Y7 MATHEMATICS
LESSON 1: PYTHAGORAS' THEOREM 1
1.
Establishing Pythagoras’ Theorem

The hypotenuse

Right-angled triangles contain one right (90°) angle. An example is shown below.
ℎ
𝑎
𝑏
–
In right-angled triangles, the longest side has a special name – the hypotenuse.
The hypotenuse is always opposite the right angle.
The right angle always says “Hi” to the hypotenuse.
–
In this triangle, the hypotenuse is the side marked ℎ.

Highlight the hypotenuse of the following triangles.
𝐵
𝑄
𝐴
𝑅
𝑃
𝐶
NOTE TO STUDENTS
The term ‘hypotenuse’ and the thereom you will learn shortly, ‘Pythagoras’ Theorem’,
only apply to right-angled triangles!
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Y7 MATHEMATICS

LESSON 1: PYTHAGORAS' THEOREM 1
Pythagoras’ theorem

Pythagoras’ theorem is named after the Greek philosopher and mathematician
Pythagoras. He was the first to offer a proof of the theorem around 569 BC–500 BC.

The theorem states:
The areas of the squares that are created by the side lengths of the two shorter
sides of a right-angled triangle will add up to fit exactly into the area of the square
formed by the side length of the hypotenuse.
𝒉𝟐 = 𝒂𝟐 + 𝒃𝟐
The square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on
the other two sides.

Pythagoras’ theorem is a very important theorem in mathematics.
–
The theorem allows you to calculate the side lengths of right angled and nonright angled triangles (by constructing perpendicular lines).
DISCUSSION QUESTION
Define perpendicular lines and draw an example.
……………………………………………………………………………………………………………
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Y7 MATHEMATICS

LESSON 1: PYTHAGORAS' THEOREM 1
Proof of Pythagoras’ theorem

Although Pythagoras’ theorem is now seen as an algebraic statement about lengths,
it is traditionally a statement about areas. Notice the key word “on”.
The square on the hypotenuse of a right-angled triangle is equal to the sum of the
squares on the other two sides.

Let’s revisit the right-angled triangle we saw at the beginning of this lesson.
–
A square on the hypotenuse would look like this. The hypotenuse is of length ℎ,
therefore the area of the square built on it would be ℎ2 .
–
Similarly, we can build squares on the two shorter sides as well.
–
The theorem states that the large square has the same area as the areas of
the small and medium squares combined.
–
This is commonly expressed in a diagram like that shown below.
ℎ2
𝑎2
𝑏2
–
There are many ways to show that this is true! Check out the following videos.
VIDEO (Length: 0.43): Wheel with liquid.
VIDEO (Length: 1:03): Puzzle pieces.
VIDEO (Length: 3:13): Origami.
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Y7 MATHEMATICS
LESSON 1: PYTHAGORAS' THEOREM 1
INVESTIGATION QUESTION
Using this statement, draw a square on each side of the right angled triangle with
hypotenuse length 5 cm, and the two other sides with length 3 cm and 4 cm.
Use a ruler.
Calculate the area of each square. Does this satisfy Pythagoras’ Theorem, ℎ2 = 𝑎2 + 𝑏 2 ?
……………………………………………………………………………………………………………
DISCUSSION QUESTION
Does it matter which short side is labelled 𝑎 and 𝑏?
[1]
……………………………………………………………………………………………………………
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Y7 MATHEMATICS
LESSON 1: PYTHAGORAS' THEOREM 1
2.
Using Pythagoras’ Theorem

