Activities
7.7 Pythagoras’ theorem
Tutor guidance notes
Aims and objectives
At the end of these activity sheets, learners should be able to:
 understand what Pythagoras’ theorem is, and recall the formula
 calculate missing sides of triangles using Pythagoras’ theorem
 calculate the perpendicular height of triangles using Pythagoras’ theorem.
These questions and activities are not intended as formal assessment. However,
the answers should be discussed with the learners as a group to ensure they
have full comprehension of the subject.
Individual teachers will use these questions and activities in different ways. They
could be used at the end of a session to recap; used individually during the
session as a ‘to-the-point’ learning tool; or used as homework. Whichever
teaching method is chosen, make sure that you set and agree upon a completion
date for each activity.
The answers given are not exhaustive and are to be used as a guide only.
Alternative, appropriate answers may be considered.
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Activities
Candidate name:
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7.7 Pythagoras’ theorem
Chapter 7: Numeracy skills
Aims and objectives
At the end of these activity sheets, you should be able to:
 understand what Pythagoras’ theorem is, and recall the formula
 calculate missing sides of triangles using Pythagoras’ theorem
 calculate the perpendicular height of triangles using Pythagoras’ theorem.
Show your workings for all calculations.
1
What is Pythagoras’ theorem used for? ___________________________
___________________________________________________________
2
Which side of a triangle is called the hypotenuse? ___________________
3
We name the hypotenuse c, and the other two sides of the triangle a and b.
State Pythagoras’ theorem using these ___________________________
4
Look at this list of square numbers and square roots. Are they all correct?
Write ‘correct’ or ‘incorrect’ next to each one, and write the correct
answers.
2² = 4
_____________________________________________
√64 = 6
_____________________________________________
5² = 50
_____________________________________________
4² = 20
_____________________________________________
√1 = 1
_____________________________________________
√49 = 7
_____________________________________________
9² = 89
_____________________________________________
√36 = 6
_____________________________________________
10² = 100
_____________________________________________
√9 = 4
_____________________________________________
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Activities
5
Calculate the missing lengths for these triangles. Give answers to one
decimal place.
a
9 cm
a
6 cm
b
b
4 cm
11 cm
c
12 cm
c
7 cm
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Activities
6
Use Pythagoras’s theorem to find the perpendicular heights, and then the
areas, of these triangles. Round numbers to one decimal place where
appropriate.
a
5 cm
h
2 cm
3 cm
b
7 cm
h
6 cm
2 cm
c
4 cm
h
4 cm
2.5 cm
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Activities
7.7 Pythagoras’ theorem
Chapter 7: Numeracy skills
Answers
1
What is Pythagoras’ theorem used for?
to find unknown lengths in right-angled triangles
2
Which side of a triangle is called the hypotenuse?
the longest side
3
We name the hypotenuse c, and the other two sides of the triangle a and b.
State Pythagoras’ theorem using these letters.
c² = a² + b²
4
Look at this list of square numbers and square roots. Are they all correct?
Write ‘correct’ or ‘incorrect’ next to each one, and write the correct
answers.
2² = 4
√64 = 6
5² = 50
4² = 20
√1 = 1
√49 = 7
9² = 89
√36 = 6
10² = 100
√9 = 4
correct
incorrect
incorrect
incorrect
correct
correct
incorrect
correct
correct
incorrect
√64 = 8
5² = 25
4² = 16
9² = 81
√9 = 3
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Activities
5
Calculate the missing lengths for these triangles. Give answers to one
decimal place.
a
a² = 9² – 6²
= 81 – 36
= 45
a = √45 = 6.7 cm
9 cm
a
6 cm
b
b
4 cm
b² = 11² – 4²
= 121 – 16
= 105
b = √105 = 10.2
cm
11 cm
c
12 cm
c
7 cm
c² = 12² + 7²
= 144 + 49
= 193
c = √193 = 13.9 cm
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Activities
6
Use Pythagoras’s theorem to find the perpendicular heights, and then the
areas, of these triangles. Round numbers to one decimal place where
appropriate.
a
5 cm
h
2 cm
h² = 5² – 3²
= 25 – 9
= 16
h = √16 = 4 cm
area =
1
2
× (2 + 3) × 4
=
1
2
×5×4
= 10 cm²
3 cm
b
h² = 7² – 6²
= 49 – 36
= 13
h = √13 = 3.6 cm
7 cm
h
6 cm
2 cm
area =
1
2
× (6 + 2) × 3.6
=
1
2
× 8 × 3.6
= 14.4 cm²
c
4 cm
h
4 cm
2.5 cm
h² = 4² – 2.5²
= 16 – 6.25
= 9.75
h = √9.75 = 3.1 cm
area =
1
2
× (4 + 2.5) × 3.1
=
1
2
× 6.5 × 3.1
= 10.1 cm²
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