using Equation 1 to substitute the value of a

(20 − c) 2 + (10) 2 = c 2

400 − 40c + c 2 + 100 = c 2

500 − 40c = 0

The length of hypotenuse = 12.5 cm

8 Answer

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

So c 2 = 12 2 = 144

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

We can assume any side to be a or b.

Let a = 5 cm, b = 9 cm.

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

So a 2 + b 2 ≠ c 2

This is a not a right angled triangle.

9 Answer

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

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

Other sides are equal. So a = b .

Applying the Pythagoras theorem:

b 2 + b 2 = (7√2) 2

2b 2 = 49(2)

Each side is 7 cm.

10 Answer

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

Given B D = 9 10 ​ D C = 27 10 ​

Applying Pythagoras Theorem to triangle BCD:

BD 2 + DC 2 = BC 2

( 9 10 ​ ) 2 + ( 27 10 ​ ) 2 = B C 2

(9 2 × 10) + (27 2 × 10) = b 2

810 + 7290 = b 2

Applying Pythagoras Theorem to triangle ABD:

BD 2 + AD 2 = AB 2

a 2 = ( 9 10 ​ ) 2 + x 2 (Equation 1)

From the figure:

AC = AD + BD

c = x + 27 10 ​ (Equation 2)

Applying Pythagoras Theorem to triangle ABC:

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

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

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

810 + 8100 = 54 10 ​ x + 7290

54 10 ​ x = 8910 − 7290 = 1620

x = 10 ​ 30 ​

Putting value of x in Equation 1:

a 2 = ( 9 10 ​ ) 2 + ( 3 10 ​ ) 2

a 2 = 810 + 90 = 900

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

30 2 + 90 2 = c 2

900 + 8100 = c 2

c = 30 10 ​

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

Difficult Pythagoras Questions (Year 10, Guided Answers) ➤

James Garfield Pythagorean Theorem (Illustration & Proof) ➤

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15 Pythagoras Theorem Questions And Practice Problems (KS3 & KS4)

Beki Christian

Pythagoras Theorem questions involve using the relationship between the sides of a right angled triangle to work out missing side lengths in triangles. Pythagoras Theorem is usually introduced towards the end of KS3 and is used to solve a variety of problems across KS4.

Here, you’ll find a selection of Pythagoras Theorem questions that demonstrate the different types of questions you are likely to encounter in KS3 and KS4, including several GCSE exam-style questions.

GCSE MATHS 2024: STAY UP TO DATE Join our email list to stay up to date with the latest news, revision lists and resources for GCSE maths 2024. We’re analysing each paper during the course of the 2024 GCSEs in order to identify the key topic areas to focus on for your revision. Thursday 16th May 2024: GCSE Maths Paper 1 2024 Analysis & Revision Topic List Monday 3rd June 2024: GCSE Maths Paper 2 2024 Analysis & Revision Topic List Monday 10th June 2024: GCSE Maths Paper 3 2024 Analysis GCSE 2024 dates GCSE 2024 results (when published) GCSE results 2023

What is Pythagoras Theorem?

Pythagoras Theorem is the geometric theorem that states that the square of the hypotenuse (longest side) of a right angled 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 labelled like this:

How to answer Pythagoras Theorem questions

• Write down the formula and substitute the values. a^2+b^2=c^2
• Work out 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.

Pythagoras Theorem in real life

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

Pythagoras Theorem in KS3

Pythagoras Theorem is usually introduced towards the end of KS3. 

The emphasis in KS3 is on students being able to:

• Correctly label a right angled triangle;
• Substitute values into the formula in order to work out the hypotenuse or one of the shorter sides.

15 Pythagoras Theorem Questions And Practice Problems Worksheet

Help your students prepare for their Maths GCSE with this free Pythagoras Theorem worksheet of 15 questions, includes answers and mark scheme

Pythagoras Theorem KS3 questions

Non-calculator questions.

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 kilometres 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.

Calculator questions

For these questions, round your answers to 3 significant figures.

3. Alex and Sam start from the same point. Alex walks 400 metres west. Sam walks x metres 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

Pythagoras Theorem in KS4

In KS4, students use Pythagoras Theorem to solve a variety of problems. Examples include:

• real life word problems
• coordinate problems
• multi-step problems
• 3D problems

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

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

Foundation GCSE Questions

Where necessary, round your answers to 3 significant figures.

