Is a triangle with sides 5, 9 and 11 right-angled?

No: 25 + 81 = 106, but 11² = 121

A 6 m ladder reaches 5.5 m up a wall. How far is its foot from the wall, to 2 decimal places?

√(36 - 30.25) = √5.75 ≈ 2.40 m

Free worksheet · Australian Curriculum

Pythagoras' theorem worksheets with answers

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Pythagoras' theorem at each year level, matched to the textbook chapters most Victorian schools use: a free printable worksheet with answers, sample questions, and an auto-marked test teachers can share.

Free Year 8 pythagoras' theorem worksheet

10 questions, then the answers with worked solutions. Printable PDF, no sign-up.

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Year 8 topics covered

Introducing Pythagoras' theorem
Using Pythagoras' theorem
Calculating the length of a shorter side

Sample Year 8 pythagoras' theorem questions

Introducing Pythagoras' theorem
A right-angled triangle has shorter sides of $5$ cm and $12$ cm. Find the value of $c^{2}$ , where $c$ is the length of the hypotenuse.
Show the answer
Key idea: Pythagoras' theorem: $c^{2} = a^{2} + b^{2}$ .
1. $c^{2} = 5^{2} + 12^{2} = 25 + 144 = 169$
Answer: $169$
Using Pythagoras' theorem
A right-angled triangle has shorter sides of length $9$ m and $12$ m. Find the length of the hypotenuse.
Show the answer
Key idea: In a right-angled triangle, $c^2 = a^2 + b^2$ , where $c$ is the hypotenuse.
1. Substitute the known sides: $c^2 = 9^2 + 12^2$
2. Evaluate: $c^2 = 81 + 144 = 225$
3. Take the square root: $c = \sqrt{225} = 15$
Answer: The hypotenuse is $15$ m.
Calculating the length of a shorter side
A right-angled triangle has a hypotenuse of $10$ m and one shorter side of $6$ m. Find the exact length of the other shorter side, then round your answer to two decimal places.
Show the answer
Key idea: Rearrange $a^2 + b^2 = c^2$ to find a shorter side: $b^2 = c^2 - a^2$ .
1. Substitute the known values: $b^2 = 10^2 - 6^2$
2. Evaluate: $b^2 = 100 - 36 = 64$
3. Take the square root: $b = \sqrt{64} = 8$
4. The exact answer is $8$ m, which rounds to $8.00$ m.
Answer: The other shorter side is exactly $8$ m (or $8.00$ m to two decimal places).
Introducing Pythagoras' theorem
A triangle has sides of $8$ cm, $15$ cm and $17$ cm. Calculate $8^{2} + 15^{2}$ .
Show the answer
Key idea: Square each number, then add.
1. $8^{2} = 64$ and $15^{2} = 225$
2. $64 + 225 = 289$ (which equals $17^{2}$ , so the triangle is right-angled)
Answer: $289$

Free Year 9 pythagoras' theorem worksheet

10 questions, then the answers with worked solutions. Printable PDF, no sign-up.

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Year 9 topics covered

Pythagoras' theorem
Finding the length of the shorter sides
Using Pythagoras' theorem to solve two-dimensional problems
Using Pythagoras' theorem to solve three-dimensional problems

Sample Year 9 pythagoras' theorem questions

Pythagoras' theorem
A right-angled triangle has shorter sides of $7$ cm and $24$ cm. Find the length of the hypotenuse.
Show the answer
Key idea: Use $c^2 = a^2 + b^2$ .
1. $c^2 = 7^2 + 24^2 = 49 + 576 = 625$
2. $c = \sqrt{625} = 25$
Answer: $25$ cm
Finding the length of the shorter sides
A right-angled triangle has a hypotenuse of $17$ cm and one shorter side of $8$ cm. Find the length of the third side.
Show the answer
Key idea: To find a shorter side, subtract: $a^2 = c^2 - b^2$ , where $c$ is the hypotenuse.
1. $a^2 = 17^2 - 8^2 = 289 - 64 = 225$
2. $a = \sqrt{225} = 15$
Answer: $15$ cm
Using Pythagoras' theorem to solve two-dimensional problems
A $4.5$ m ladder leans against a vertical wall on flat ground. The foot of the ladder is $1.2$ m from the wall. How far up the wall does the ladder reach? Give your answer correct to $2$ decimal places.
Show the answer
Key idea: The ladder, wall and ground form a right-angled triangle with the ladder as the hypotenuse.
1. $h^2 = 4.5^2 - 1.2^2 = 20.25 - 1.44 = 18.81$
2. $h = \sqrt{18.81} = 4.337\ldots \approx 4.34$
Answer: $4.34$ m
Using Pythagoras' theorem to solve three-dimensional problems
A rectangular box has length $12$ cm, width $5$ cm and height $8$ cm. Find the length of the longest diagonal of the box, correct to one decimal place.
Show the answer
Key idea: The longest diagonal of a rectangular prism goes from one corner to the opposite corner. Use Pythagoras twice: first on the base, then use that result with the height.
1. Find the diagonal $d$ of the base using Pythagoras:\
$\begin{gathered}d^2 = 12^2 + 5^2 = 144 + 25 = 169 \\ d = 13 \text{ cm}\end{gathered}$
1. Use $d$ and the height to find the space diagonal $D$ :\
$\begin{gathered}D^2 = d^2 + 8^2 = 169 + 64 = 233 \\ D = \sqrt{233} \approx 15.3 \text{ cm}\end{gathered}$
Answer: The longest diagonal is approximately $15.3$ cm.

