21
528 Shape 6
Pythagoras’
theorem
4
22
Shape
6
This chapter will show you how to:
✔ understand Pythagoras’ theorem
✔ calculate the longest side (the hypotenuse) of a rightangled triangle
✔ calculate one of the shorter sides of a right-angled
triangle
✔ apply Pythagoras’ theorem to solving ‘real’ problems
21.1 Pythagoras’ theorem
You will need to know:
● how to use your calculator
● how to square numbers
● how to find square roots
Key words:
Pythagoras’ theorem
right-angled triangle
hypotenuse
Pythagoras’ theorem only applies to right-angled
triangles .
A right-angled triangle has one angle of 90°.
shows a right angle.
This right-angled triangle has sides a 4 cm, b 3 cm
and c 5 cm.
c5
a3
c is the hypotenuse.
This is the right angle.
b4
In a right-angled triangle the longest side is called
the hypotenuse .
The hypotenuse is always opposite the right angle.
4
3
9
a2b
11 Pythagoras’ theorem
529
You can construct a square on each side of the triangle.
Area C 5 5 52 25
C
A
a3
c5
Area A 3 3 32 9
b4
B
Area B 4 4 42 16
Area A Area B 9 cm2 16 cm2 25 cm2 Area C
Area A Area B Area C
42 52
32 In other words a2 b2 c2
Use squared paper to check
this using different
triangles.
This leads to Pythagoras’ theorem:
In any right-angled triangle the square of the
hypotenuse (c2) is equal to the sum of the squares on
the other two sides (a2 b2).
For a right-angled triangle
with sides of lengths a, b and c,
a
where c is the hypotenuse,
Pythagoras’ theorem states
that a2 b2 c2
‘Sum’ means ‘add’.
c
b
Exercise 21A
1 In the following triangles, write the letter that
represents the hypotenuse.
(a)
b
c
a
(b)
(c)
i
j
g
f
k
h
530 Shape 6
2 For the triangles above, write out the formula using
Pythagoras’ theorem.
Make the longest side (the hypotenuse) the subject of
the formula.
The subject appears on its
own on one side of the
equals sign.
21.2 Finding the hypotenuse
You can use Pythagoras’ theorem to find the length of the
hypotenuse.
Example 1
A
Work out the length
of the hypotenuse
(the side marked x) in
this right-angled triangle.
x
5 cm
C
a2 b2 c2
c a b
2
2
2
B
12 cm
Write out Pythagoras’
theorem and then
substitute in the values
given.
c
a
x2 122 52
b
x2 144 25
x2 169
x 169
x 13 cm
Example 2
In the triangle PQR angle Q 90°,
QP 4 cm and QR 7 cm.
Calculate y, the length of PR,
to 1 decimal place.
c2 a2 b2
y2 72 42
a
c
y2 49 16
y2 65
y 65
y 8.06 8.1 cm (to 1 dp)
b
Q
4 cm
7 cm
y
R
P
PR is the hypotenuse,
because it is opposite the
right angle.
Use your calculator to find
the square root.
Always put the units in
your answer.
11 Pythagoras’ theorem
531
Example 3
In this isosceles triangle EF EG, M is the mid-point of
the base line FG and EM 3.2 cm.
Work out the slant length of the triangle (marked x in the
diagram) to 2 dp.
E
x
G
Split the triangle EFG into
two right-angled triangles:
3.2 cm
x
F
M
8.4 cm
Use either one to find x.
E
c2 a2 b2
x
x 4.2 3.2
2
2
c
2
x 17.64 10.24
b
G
2
a
y2 27.88
3.2 cm
M
1
2
GM 8.4 cm 4.2 cm.
y 27.88
On a calculator
27.88
5.280 15 …
which is 5.28 to 2 dp.
x 5.28 cm (to 2 dp)
Exercise 21B
1 Calculate the length marked with a letter in each
triangle.
P
A
(a)
(b)
(c)
12 cm
r
9 cm
6 cm
t
Q
s
C
12 cm
B
8 cm
16 cm
R
2 Calculate the lengths marked with letters to 1 dp.
(a)
(b) 4 cm
(c)
a
b
6 cm
11 cm
(d)
8 cm
14 cm
9 cm
5 cm
d
c
8 cm
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3 A rectangular gate measures
3.5 m long by 1.5 m high.
Work out the length of wood
needed to make the diagonal.
3.5 m
1.5 m
4 An extendable ladder rests against a
vertical wall just below a window.
PR 2.6 m and the foot of the ladder is
1.2 m away from the wall (RQ).
How long does the ladder need to be to
reach the window?
P
R
Q
5 Fiona is flying her kite in a strong breeze. The kite is
flying at a height of 18 m and is 20 m away
horizontally from Fiona. How long is the kite of string?
UAM
For questions 5 and 6,
sketch a diagram and
label it.