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 centimetres.

7. ABC is an isosceles triangle.

Work out the height of the triangle.

8. ABCD is an isosceles trapezium.

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 Pythagoras Theorem doesn’t work.

Right angled triangle because Pythagoras 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:}

Triangle \text{ADE:}

Higher GCSE Questions

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 cannot have a negative side length so x=12

Pythagoras Theorem is used to work out a missing length in a right angled 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). Pythagoras Theorem is a^2+b^2=c^2. Substitute the values you know into Pythagoras Theorem and solve to find the missing side. For a more detailed explanation, including a video and worked examples, see: Pythagoras Theorem .

The hypotenuse of a right angled 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 angled triangle and you know two of the sides, you can use Pythagoras 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 Pythagoras Theorem questions and resources?

Third Space Learning’s free GCSE maths resource library contains detailed lessons with step-by-step instructions on how to solve Pythagoras Theorem problems, as well as maths worksheets with practice questions and more GCSE exam questions, based on past Edexcel, OCR and AQA exam questions.

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Pythagoras’ Theorem: Practice Questions with Answers & Explanations [Year 10 Maths]

In Year 10 standard maths students are expected to master basic trigonometry and understand Pythagoras’ theorem. Today we’ll dive into the practical side of Pythagorean calculations, exploring example questions and how to work out their answers. We will take things step by step, with writen explanations as we go.

In previous blog posts we have covered:

• An introduction to basic trigonometry
• What is Pythagoras’ Theorem
• 6 Steps to Apply Pythagoras’ Theorem

If you need to revise any of those topics it is best to do so before exploring these example questions.

Let’s get started:

Example 1: Finding the Hypotenuse Length

Question:  In a right-angled triangle, one side measures 5cm, and the other side measures 12cm. What is the length of the hypotenuse?

Solution:  

Example 2: Calculating a Missing Side

Question:  In a right-angled triangle with a hypotenuse of length 17 cm and one side measuring 8 cm, what is the length of the other side?

Example 3: Real word example without a diagram

Question:  If a ladder of length 5m is resting on a wall of length 4m, then find the distance between the foot of the ladder and the bottom of the wall.

We can assume the wall meets the ground at a right angle so Pythagoras’ theorem will apply to the triangle created by the wall, the ground, and the ladder.

The wall height will be one side, the ladder will be the hypotenuse, and we are looking for the distance between the wall and the foot of the ladder (the other side of the triangle).

You may find it helpful to create a sketch to better understand the problem. Or you can simply jump right in to the equation.

How did you go?

Often when we are struggling with a new concept, seeing it broken down step by step can unlock an understanding we hadn’t been able to grasp before.

If there are things you don’t understand in these examples, seeing the step by step will also allow you to ask questions from you teacher, peers or tutor and point to exactly where you are getting lost. If you still need help we recommend a tutor to talk you through your understanding and help you on your way.

Need More Help?

Year 10 Maths Tutoring

Master Coaching offers  one on one and small group tutoring for year 10 m aths . We are located in Hurstville Sydney, and also offer  online tutoring  to students across NSW. We offer one free trial class per family. Get in contact with us to find out more.

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Word problems on Pythagorean Theorem

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

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

Three cases of word problems on Pythagorean Theorem :

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

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

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

Word problems using the Pythagorean Theorem:

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

⇒ 200x = 10000 + 3600

⇒ 200x = 13600

⇒ x = 13600/200

Therefore, distance between X and Y = 68 meters.

Therefore, length of each side is 8 cm.

Using the formula solve more word problems on Pythagorean Theorem.

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

In a rectangle, each angle measures 90°.

Therefore PSR is right angled at S

Using Pythagoras theorem, we get

⇒ PS = √6400

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

                                                          = 2 (150 + 80) m

                                                          = 2 (230) m

                                                          = 460 m

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

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

According to Pythagorean Theorem,

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

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

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

Draw DE ┴ AB

Then AE = AB – EB but EB = BC

Therefore AE = 34 m - 29 m = 5 m

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

⇒ AD = √169

Therefore the distance between their tops = 13 m.

