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
Student book sections
Foundation
Higher
Linear
31.1, 31.2, 31.3
19.1, 19.2
L
Modular
38.1, 38.2, 38.3
29.1, 29.2
Course Companion
Chapter 24
Chapter 14
Marks
Marks
lost (%) available
Possible errors
Not using Pythagoras’ theorem
There is no trigonometry on the Foundation tier, so that questions involving the calculation of a length in a
right-angled will require the use of Pythagoras’ theorem.
When faced with a right-angled triangle, students can also make the mistake of thinking that they have to
find its area.
Not knowing whether to add or subtract in Pythagoras’ theorem
When using Pythagoras’ theorem, students should first of all identify the hypotenuse of the triangle as the side
opposite the right angle and write down Pythagoras’ theorem as it applies to the given triangle.
To work out the length of YZ in this triangle, students
should identify YZ as the hypotenuse and write down:
YZ 2 = 3.22 + 1.72
YZ 2 = 10.24 + 2.89
YZ 2 = 13.13
X
13.13 = 3.6235... = 3.62 cm (to 2 d.p.)
To work out the length of AB in this triangle, students
should identify AC as the hypotenuse and write down
and then rearrange the equation to get:
Y
1.7 cm
Students sometimes leave this as their answer and do not find the square root.
In this case, YZ =
3.2 cm
Z
Diagram NOT
accurately drawn
A
12.62 = AB 2 + 4.72
158.76 = AB 2 + 22.09
12.6 cm
AB 2 = 158.76 – 22.09
AB 2 = 136.67
AB = 136.67 = 11.690... = 11.7 cm (to 1 d.p.)
B
4.7 cm
C
In this case, students should check that their answer is
not more than 12.6 cm, as the hypotenuse is the longest side in the triangle.
Using this approach means that the first line in the working always has a ‘+’ sign, so that students will gain a
method mark before they have to decide whether to add or subtract.
In questions requiring the use of Pythagoras’ theorem, students sometimes add or subtract the lengths, without
first squaring them.
Making a scale drawing or taking measurements from the diagram in the question
The instructions in a question to “Work out” or “Calculate” mean that working or a calculation is required.
Students must not try to answer the question by making a scale drawing. Similarly, the statement beside each
diagram, “Diagram NOT accurately drawn”, means that the length required cannot be found by measuring the
length on the diagram.
Rounding errors and premature approximation
When the answer has to be corrected to a given degree of accuracy, students should be advised to write down
an unrounded answer, showing at least 4 figures of their calculator display, before correcting their answer to the
required degree of accuracy. Also, students should work with the figures on their calculator display, rounding only
at the final stage.
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Now try these
When the length the of side that has been worked out is not an exact value, the degree of accuracy required is
given in the question.
Questions 1, 2, 5 and 6 – routine calculation of the length of the hypotenuse of a right-angled triangle. In
Question 2, the triangle is in a ‘non-familiar’ position
Questions 3, 4 and 7 – routine calculation of the length of a side which is not the hypotenuse. In Question 4, the
triangle is in a ‘non-familiar’ position.
Questions 8–10 – are more involved. However, a diagram is given in each case, and it is unlikely that, at this level,
a question without a diagram will be set.
Question 8 – an isosceles triangle. All the information that is needed is given in the question.
Question 9 – set in a context. The numbers involved are larger than in the other questions. The length of the
hypotenuse has to be found correct to the nearest kilometre.
Question 10 – set in a context. The length of a side, which is not the hypotenuse, has to be found correct to
2 decimal places.
Question 5, 6, 7, 8, 9 and 10 are past GCSE questions.
METHOD
Picking up marks in the exam
Finding the hypotenuse
Example:
Q6
Diagram NOT
accurately drawn
P
5.7 cm
Q
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.
R
9.3 cm
METHOD
(1388 November 2005)
1 mark for finding the square
root of (5.72 + 9.32).
PR2 = 5.72 + 9.32
PR2 = 32.49 + 86.49
PR2 = 118.98
Write down at least 4 figures
of the calculator display.
PR = 118.98 = 10.907795...
PR = 10.9 cm (to 1 d.p.)
1 mark for the correct answer.
Solving problems using Pythagoras’ theorem
Example:
Q8
1 mark for this working
showing use of Pythagoras’
theorem.
ABC is a triangle
AB = AC = 13 cm.
BC = 10 cm.
M is the midpoint of BC.
Angle AMC = 90°.
Work out the length of AM.
A
METHOD
Diagram NOT
accurately drawn
13 cm
B
1 mark for realising that
BM = 5 cm. (This may be
marked on the diagram or
implied in later working.)
13 cm
M
10 cm
C
(4400 November 2006)
BM
132
169
AM2
AM2
AM
AM
= 5 cm
= 52 + AM2
= 25 + AM2
= 169 – 25
= 144
= 144
= 12 cm
METHOD
1 mark for this rearrangement.
METHOD
1 mark for finding the square
root of (132 – 52).
1 mark for the correct answer.
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