year 8 maths

3D
Area of special quadrilaterals
137
Example 9
Finding areas of special quadrilaterals
Find the area of these shapes.
a
3 m 8 m
b
10 cm
20 cm
c
3 mm
5 mm
11 mm
SOLUTION
a A = bh
= 8 × 3
= 24 m 2
EXPLANATION
The height is measured at right angles to the base.
b
A = 1 2 xy
= 1 × 10 × 20
2
= 100 cm 2
Use the formula A = 1 xy since both diagonals are given.
2
This formula can also be used for a rhombus.
c A = 1 h(a + b)
2
= 1 × 5 × (11 + 3)
2
The two parallel sides are 11 mm and 3 mm in length.
The perpendicular height is 5 mm .
= 1 2 × 5 × 14
= 35 mm 2
Exercise 3D
UNDERSTANDING AND FLUENCY
1, 2, 3(½), 4 1, 3(½), 4
3(½), 4
1 Find the value of A using these formulas and given values.
a A = bh (b = 2, h = 3) b A = 1 xy (x = 5, y = 12)
2
c A = 1 2 h(a + b) (a = 2, b = 7, h = 3) d A = 1 h(a + b) (a = 7, b = 4, h = 6)
2
2 Complete these sentences.
a A perpendicular angle is __________ degrees.
b In a parallelogram, you find the area using a base and the __________________.
c The two diagonals in a kite or a rhombus are __________________.
d To find the area of a trapezium you multiply 1 by the sum of the two __________________
2
sides and then by the __________________ height.
e
The two special quadrilaterals that have the same area formula using diagonal lengths x and y
are the __________________ and the __________________.
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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