Free Year 10 pythagoras' theorem worksheet

10 questions, then the answers with worked solutions. Printable PDF, no sign-up.

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Year 10 topics covered

Pythagoras' theorem including three-dimensional problems

Sample Year 10 pythagoras' theorem questions

Pythagoras' theorem including three-dimensional problems
In $\triangle ABC$ , $AB = AC$ and $M$ is the midpoint of $BC$ . It can be proved that $\triangle ABM \equiv \triangle ACM$ by SSS. If $AB = 10$ cm and $BC = 12$ cm, what is the length of $AM$ in centimetres, correct to 2 decimal places?
Show the answer
Key idea: Because $\triangle ABM \equiv \triangle ACM$ (SSS), $AM$ is the perpendicular bisector of $BC$ , so $\angle AMB = 90°$ .
1. $BM = \dfrac{BC}{2} = \dfrac{12}{2} = 6$ cm
2. Apply Pythagoras' theorem in $\triangle ABM$ : $AM^2 + BM^2 = AB^2$
3. $AM^2 = AB^2 - BM^2 = 10^2 - 6^2 = 100 - 36 = 64$
4. $AM = \sqrt{64} = 8$ cm
Answer: $8.00$ cm
Pythagoras' theorem including three-dimensional problems
A rectangular room is 6 m long, 4 m wide and 3 m high. A cable is to be run in a straight line from one bottom corner of the room to the opposite top corner. Find the length of the cable and the angle it makes with the floor, correct to one decimal place.
Show the answer
Key idea: First use Pythagoras' theorem to find the diagonal of the floor, then use that diagonal and the height to find the length of the cable and the angle with the floor.
1. Find the floor diagonal $d$ :\
$d = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52}$
1. Find the cable length $L$ using the height of 3 m:\
$L = \sqrt{(\sqrt{52})^2 + 3^2} = \sqrt{52 + 9} = \sqrt{61} \approx 7.8 \text{ m}$
1. Find the angle $\theta$ the cable makes with the floor (opposite $= 3$ m, adjacent $= \sqrt{52}$ m):\
$\tan(\theta) = \frac{3}{\sqrt{52}}$
1. Solve:\
$\theta = \tan^{-1}\!\left(\frac{3}{\sqrt{52}}\right) \approx 22.6°$
Answer: The cable is approximately $7.8$ m long and makes an angle of approximately $22.6°$ with the floor.
Pythagoras' theorem including three-dimensional problems
A rectangular prism has length $12$ cm, width $5$ cm and height $4$ cm. Find the length of the space diagonal (the diagonal from one corner to the opposite corner through the inside of the prism). Give your answer correct to two decimal places.
Show the answer
Key idea: The space diagonal $d$ can be found using the 3D extension of Pythagoras' theorem: $d = \sqrt{l^2 + w^2 + h^2}$ .
1. Write the formula and substitute the dimensions:\
$d = \sqrt{12^2 + 5^2 + 4^2}$
1. Evaluate each square:\
$d = \sqrt{144 + 25 + 16}$
1. Add and take the square root:\
$d = \sqrt{185} \approx 13.60 \text{ cm}$
Answer: The space diagonal is approximately $13.60$ cm.
Pythagoras' theorem including three-dimensional problems
A rectangular prism has length $9$ cm, width $4$ cm and height $6$ cm. A straight rod is placed along the space diagonal of the prism. Find the length of the rod correct to 2 decimal places. Give your answer in centimetres.
Show the answer
Key idea: The space diagonal of a rectangular prism with dimensions $l$ , $w$ and $h$ is $d = \sqrt{l^2 + w^2 + h^2}$ .
1. Identify the dimensions: $l = 9$ cm, $w = 4$ cm, $h = 6$ cm.
2. Find the base diagonal: $d_{\text{base}} = \sqrt{9^2 + 4^2} = \sqrt{81 + 16} = \sqrt{97}$ cm.
3. Use $d_{\text{base}}$ and the height to find the space diagonal: $d = \sqrt{(\sqrt{97})^2 + 6^2} = \sqrt{97 + 36} = \sqrt{133}$ .
4. Evaluate: $d = \sqrt{133} \approx 11.53$ cm (correct to 2 decimal places).
Answer: $11.53$ cm

Topic guide

Pythagoras' theorem, explained

One rule, used forwards to find the hypotenuse, backwards to find a shorter side, and as a test for right angles.