6 A boat sails due east for 24 km then due south for 10 km.
How far is the boat from its starting point?
N
W
E
S
Key words:
subject
21.3 Finding the length of a shorter
side
In Pythagoras’ theorem: a2 b2 c2.
To work out the length of the shorter side a,
make a, the subject of the equation.
a2 b2 c2
a2 b2 b2 c2 b2
a2 c2 b2
c
a
For help with
changing the subject
of an equation see
Chapter 7.
b
Subtract b2 from both sides
of the equation.
You can make b the subject in the same way.
b2 c2 a2
You can calculate the lengths of the shorter
c
sides of a right-angled triangle using
2
2
2
● a c b
● b2 c2 a2
b
a
11 Pythagoras’ theorem
533
Example 4
In this triangle b 4 cm and c 7 cm.
Calculate the length a (to 1 dp).
c
b
c is the longest side (the
hypotenuse).
a
To find a Shorter side you
Subtract squares of sides.
Using the Pythagoras equation:
a2 c2 b2
If you add by mistake you
get
a2 72 42
a2 49 16
a2 49 16
a2 65
a 65
8.06 …
a2 33
a 33
This is impossible because
it is longer than the
hypotenuse.
a 5.744 5.7 cm (1 dp)
Example 5
The end face of a chocolate bar is in the shape of an
isosceles triangle EFG. The mid-point of side EG is H.
Using the dimensions shown on the diagram work out
the base length EG of the chocolate bar (to 1 dp).
F
2.7 cm
E
a
1.8 cm
H
G
Triangle EHF is a right-angled triangle, so use Pythagoras’
theorem.
F
a2 c2 b2
a2 2.72 1.82
a2 7.29 3.24
a2 4.05
a 4.05
a 2.012
EG 2a 4.024 cm 4.0 cm (to 1 dp)
2.7 cm
c
E
a
1.8 cm
b
a
H
Shorter side → Subtract.
Multiply by 2 on your
calculator before rounding
your final answer.
534 Shape 6
Exercise 21C
1 Calculate the lengths marked with letters in these triangles.
Give your answers to 1 d.p.
(a)
(b)
(c)
4 cm
2 cm
9 cm
13 cm
7 cm
y
UAM
z
6 cm
x
2 Calculate the lengths (a) XY
Give your answers to 1 dp.
(b) QR.
Y
Q
20.2 cm
6.2 cm
X
P
16.4 cm
R
5.8 cm
Z
UAM
UAM
UAM
3 A children’s slide is 3.6 m long.
The vertical height of the slide above the ground is 2.1 m.
Work out the horizontal distance between each end
of the slide.
4 Jotinder walks 120 m up a gentle slope. From the map he
sees he has walked a horizontal distance of only 95 m.
Work out the height he has climbed.
5 A ladder 5.2 metres long is
leaning against a wall.
The top of the ladder is
4.6 metres up the wall.
How far from the bottom of the
wall is the foot of the ladder?
For questions 3 to 6, sketch
a diagram and label it.
11 Pythagoras’ theorem
UAM
535
6 A ship sails 46 km North East.
Then it changes course and sails 62 km South East.
How far is the ship from its starting point?
7 Work out the lengths c and d on this diagram.
3 cm
4 cm
c
d
13 cm
8 The diagram shows the
cross-section of a shed.
Work out the width of
the shed, w, to 2 dp.
3.8 m
2.8 m
2.2 m
w
Examination questions
1 The diagram shows a right-angled triangle ABC.
AB 10 cm and AC 15 cm.
Calculate the length of BC.
Leave your answer as a square root.
A
C
Not drawn
to scale
15 cm
10 cm
B
(3 marks) C
2 A support for a flagpole is attached at a height
of 3 m and is fixed to the ground at a distance
of 1.2 m from the base.
Calculate the length of the support
(marked x on the diagram).
x
3m
Not drawn
to scale
1.2 m
(3 marks) C
536 Shape 6
3 A balloon is held to the ground by a
cable at point O.
The balloon is 30 m horizontally
from point O.
The balloon is held at a height of 75 m.
How long is the cable?
75 m
Not drawn
to scale
O
30 m
(3 marks) C
4 A square of side length a has a diagonal of 15 cm.
Calculate the value of a to 1 d.p.
a
Not drawn
to scale
a
(4 marks) C
Summary of key points
A right-angled triangle has one angle of 90° angle.
In a right-angled triangle the longest side is called the
hypotenuse.
The hypotenuse is always opposite the right-angle.
For a right-angled triangle with sides of lengths
a, b and c, where c is the hypotenuse,
Pythagoras’ theorem states that
c
a
a2 b2 c2.
You can calculate the lengths of the shorter sides
using
a2 c2 b2
b2 c2 a2
This work is grade C.
b