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1B_Ch7(1)
7.2 Surface Areas and Volumes
of Prisms
1B_Ch7(2)
A What is a Prism?
B Total Surface Areas
of Prisms
C Volumes of Prisms
Index
7.1 Areas of Simple Polygons
1B_Ch7(3)
Areas of Simple Polygons
‧ The areas of some polygons can be found by dividing
them into simple plane figures including triangles,
squares, rectangles, parallelograms or trapeziums. The
areas of these simple plane figures can be found easily
by formulas.
Index
7.1 Areas of Simple Polygons
1B_Ch7(4)
 Example
Areas of Simple Polygons
‧
Then the areas of the polygons can be obtained by:
1. adding up the areas of the simple figures;
2. considering the difference between the areas of
the simple figures.
Index
7.1 Areas of Simple Polygons
1B_Ch7(5)
Divide the polygons into simple figures following the information
in the bracket.
(a)
(b)
(3 triangles)
(1 triangle, 2 trapeziums)
Index
7.1 Areas of Simple Polygons
1B_Ch7(6)
Find the area of the pentagon
A 3 cm B
ABCDE in the figure.
C
8 cm
4 cm
E
A 3 cm B
C
8 cm
7 cm
D
【 The dotted line joining B and F in the
figure divides the pentagon into one
rectangle and one trapezium as shown. 】
4 cm
E
D
F
3 cm
4 cm
Index
7.1 Areas of Simple Polygons
1B_Ch7(7)
 Back to Question
Area of rectangle ABFE
A 3 cm B
= 8 × 3 cm2
= 24 cm2
C
8 cm
4 cm
Area of trapezium BCDF
1
= (8 + 4) × 4 cm2
2
= 24 cm2
E
D
F
3 cm
4 cm
Area of pentagon ABCDE
Fulfill Exercise Objective
= (24 + 24) cm2
 Use addition of areas to find the
area of a polygon.
= 48 cm2
Index
7.1 Areas of Simple Polygons
1B_Ch7(8)
Mrs Yeung wants to rent a shop in a shopping mall and
she wants the floor area of the shop to be at least 60 m2.
There are three shops available for rent and their sizes
are shown in the floor plan on the right. Which shop
meets Mrs Yeung’s requirement?
9 m Shop
A
Floor Plan
7m
4.5 m
4m
5m
6m
Shop B 4 m
4m
Shop C
13 m
Index
7.1 Areas of Simple Polygons
1B_Ch7(9)
 Back to Question
1
Area of shop A =  7  4.5   4.5  2  m2
2


= (31.5 + 4.5) m2
Area of shop B = (9 × 4.5 + 6 × 4) m2
= 64.5 m2
9m
7m
4.5 m
Height of triangle = (9 – 7) m
=2m
= 36 m2
= (40.5 + 24)
2m
4.5 m
5m
6m
m2
4m
4m
Length of the left rectangle
= (5 + 4) m = 9 m
Index
7.1 Areas of Simple Polygons
1B_Ch7(10)
 Back to Question
4m
Area of shop C = (4 × 4 + 4 × 9) m2
= (16 + 36) m2
= 52 m2
4m
9m
4m
13 m
Height of parallelogram
= (13 – 4) m = 9 m
∴ Shop B meets Mrs Yeung’s requirement.
Fulfill Exercise Objective
 Application questions.
Index
7.1 Areas of Simple Polygons
1B_Ch7(11)
4m
There is a wardrobe in Suk-yee’s
room. The back of the wardrobe
rests against one side of the wall as
shown in the diagram.
Now Suk-yee needs to paint the wall
Wall
1m
3.5 m
2m
wardrobe
(but not that part of the wall covered by the wardrobe).
(a) Find the area of the wall that Suk-yee needs to
paint.
(b) If one litre of paint can cover an area of 2 m2,
how many litres of paint does Suk-yee need?
Index
7.1 Areas of Simple Polygons
1B_Ch7(12)
 Back to Question
(a) Area of the wall = 4 × 3.5 m2
= 14 m2
Area of the wall covered by the wardrobe = 1 × 2 m2
= 2 m2
Area of the wall that Suk-yee needs to paint = (14 – 2) m2
= 12 m2
12
litres
2
= 6 litres
(b) Volume of paint required =
Fulfill Exercise Objective
 Application questions.
Index
7.1 Areas of Simple Polygons
1B_Ch7(13)
(a) From a large piece of rectangular paper of
dimensions 1.2 m × 1.5 m, Jenny cuts out a letter
‘E’ as shown by shading in the following figure.
Find the area of the letter ‘E’.  Soln
1.2 m
0.4 m
1.5 m
0.3 m
0.3 m
0.3 m
0.3 m
0.4 m
0.3 m
Index
7.1 Areas of Simple Polygons
1B_Ch7(14)
(b) Joseph has an identical letter ‘E’. Jenny and Joseph
put the two letters together to form a Chinese
character and they want to cover the Chinese
character with coloured paper. If the cost of each m2
of coloured paper is $15, find the total cost of
coloured paper for covering the Chinese character.
 Soln
( Note: If the area of coloured paper required is less than 1 m2,
the cost is still $15. )
Index
7.1 Areas of Simple Polygons
1B_Ch7(15)
 Back to Question
(a) Area of the rectangular paper
= 1.2 × 1.5 m2
= 1.8 m2
Total area of the two white rectangles
= (0.4 × 0.3) × 2 m2
= 0.24 m2
Area of the shaded letter ‘E’
= (1.8 – 0.24) m2
= 1.56 m2
Index
7.1 Areas of Simple Polygons
1B_Ch7(16)
 Back to Question
(b) Area of the Chinese character formed
= 2 × 1.56 m2
= 3.12 m2
i.e. The area is (3 + 0.12) m2.
