33
Pythagoras’ theorem
SKILL
EXAM FACTS
Use Pythagoras’ theorem to find one side of a
right-angled triangle, given the lengths of the
other two sides
H
M
L
KEY FACTS
• In a right-angled triangle the side opposite the right
angle (the longest side of the triangle) is called the
hypotenuse of the triangle.
c
b
Marks
lost (%)
Marks
available
• In the diagram the length of the hypotenuse is c.
• Pythagoras’ theorem states that the area of the
square on the hypotenuse is equal to the sum of the
areas of the squares on the other two sides.
a
D
• That is c2 = a2 + b2
• In triangle DEF, Pythagoras’ theorem gives
DE2 = EF 2 + DF 2
F
E
• DE2 means that the length of the side DE is squared.
Getting it right
EXAM TIP
D
In triangle DEF
6.4 cm
FE = 8.7 cm,
F
DF = 6.4 cm,
8.7 cm
Angle DFE = 90°.
Calculate the length of DE.
Give your answer correct to 1 decimal place.
DE2 = EF2 + DF2
DE 2 = 8.72 + 6.42
DE 2 = 75.69 + 40.96
96
AIMH_C33.indd 96
Diagram NOT
E accurately drawn
“Diagram NOT accurately
drawn” means that taking
measurements from the
diagram will not give the
correct answer.
Identify the hypotenuse
of the triangle (the side
opposite the right-angle).
Then write down Pythagoras’
theorem for the triangle.
Substitute the given lengths.
You would get 1 mark
for this.
Pythagoras’ theorem
13/6/07 09:00:29
WARNING
DE2 = 116.65
DE =
!
A common error is to fail to
find the square root and give
the answer as 116.65
116.65 = 10.80046…
DE = 10.8 cm
Remember to round your
answer to 1 decimal place
and write the units.
PQR is a right-angled triangle.
Angle PQR = 90°.
QR = 15 cm.
PR = 19 cm.
Work out the length of PQ.
Give your answer correct to 1 decimal place.
P
Q
19 cm
Diagram NOT
accurately drawn
15 cm
WARNING
R
(1388 November 2005)
PR2
192
361
361 − 225
PQ2
PQ
= PQ2 + QR2
= PQ2 + 152
= PQ2 + 225
= PQ2
= 136
= 136 = 11.6619...
PQ = 11.7 cm
!
The side to be found is not
opposite the right-angle so it
is NOT the hypotenuse.
A common error is to write
incorrectly PQ2 = 152 + 192
This gives PQ = 24.2 which
is not sensible, as PQ
must be shorter than the
hypotenuse, PR.
Now try these
In Questions 1–4, work out the lengths of the sides marked with letters.
The diagrams are not accurately drawn.
Give each answer correct to 1 decimal place.
1
7 cm
a
12 cm
2
3
10.3 cm
9 cm
4 cm
b
4.6 cm
c
5.8 cm
4
d
13.6 cm
Pythagoras’ theorem
AIMH_C33.indd 97
97
13/6/07 09:00:32
P
5 PQR is a right-angled triangle.
PR = 6 cm. QR = 4 cm
Work out the length of PQ.
Give your answer correct to 1 decimal place.
Diagram NOT
accurately drawn
6 cm
Q
6 In triangle PQR
QR = 9.3 cm. PQ = 5.7 cm. Angle PQR = 90°.
Calculate the length of PR.
Give your answer correct to 1 decimal place.
4 cm
R
(1387 June 2006)
P
Diagram NOT
accurately drawn
5.7 cm
Q
R
9.3 cm
(1388 November 2005)
Diagram NOT
accurately drawn
7 Work out the value of x.
7.5 cm
x cm
A
7.2 cm
(4400 May 2006)
8 ABC is a triangle.
AB = AC = 13 cm.
BC = 10 cm.
M is the midpoint of BC.
Angle AMC = 90°.
13 cm
13 cm
Diagram NOT
accurately drawn
B
M
10 cm
Work out the length of AM.
9 The diagram shows three cities.
Norwich is 168 km due East of Leicester.
York is 157 km due North of Leicester.
Calculate the distance between Norwich and York.
Give your answer correct to the nearest kilometre.
C
(4400 November 2006)
York
Diagram NOT
accurately drawn
157 km
Leicester
Norwich
168 km
N
(1387 November 2006)
10 The diagram shows the positions of three
telephone masts A, B and C.
Mast C is 5 kilometres due East of Mast B.
Mast A is due North of Mast B and
8 kilometres from Mast C.
A
Diagram NOT
accurately drawn
N
8 km
Calculate the distance of A from B.
Give your answer in kilometres, correct to
2 decimal places.
B
98
Pythagoras’ theorem
AIMH_C33.indd 98
5 km
C
(1385 June 1999)
Pythagoras’ theorem
98
13/6/07 09:00:33