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

Congruent Shapes

Congruent Line-segments

Congruent Angles

Congruent Triangles

Conditions for the Congruence of Triangles

Side Side Side Congruence

Side Angle Side Congruence

Angle Side Angle Congruence

Angle Angle Side Congruence

Right Angle Hypotenuse Side congruence

Pythagorean Theorem

Proof of Pythagorean Theorem

Converse of Pythagorean Theorem

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Problems With Pythagoras & Trigonometry - GCSE Questions - Higher - AQA

Subject: Mathematics

Age range: 14-16

Resource type: Worksheet/Activity

Last updated

14 November 2019

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Cloned/Copied questions from previous 9-1 AQA GCSE exams. In two sizes, pdf and ppt.

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Pythagoras and Trigonometry Collection

Interesting teaching ideas for introducing and revising pythagoras’ theorem and trigonometry.

Pythagoras’ theorem and trigonometry are two of those classic topics that pupils revisit year-on-year. This is partly because these topics come in many forms and interesting contexts, from basic Pythagoras and Soh-Cah-Toa, to graphs of trigonometric functions and calculus.

With the increased emphasis on ratio in the new GCSE specification, trigonometry is likely to start appearing in non-calculator papers as well. As such, it is more important than ever for students to get to grips with this crucial and fascinating area of mathematics.

Fortunately, TES Maths is here to help!

Below is a selection of 15 top Pythagoras and trigonometry resources, all uploaded by the talented members of the TES Maths community. Why not use them to give you some new ideas for teaching these topics across all year groups?

Craig Barton, TES Maths adviser

• Perigal’s dissection Encourage pupils to stumble upon Pythagoras’ theorem by themselves by first introducing them to Perigal’s cut-and-shift proof.
• Introduction to Pythagoras’ theorem This well-differentiated lesson neatly introduces the famous theorem and the correct use of the calculator in solving Pythagoras-related problems.
• Interactive Pythagoras lesson Culminating in a treasure hunt, this action-packed lesson starts with an element of discovery before progressing on to using Pythagoras in real-life contexts.
• Have I got hypotenuse for you? Revise the key aspects of Pythagoras in a fun and effective way with this select-and-reveal starter and plenary activity.
• Trigonometry investigation Give your students the opportunity to discover the relationship between trigonometry and ratio in this fully-resourced investigation.
• Introduction to trigonometry Introduce sin, cos and tan for the first time with this clear, well-structured lesson, which can be easily adapted to suit your class’ needs.
• Soh-Cah-Toa Crammed full of differentiated activities, this complete lesson on trigonometry uses Bloom’s taxonomy questioning to challenge pupils.
• Batman-themed word problems Help Gotham City’s Police Commissioner catch criminals while recapping existing learning on trigonometry with these engaging problems.
• Pirate problem-solving Revise trigonometry, Pythagoras’ theorem and even geometry in this themed activity, in which pupils are tasked with finding Admiral Angle’s buried treasure.
• Pythagoras or trigonometry? Clarify the process of deciding whether a question requires the use of Pythagoras or Soh-Cah-Toa with this handy flow chart.
• 3D trigonometry Stretch more able learners with this innovative task, which is ideal for pair work and contains plenty of opportunities for peer and self-assessment.
• Finding exact trigonometric values Use these clear worksheets to encourage students to become familiar with ‘special triangles’, in order to be able to calculate exact trigonometric values without a calculator.
• Sine and cosine rule lesson This step-by-step lesson plan with precise explanations and well-worked examples is ideal for KS4 classes.
• Sine and cosine sorting exercise Put your pupils’ understanding of when to use the sine and cosine rule to the test with this versatile resource, with plenty of AfL opportunities.
• Pythagoras and trigonometry revision booklet Containing lots of exam-style questions, this comprehensive workbook has plenty of space for students to show their working and explain their reasoning.

2 thoughts on “ Pythagoras and Trigonometry Collection ”

Hardly unique! But thanks for sharing. Your website is a really useful collection of resources.

Good point. That was the TES editor’s choice of title. I have changed to “interesting”!

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First teaching 2021

Last exams 2024

Trigonometry ( CIE IGCSE Maths: Core )

Topic questions.

The diagram shows a right-angled triangle.

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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