1. The theorem

In a right-angled triangle with shorter sides $a$ and $b$ and hypotenuse $c$ :

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

The hypotenuse is the longest side, opposite the right angle. Label it first, every time.

2. Finding the hypotenuse

Square the two shorter sides, add, then take the square root:

$c^2 = 6^2 + 8^2 = 36 + 64 = 100, c = 10$

If the answer is not exact, round at the end: with sides $5$ and $7$ ,

$c = \sqrt{25 + 49} = \sqrt{74} \approx 8.60$

3. Finding a shorter side

Subtract instead of adding:

$b^2 = c^2 - a^2 = 13^2 - 5^2 = 169 - 25 = 144, b = 12$

With a hypotenuse of $10$ and a side of $7$ : $b = \sqrt{100 - 49} = \sqrt{51} \approx 7.14$ . A shorter side must come out smaller than the hypotenuse.

4. Pythagorean triples

Some whole-number sets fit exactly. Worth knowing:

$3, 4, 5 5, 12, 13 8, 15, 17 7, 24, 25$

Any multiple of a triple is also a triple, so $6, 8, 10$ and $9, 12, 15$ work too.

5. Testing for a right angle

A triangle is right-angled only if the square of the longest side equals the sum of the squares of the other two. For $9, 40, 41$ :

$9^2 + 40^2 = 81 + 1600 = 1681 = 41^2$

so it is right-angled. For $6, 7, 9$ : $36 + 49 = 85$ but $9^2 = 81$ , so it is not.

6. Worded problems

A $5$ m ladder leans against a wall with its foot $1.5$ m from the wall. How high up the wall does it reach?

$h = \sqrt{5^2 - 1.5^2} = \sqrt{22.75} \approx 4.77 \text{ m}$

Draw the triangle, mark the right angle, and decide whether the unknown is the hypotenuse.

Quick check: 10 questions

Try each one, then open the answer.

Find the hypotenuse when the shorter sides are $3$ cm and $4$ cm.
Answer
$5$ cm
Find the hypotenuse when the shorter sides are $9$ m and $12$ m.
Answer
$15$ m
The hypotenuse is $17$ cm and one side is $8$ cm. Find the other side.
Answer
$\sqrt{289 - 64} = 15$ cm
Find the hypotenuse when the shorter sides are $4$ cm and $6$ cm, to 2 decimal places.
Answer
$\sqrt{52} \approx 7.21$ cm
The hypotenuse is $12$ m and one side is $5$ m. Find the other side to 2 decimal places.
Answer
$\sqrt{119} \approx 10.91$ m
Is a triangle with sides $7$ , $24$ and $25$ right-angled?
Answer
Yes: $49 + 576 = 625 = 25^2$
Is a triangle with sides $5$ , $9$ and $11$ right-angled?
Answer
No: $25 + 81 = 106$ , but $11^2 = 121$
Find the length of the diagonal of a $20$ cm by $15$ cm rectangle.
Answer
$\sqrt{400 + 225} = 25$ cm
Mia walks $7$ km north, then $4$ km east. How far is she from her starting point, to 2 decimal places?
Answer
$\sqrt{65} \approx 8.06$ km
A $6$ m ladder reaches $5.5$ m up a wall. How far is its foot from the wall, to 2 decimal places?
Answer
$\sqrt{36 - 30.25} = \sqrt{5.75} \approx 2.40$ m

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These topic pages cover this skill, with worked examples, common mistakes and a question to try.

Year 8 worksheetPythagoras' theorem Year 9 worksheetPythagoras (applied)Year 9 worksheetTrigonometry Year 10A worksheetPythagoras and trigonometry in 3D

Questions

What is Pythagoras' theorem?

In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: c squared equals a squared plus b squared. It only works for right-angled triangles.

How do you find a shorter side with Pythagoras?

Square the hypotenuse, subtract the square of the known shorter side, then take the square root. Check your answer is smaller than the hypotenuse.

Which year level learns Pythagoras' theorem?

Pythagoras is introduced in Year 8 or Year 9, depending on the school, and is used again in Year 9 and 10 trigonometry, coordinate geometry and 3D problems.

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Pythagoras' theorem, explained

1. The theorem

2. Finding the hypotenuse

3. Finding a shorter side

4. Pythagorean triples

5. Testing for a right angle

6. Worded problems

Quick check: 10 questions

Generate a free maths worksheet

Practice built from mistakes.

Topic worksheets with worked examples

Questions

What is Pythagoras' theorem?

How do you find a shorter side with Pythagoras?

Which year level learns Pythagoras' theorem?

More skill worksheets

Better maths marks start with the next wrong answer.