3 m2 of coloured paper will cost $3 × 15 and 0.12 m2 of
coloured paper will still cost $15.
∴ Total cost = $(3 × 15 + 15)
= $60
Fulfill Exercise Objective
 Application questions.
 Key Concept 7.1.1
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(17)
 Example
A) What is a Prism?
1. A solid with uniform cross-sections (which are polygons)
is called a prism.
2. The perpendicular distance between
base
the two parallel bases is called the
height
height of the prism. The faces of a
prism other than the two parallel
bases are called lateral faces.
lateral
faces
base
 Index 7.2
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(18)
In each of the following solids, one of its cross-sections is indicated
with a coloured plane. Determine which one of these solids is a
prism.
Solid A
Solid A
Solid B
Solid C
 Key Concept 7.2.1
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(19)
 Example
B) Total Surface Areas of Prisms
‧
For any prism,
total surface area
= areas of the two bases + total area of all lateral faces.
 Index 7.2
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(20)
8 cm
Find the total surface area of a
8 cm
square prism (cube).
8 cm
【 A cube is a prism with 6 equal faces. 】
Total surface area = 8 × 8 × 6 cm2
= 384 cm2
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(21)
The figure shows a paperweight made with metal sheets.
It is in the shape of a triangular prism. If each m2 of the
metal sheet costs $250, what is the total cost of making
100 such paperweights?
6 cm
8 cm
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(22)
 Back to Question
Total surface area of one paperweight
 68
 2
= 2
  10 15  8 15  6 15 cm
  2 

= 408 cm2
Total surface area of 100 paperweights = 100 × 408 cm2
= 40 800 cm2
= 4.08 m2
∴ Total cost of making 100 paperweights = $250 × 4.08
= $1 020
Fulfill Exercise Objective
 Application questions.
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(23)
The figures below show the floor plan, the threedimensional plan and the dimensions of a flat. The
total surface area of the ceiling, floor and the walls
(including door and window) is 220 m2. Find the
height (h m) of the flat.
9.5 m
9.5 m
6m
window
4.2 m
6m
door
4.5 m
floor plan
hm
4.2 m
4.5 m
three-dimensional plan
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(24)
 Back to Question
9.5 m
Area of the floor = (6 × 4.5 + 5 × 4.2) m2
= 48 m2
Area of the ceiling = area of the floor
4.2 m
6m
5m
1.8 m
4.5 m
= 48 m2
Total area of the walls = (6 + 4.5 + 1.8 + 5 + 4.2 + 9.5) × h m2
= 31h m2
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(25)
 Back to Question
Total surface area of the flat = (2 × 48 + 31h) m2
= (96 + 31h) m2
Therefore
96 + 31h = 220
31h = 124
h =4
∴ The height of the flat is 4 m.
Fulfill Exercise Objective
 Application questions.
 Key Concept 7.2.2
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(26)
 Example
C) Volumes of Prisms
‧
Volume of prism = area of base × height
 Index 7.2
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(27)
Find the volumes of the rectangular prism and the cube as shown.
(a)
(b)
3.1 cm
3m
2.5 m
4m
3.1 cm
3.1 cm
(a) Volume of the rectangular prism = (4 × 2.5) × 3 m3
= 30 m3
(b) Volume of the cube = (3.1 × 3.1) × 3.1 cm3
= 29.791 cm3
Index
7.2 Surface Areas and Volumes of Prisms
The figure shows a victorystand in the shape of a prism.
Find its volume.
1B_Ch7(28)
0.8 m
0.4 m
0.5 m
0.4 m
2.4 m
Area of base = (0.8 × 0.4 + 2.4 × 0.4) m2
= (0.32 + 0.96) m2
0.4 m
2
= 1.28 m
0.8 m
0.4 m
2.4 m
∴ Volume of the victory-stand = 1.28 × 0.5 m3
= 0.64 m3
Fulfill Exercise Objective
 Application questions.
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(29)
The figure shows the dimensions of a container which is in
the shape of a prism. The base of the prism (the shaded
part) is a pentagon made up of a rectangle and a triangle.
Find the volume of the container if its total surface area is
10.8 m2.
1.8 m
1m
1.2 m
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(30)
 Back to Question
Total area of all the lateral faces
= [1.8 × 1.2 + 2 × (1.8 × 1) + 2 × (1.8 × 0.65)] m2
= 8.1 m2
Total area of the two bases = (10.8 – 8.1) m2
= 2.7 m2
Area of the base = (2.7 ÷ 2) m2
= 1.35 m2
∴ Volume of the container
Fulfill Exercise Objective
 Application questions.
= 1.35 × 1.8 m3
= 2.43 m3
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(31)
The figure shows the dimensions of a swimming pool
which is in the shape of a prism. The water level is
originally 40 cm below the top edges of the pool. But
this water level is considered too low and so water is
added until the depth at the shallow end has raised by
25%. What is the volume of water in the pool now?
50 m
25 m
1.2 m
5m
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(32)
 Back to Question
Original depth of water at the
top
0.4 m
shallow end
= (1.2 – 0.4) m
5m
1.2 m
bottom
= 0.8 m
New depth of water at the shallow end = 0.8 × (1 + 25%) m
=1m
New depth of water at the deep end = [5 – (1.2 – 1)] m
= 4.8 m
Index
7.2 Surface Areas and Volumes of Prisms
1B_Ch7(33)
 Back to Question
The water in the pool takes up the
shape of a prism. The cross-sections
of the water and the pool are shown on
50 m
1m
the right.
1
Area of base = (1  4.8)  50 m 2
2
= 145 m2
∴ Volume of water in the pool
= 145 × 25 m3
= 3 625 m3
Fulfill Exercise Objective
 Application questions.
 Key Concept 7.2.3
Index