year 8 maths

120 Chapter 3 Measurement and Pythagoras’ theorem
Example 1
Converting length measurements
Convert these lengths to the units shown in the brackets.
a 5.2 cm (mm) b 85 000 cm ( km)
SOLUTION
a 5.2 cm = 5.2 × 10
= 52 mm
b 85 000 cm = 85 000 ÷ 100 ÷ 1000
= 0.85 km
EXPLANATION
1 cm = 10 mm so multiply by 10 .
×10
cm mm
1 m = 100 cm and 1 km = 1000 m , so divide by
100 and 1000 .
Example 2
Finding perimeters
Find the perimeter of this shape.
4 cm
3 cm
SOLUTION
P = 2 × (3 + 3) + 2 × 4
= 12 + 8
= 20 cm
EXPLANATION
6 cm
4 cm 3 cm
4 cm
3 cm
Example 3
Finding an unknown length
Find the unknown value x in this triangle if the perimeter is 19 cm .
x cm
P = 19 cm
5 cm
SOLUTION
2x + 5 = 19
2x = 14
x = 7
EXPLANATION
2x + 5 makes up the perimeter.
2x is the difference between 19 and 5 .
If 2x = 14 then x = 7 since 2 × 7 = 14 .
Cambridge Maths NSW
Stage 4 Year 8 Second edition
ISBN 978-1-108-46627-1 © Palmer et al. 2018
Cambridge University Press
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