The Pythagoras Theorem Made Easy with Free Practice Test

Learn the Pythagorean Theorem and solve right-angled triangle problems easily. This article includes the formula, examples, and practice with common Pythagorean triples to boost your math skills and exam confidence.
The Pythagoras Theorem

Table of Contents

Interactive maths practice

Pythagoras’ Theorem Practice Questions

Practise identifying the hypotenuse, finding a missing side and applying Pythagoras’ theorem to real-life problems. Every question includes a labelled diagram, an instant answer check and complete working.

a2 + b2 = c2 c is the hypotenuse, the longest side opposite the right angle.
  • Identify the sides: locate the right angle and the hypotenuse
  • Find the hypotenuse: add the squares of the two shorter sides
  • Find a shorter side: subtract the known shorter-side square from the hypotenuse square
  • Apply the theorem: solve ladder, ramp, rectangle, road and rope problems
  • Check reasonableness: the hypotenuse must be longer than either shorter side
  • 3 progressive levels
  • 15 illustrated questions
  • Instant answer checks
  • Hints and full working
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Enter the number only or include the correct unit. Your answers and checked results are saved automatically in this browser.

Level 1 Finding the Hypotenuse 5 questions · Foundation 0/5

Tip: The hypotenuse is the longest side and is always opposite the right angle. Use a² + b² = c², where c is the hypotenuse.
  1. Find the hypotenuse of a right-angled triangle with shorter sides 3 cm and 4 cm.

    Right-angled triangle The hypotenuse is opposite the right angle. Diagram not to scale. 3 cm 4 cm c = ?
    The hypotenuse is opposite the right angle. Diagram not to scale.

    View hint and full working

    Hint: Square both shorter sides, add the results, then take the positive square root.

    Working
    a² + b² = c²
    3² + 4² = c²
    9 + 16 = c²
    25 = c²
    c = √25 = 5 cm
    Answer: 5 cm
  2. Find the hypotenuse of a right-angled triangle with shorter sides 5 m and 12 m.

    Right-angled triangle The diagonal side is the hypotenuse. Diagram not to scale. 5 m 12 m c = ?
    The diagonal side is the hypotenuse. Diagram not to scale.

    View hint and full working

    Hint: Use the two shorter sides in a² + b² = c².

    Working
    5² + 12² = c²
    25 + 144 = c²
    169 = c²
    c = √169 = 13 m
    Answer: 13 m
  3. The perpendicular sides of a right-angled triangle are 6 cm and 8 cm. Find the hypotenuse.

    Right-angled triangle Use the two perpendicular sides. Diagram not to scale. 6 cm 8 cm c = ?
    Use the two perpendicular sides. Diagram not to scale.

    View hint and full working

    Hint: The perpendicular sides are the two shorter sides.

    Working
    6² + 8² = c²
    36 + 64 = c²
    100 = c²
    c = √100 = 10 cm
    Answer: 10 cm
  4. Find the hypotenuse of a right-angled triangle with shorter sides 7 cm and 24 cm.

    Right-angled triangle The drawing is not to scale. Diagram not to scale. 7 cm 24 cm c = ?
    The drawing is not to scale. Diagram not to scale.

    View hint and full working

    Hint: Substitute 7 and 24 as the shorter sides.

    Working
    7² + 24² = c²
    49 + 576 = c²
    625 = c²
    c = √625 = 25 cm
    Answer: 25 cm
  5. A ladder stands 9 m from a wall and reaches 12 m up the wall. Find the length of the ladder.

    Right-angled triangle The ladder is opposite the right angle. Diagram not to scale. 9 m 12 m ladder = ?
    The ladder is opposite the right angle. Diagram not to scale.

    View hint and full working

    Hint: The wall and ground form the right angle, so the ladder is the hypotenuse.

    Working
    9² + 12² = L²
    81 + 144 = L²
    225 = L²
    L = √225 = 15 m
    Answer: 15 m

Level 2 Finding a Missing Shorter Side 5 questions · Rearranging 0/5

Tip: When the hypotenuse is known, subtract the square of the known shorter side: a² = c² − b². Take the positive square root because a length cannot be negative.
  1. The hypotenuse is 13 cm and one shorter side is 5 cm. Find the other shorter side.

    Right-angled triangle The known longest side is the hypotenuse. Diagram not to scale. a = ? 5 cm 13 cm
    The known longest side is the hypotenuse. Diagram not to scale.

    View hint and full working

    Hint: Subtract 5² from 13² before taking the square root.