Finding the hypotenuse

Pythagoras’ theorem is essentially an equation, 𝒉𝟐 = 𝒂𝟐 + 𝒃𝟐 .
Step 1:
Identify the hypotenuse ℎ .
Step 2:
Substitute the given values of the triangle into the formula ℎ2 = 𝑎2 + 𝑏 2
Step 3:
Solve for ℎ
ℎ = √𝑎2 + 𝑏 2
NOTE TO STUDENTS:
Don’t forget to write your measuring unit and to square root your answer at the end! This is
a very common step that students forget to do.
Example 1:
Find the length of the hypotenuse of the following triangle.
6 𝑐𝑚
8 𝑐𝑚
𝑥
Solution:
Identify the hypotenuse and the shorter sides
The hypotenuse is 𝑥 and the two shorter sides are 𝑎 = 6 and 𝑏 = 8.
Substitue the values into Pythagoras’ theorem and solve
ℎ2 = 𝑎2 + 𝑏 2
(Pythagoras’ theorem)
𝑥 2 = 62 + 82
= 36 + 64
= 100
∴ 𝑥 = √100
= 10 cm
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Y7 MATHEMATICS
LESSON 1: PYTHAGORAS' THEOREM 1
Example 2:
Find the length of the hypotenuse of the following triangle.
𝑙
𝑚
Solution:
𝑛
Identify the hypotenuse and the shorter sides
The hypotenuse is 𝑙 and the two shorter sides are 𝑚 and 𝑛.
Substitue the values into Pythagoras’ theorem and solve
ℎ2 = 𝑎2 + 𝑏 2
[2]
(Pythagoras’ theorem)
𝑙 2 = 𝑚2 + 𝑛2
∴𝑙=
NOTE TO STUDENTS
As can be seen in this example, the hypotenuse will not always be labelled with the
pronumeral ℎ!
Make sure you re-write Pythagoras’ theorem in terms of the pronumerals given to
you in the diagram.
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Y7 MATHEMATICS
LESSON 1: PYTHAGORAS' THEOREM 1
Concept Check 2.1
Find the exact length of the hypotenuse of the following right-angled triangles, by expressing in the
form ℎ2 = 𝑎2 + 𝑏 2 . Leave your answer with the √
(a)
sign.
[3]
…………………………………………………………………
𝐴
7 𝑐𝑚
…………………………………………………………………
…………………………………………………………………
𝑥 𝑐𝑚
…………………………………………………………………
…………………………………………………………………
𝐵
24 𝑐𝑚
𝐶
…………………………………………………………………
…………………………………………………………………
(b)
…………………………………………………………………
[4]
3 𝑚𝑚
…………………………………………………………………
…………………………………………………………………
4 𝑚𝑚
…………………………………………………………………
𝑘
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
(c)
[5]
…………………………………………………………………
…………………………………………………………………
𝑦
15 𝑚
…………………………………………………………………
…………………………………………………………………
8𝑚
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
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Y7 MATHEMATICS
LESSON 1: PYTHAGORAS' THEOREM 1
Concept Check 2.2
In the diagram, 𝑃𝑄𝑅𝑆 is a rectangle with sides 9 cm and 7 cm. Find the length of the diagonal
(a)
𝑆𝑄 to the nearest mm.
[6]
…………………………………………………………………
𝑄
𝑃
…………………………………………………………………
7 𝑐𝑚
…………………………………………………………………
…………………………………………………………………
𝑆
𝑅
9 𝑐𝑚
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
𝐷𝐸𝐹𝐺 is a trapezium. Find the length of 𝐸𝐹 to 1 decimal place. [7]
(b)
12 𝑐𝑚
𝐷
…………………………………………………………………
𝐸
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
𝐹
18 𝑐𝑚
𝐺
…………………………………………………………………
…………………………………………………………………
(c)
𝐴𝐵𝐶𝐷 is a square of side length 8 cm. Find the exact length of the diagonal 𝐴𝐶.
𝐴
[8]
𝐵
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
𝐷
𝐶
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
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Y7 MATHEMATICS

LESSON 1: PYTHAGORAS' THEOREM 1
Finding a shorter side

To find the length of a shorter side, you must know the length of the hypotenuse and
the other short side.

Again, you are substituting values and solving. In this case, however, there is an
additional step in the solution as terms need to be re-arranged by changing the
subject.
INVESTIGATION QUESTION (Changing the Subject)
(a)
If the formula ℎ2 = 𝑎2 + 𝑏 2, find the equation where 𝑎 is the subject.
𝒉𝟐 = 𝒂𝟐 + 𝒃𝟐
−𝒃𝟐
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………
(b)
If the formula ℎ2 = 𝑎2 + 𝑏 2, find the equation where 𝑏 is the subject.
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………

Using this process of changing the subject, we can now apply this to specific
Pythagoras questions, to solve the length of the other sides.
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Y7 MATHEMATICS
LESSON 1: PYTHAGORAS' THEOREM 1
Example:
Find the length of the hypotenuse of the following triangle.
𝑥 𝑐𝑚
8 𝑐𝑚
13 𝑐𝑚
Solution:
Identify the hypotenuse and the shorter sides
The hypotenuse is 13 𝑐𝑚 and the two shorter sides are 𝑥 and 8 𝑐𝑚
Substitue the values into Pythagoras’ theorem and solve
132 = 𝑥 2 + 82
169 = 𝑥 2 + 64
𝑥 2 = 169 − 64
= 105
∴ 𝑥 = √105 𝑐𝑚
Concept Check 2.3
Find the exact length (leave your answer with the √
sign) of the missing side of the following right-
angled triangles.
(a)
[9]
…………………………………………………………………
…………………………………………………………………
13 𝑐𝑚
𝑘
…………………………………………………………………
…………………………………………………………………
16 𝑐𝑚
…………………………………………………………………
…………………………………………………………………
(b)
𝑥 2 + 162 = 192 , measured in centimetres. Solve for 𝑥.
……………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………
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Y7 MATHEMATICS
(c)
LESSON 1: PYTHAGORAS' THEOREM 1
Hypotenuse = 10𝑚𝑚
Short side = 6𝑚𝑚
…………………………………………………………………
Other side= 𝑏 𝑚𝑚
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
(d)
[10]
…………………………………………………………………
8 𝑚𝑚
25 𝑚𝑚
𝑛
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
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