    Working
    a² + 5² = 13²
    a² = 13² − 5²
    a² = 169 − 25
    a² = 144
    a = √144 = 12 cm
    Answer: 12 cm
  2. A right-angled triangle has hypotenuse 10 cm and one shorter side 6 cm. Find the other shorter side.

    Right-angled triangle Subtract the known leg squared from the hypotenuse squared. Diagram not to scale. x = ? 6 cm 10 cm
    Subtract the known leg squared from the hypotenuse squared. Diagram not to scale.

    View hint and full working

    Hint: Use x² = 10² − 6².

    Working
    x² + 6² = 10²
    x² = 100 − 36
    x² = 64
    x = √64 = 8 cm
    Answer: 8 cm
  3. One shorter side of a right-angled triangle is 9 m and the hypotenuse is 15 m. Find the other shorter side.

    Right-angled triangle The 15 m side is opposite the right angle. Diagram not to scale. x = ? 9 m 15 m
    The 15 m side is opposite the right angle. Diagram not to scale.

    View hint and full working

    Hint: The missing side is not the hypotenuse, so use subtraction.

    Working
    x² + 9² = 15²
    x² = 225 − 81
    x² = 144
    x = √144 = 12 m
    Answer: 12 m
  4. The hypotenuse is 17 cm and one shorter side is 15 cm. Find the other shorter side.

    Right-angled triangle Identify the 17 cm side as the hypotenuse. Diagram not to scale. x = ? 15 cm 17 cm
    Identify the 17 cm side as the hypotenuse. Diagram not to scale.

    View hint and full working

    Hint: Calculate 17² − 15².

    Working
    x² + 15² = 17²
    x² = 289 − 225
    x² = 64
    x = √64 = 8 cm
    Answer: 8 cm
  5. A right-angled triangle has shorter sides 20 m and 21 m. Find the hypotenuse.

    Right-angled triangle Add the squares because the hypotenuse is missing. Diagram not to scale. 20 m 21 m c = ?
    Add the squares because the hypotenuse is missing. Diagram not to scale.

    View hint and full working

    Hint: This question returns to finding the hypotenuse, so add the squares.

    Working
    20² + 21² = c²
    400 + 441 = c²
    841 = c²
    c = √841 = 29 m
    Answer: 29 m

Level 3 Applying Pythagoras’ Theorem 5 questions · Word problems 0/5

Tip: Sketch or identify the right-angled triangle first. Label the longest side as the hypotenuse, choose addition or subtraction, and include the correct unit in the final answer.
  1. A rectangular field is 24 m long and 7 m wide. Find the length of its diagonal.

    Rectangle with a diagonal The diagonal forms a right-angled triangle. Diagram not to scale. 24 m 7 m d = ?
    The diagonal forms a right-angled triangle. Diagram not to scale.

    View hint and full working

    Hint: The rectangle’s diagonal is the hypotenuse of a right-angled triangle.

    Working
    d² = 24² + 7²
    d² = 576 + 49
    d² = 625
    d = √625 = 25 m
    Answer: 25 m
  2. A television screen is 40 cm wide and 30 cm high. Find its diagonal length.

    Rectangle with a diagonal Screen sizes are measured diagonally. Diagram not to scale. 40 cm 30 cm d = ?
    Screen sizes are measured diagonally. Diagram not to scale.

    View hint and full working

    Hint: Use the screen width and height as the shorter sides.

    Working
    d² = 40² + 30²
    d² = 1600 + 900
    d² = 2500
    d = √2500 = 50 cm
    Answer: 50 cm
  3. A ramp is 13 m long and reaches a vertical height of 5 m. How far is the base of the ramp from the wall?

    Right-angled triangle The ramp is the hypotenuse. Diagram not to scale. base = ? 5 m 13 m ramp
    The ramp is the hypotenuse. Diagram not to scale.

    View hint and full working

    Hint: The ramp is the hypotenuse, so subtract 5² from 13².

    Working
    x² + 5² = 13²
    x² = 169 − 25
    x² = 144
    x = √144 = 12 m
    Answer: 12 m
  4. Two perpendicular roads extend 16 km and 12 km from the same intersection. Find the straight-line distance between their endpoints.

    Right-angled triangle The direct route is the hypotenuse. Diagram not to scale. 16 km 12 km d = ?
    The direct route is the hypotenuse. Diagram not to scale.

    View hint and full working

    Hint: The roads are perpendicular, so they form the shorter sides of a right-angled triangle.

    Working
    d² = 16² + 12²
    d² = 256 + 144
    d² = 400
    d = √400 = 20 km
    Answer: 20 km
  5. A flagpole is 18 m tall. A rope runs from its top to a point 24 m from the base. Find the rope’s length.

    Right-angled triangle The rope is the hypotenuse. Diagram not to scale. 24 m 18 m rope = ?
    The rope is the hypotenuse. Diagram not to scale.

    View hint and full working

    Hint: The vertical pole and horizontal ground form the shorter sides.

    Working
    r² = 18² + 24²
    r² = 324 + 576
    r² = 900
    r = √900 = 30 m
    Answer: 30 m
NSW Year 8–9 maths

Pythagoras’ Theorem Explained Step by Step

Pythagoras’ theorem connects the three side lengths of a right-angled triangle. When two sides are known, the third side can be calculated. The rule applies only when the triangle contains a 90° angle, and the hypotenuse is always the longest side opposite that angle.

90° The theorem requires a right angle.
2 sides Two known lengths determine the third side.
1 final root Take the positive square root to obtain a length.

Pythagoras’ Theorem in NSW Year 8 and Year 9 Maths

The NSW Mathematics 7–10 syllabus includes Pythagoras and trigonometry as a core focus area. Schools organise content through their own scope and sequence, so the exact term can differ. In my tutoring experience, students around Telopea, Dundas, Oatlands, Ermington, Rydalmere and Carlingford most commonly meet Pythagoras’ theorem in Year 8, while Year 9 students often use it for revision and more applied questions.

What I notice in local tutoring sessions

Most students can identify the hypotenuse and complete the squaring, addition or subtraction. The two recurring difficulties are more specific:

  • They calculate a value such as c² = 25 but forget to take the final square root.
  • They become unsure when the question asks for a shorter side and subtraction is required.

Older students are usually comfortable with the calculation itself. Their main challenge is recognising that an unfamiliar diagram, map or word problem contains a right-angled triangle and may require Pythagoras’ theorem.

The Pythagoras Rule and the Hypotenuse

For a right-angled triangle, let a and b be the two shorter sides and let c be the hypotenuse.

a² + b² = c²

c must represent the hypotenuse. The letters a and b can be swapped, but c cannot be assigned to a shorter side when using this standard form.

The hypotenuse

The longest side of the right-angled triangle. It sits directly opposite the 90° angle.

The two shorter sides

These sides meet to form the right angle. They are sometimes called the legs of the triangle.

Fast check: Find the small square marking the right angle. The side that does not touch that square is the hypotenuse.

How to Find the Hypotenuse

When the unknown side is the hypotenuse, square the two shorter sides, add them and take the positive square root.

c = √(a² + b²)
Worked example: shorter sides of 3 cm and 4 cm
c² = 3² + 4²
c² = 9 + 16 = 25
c = √25 = 5

Answer: The hypotenuse is 5 cm.

The answer should be longer than either shorter side:

c > max(a, b)

How to Find a Missing Shorter Side

This is the form that causes the most confusion. The longest side is already known, so subtract the square of the known shorter side from the square of the hypotenuse.

a = √(c² − b²)
Worked example: hypotenuse 13 cm and one shorter side 5 cm
x² + 5² = 13²
x² = 13² − 5²
x² = 169 − 25 = 144
x = √144 = 12

Answer: The missing shorter side is 12 cm.

Do not add in every question. Add the two shorter-side squares when finding the hypotenuse. Subtract from the hypotenuse square when finding a shorter side.

The Step Students Most Often Forget: Take the Square Root

When the working reaches c² = 25, the value 25 is the square of the length. It is not the final side length.

Incorrect stopping point
c² = 25

Correct final step

c = √25 = 5

A side length is positive, so use the positive square root. Writing the square-root line separately makes the final step harder to miss.

A useful habit: After every Pythagoras calculation, ask, “Have I found the side length, or only the square of the side length?”

How to Recognise a Pythagoras Word Problem

A question may not use the word “Pythagoras”. Look for a hidden right-angled triangle created by perpendicular directions, a rectangle diagonal, a wall and the ground, or horizontal and vertical distances.

Find or draw the right angle

Look for a 90° marker, perpendicular lines, north–east movement, a rectangle corner, a vertical wall or level ground.

Identify the hypotenuse

It is opposite the right angle and is usually the diagonal or direct straight-line distance.

Decide between addition and subtraction

Add when the hypotenuse is missing. Subtract when a shorter side is missing.

Finish with the square root and unit

Round only when the question requests it, and state cm, m, km or the relevant unit.

Map-style example

A person travels 6 km east and then 8 km north. The east and north directions are perpendicular, so the direct distance back to the starting point forms the hypotenuse.

d² = 6² + 8²
d = √100 = 10

Direct distance: 10 km.

This same structure appears in questions involving ladders, ramps, television screens, sports fields, building diagonals and the shortest distance between two locations.

Square Roots, Surds and Rounding

Not every right-angled triangle has a whole-number answer. When the square root does not simplify to an integer, the answer may be left as an exact surd or written as a decimal, depending on the instruction.

Example: shorter sides of 7 cm and 9 cm
c = √(7² + 9²)
c = √130 ≈ 11.4

The exact length is √130 cm. Correct to 1 decimal place, it is 11.4 cm.

  • Keep the calculator value unrounded during the working.
  • Round only the final answer.
  • Use the number of decimal places or significant figures stated in the question.
  • Include the unit after the rounded answer.

Pythagorean Triples: Useful, but Not a Memorisation Test

A Pythagorean triple is a set of three whole numbers that satisfies the theorem. Common examples include:

(3, 4, 5), (5, 12, 13), (8, 15, 17)

Recognising a familiar triple can make checking faster, and multiples also work—for example, 6, 8 and 10 are double 3, 4 and 5. However, students do not need to memorise a long list. Understanding how to apply the theorem is more reliable than depending on memory.

The Converse of Pythagoras’ Theorem

The converse works in the reverse direction. If the square of the longest side equals the sum of the squares of the other two sides, the triangle is right-angled.

If a² + b² = c², the triangle is right-angled.
Example: side lengths 6, 8 and 10
6² + 8² = 36 + 64 = 100 = 10²

Therefore, the triangle is right-angled.

Always test the longest side as c.

How Pythagoras Connects to Number Planes and Trigonometry

Pythagoras’ theorem becomes part of later topics rather than disappearing after Year 8.

Distance on a number plane

The horizontal and vertical coordinate changes form the shorter sides of a right-angled triangle. This produces the distance formula:

d = √((x₂ − x₁)² + (y₂ − y₁)²)

Right-angled trigonometry

Trigonometric ratios use opposite, adjacent and hypotenuse. Pythagoras may be used first to find a missing side before applying sine, cosine or tangent.

Explore the number plane and distance guide or continue to the NSW trigonometry guide.

A Reliable Exam Method

Step What to do Quick check
1. Recognise Confirm that the triangle is right-angled. Can you identify a 90° angle?
2. Label Mark the hypotenuse as c. Is c opposite the right angle?
3. Choose Add for a missing hypotenuse; subtract for a missing shorter side. Which side is unknown?
4. Calculate Square, add or subtract, then take the positive square root. Did you include the final √ step?
5. Present Round as requested and include units. Is the hypotenuse the longest side?

Pythagoras Revision Checklist

  • I use the theorem only with a right-angled triangle.
  • I identify the hypotenuse before substituting numbers.
  • I add squares when finding the hypotenuse.
  • I subtract from the hypotenuse square when finding a shorter side.
  • I take the positive square root at the end.
  • I round only when instructed and include the unit.
  • I check whether my answer is reasonable.

Frequently Asked Questions

When can Pythagoras’ theorem be used?

It can be used when a triangle is right-angled, two side lengths are known and the third side is required.

Why must I take a square root at the end?

The formula first produces the square of the unknown length. Taking the positive square root converts that squared value into the actual side length.

How do I know whether to add or subtract?

Add the two shorter-side squares when the hypotenuse is unknown. Subtract the known shorter-side square from the hypotenuse square when a shorter side is unknown.

Does the hypotenuse always have to be called c?

No. A diagram may use any letter. However, in the standard formula a² + b² = c², c represents the hypotenuse. Identify the side first rather than relying only on its letter.

Do students need to memorise Pythagorean triples?

Recognising common triples can save time, but understanding the formula is more important. The theorem works even when the side lengths are unfamiliar or the answer is a decimal.

Related Aussie Math Tutor NSW Resources

Trusted external references

For official curriculum information, see the NSW Mathematics K–10 syllabus overview and the NSW Mathematics glossary. Additional explanations and practice are available through Khan Academy’s Pythagorean theorem unit and Maths Is Fun.

Still Unsure When to Use Pythagoras?

Some students can follow a familiar example but become stuck when the missing side changes or the right-angled triangle is hidden inside a word problem. Patient, step-by-step support can strengthen the method and the recognition skills needed for school assessments.

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