1b_ch7(1). 7.2surface areas and volumes of prisms a what is a prism? index 1b_ch7(2) b total surface...
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7.2 Surface Areas and Volumes of Prisms
A What is a Prism?
Index
1B_Ch7(2)
B Total Surface Areas of
Prisms
C Volumes of Prisms
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.
7.1 Areas of Simple Polygons
Index
1B_Ch7(3)
Areas of Simple Polygons
‧ Then the areas of the polygons can be obtained by:
7.1 Areas of Simple Polygons
Index
Example
1B_Ch7(4)
1. adding up the areas of the simple figures;
2. considering the difference between the areas of
the simple figures.
Divide the polygons into simple figures following the
information in the bracket.
Index
7.1 Areas of Simple Polygons 1B_Ch7(5)
(a) (b)
(3 triangles) (1 triangle, 2 trapeziums)
Find the area of the pentagon
ABCDE in the figure.
Index
7.1 Areas of Simple Polygons 1B_Ch7(6)
3 cmA B
C
DE
4 cm
8 cm
7 cm
【 The dotted line joining B and F in the figure divides the pentagon into one rectangle and one trapezium as shown. 】
3 cmA B
C
DE
4 cm
8 cm
3 cm 4 cm
F
Index
7.1 Areas of Simple Polygons 1B_Ch7(7)
3 cmA B
C
DE
4 cm
8 cm
3 cm 4 cm
F
= (24 + 24) cm2
= 48 cm2
Area of rectangle ABFE
= 8 × 3 cm2
= 24 cm2
Area of trapezium BCDF
= 24 cm2
Area of pentagon ABCDE Fulfill Exercise Objective
Use addition of areas to find the
area of a polygon.
Back to Question
= (8 + 4) × 4 cm2
21
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?
Index
7.1 Areas of Simple Polygons 1B_Ch7(8)
Shop A
Shop B Shop C
4.5 m 4 m
4 m5 m
6 m
4 m
13 m
Floor Plan9 m 7 m
Index
7.1 Areas of Simple Polygons 1B_Ch7(9)
Area of shop A 2 m
4.5 m
9 m 7 m
Height of triangle = (9 – 7) m = 2 m
4.5 m
5 m
6 m
4 m4 m
Length of the left rectangle= (5 + 4) m = 9 m
= m2
25.4
21
5.47
= (31.5 + 4.5) m2
= 36 m2
Area of shop B = (9 × 4.5 + 6 × 4) m2
= (40.5 + 24) m2
= 64.5 m2
Back to Question
Index
7.1 Areas of Simple Polygons 1B_Ch7(10)
Fulfill Exercise Objective
Application questions.
4 m
4 m
4 m
9 m
13 m
Height of parallelogram= (13 – 4) m = 9 m
Area of shop C = (4 × 4 + 4 × 9) m2
= (16 + 36) m2
= 52 m2
∴ Shop B meets Mrs Yeung’s requirement.
Back to Question
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.
Index
7.1 Areas of Simple Polygons 1B_Ch7(11)
(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?
3.5 m
4 m
wardrobe
Wall1 m
2 m
Now Suk-yee needs to paint the wall
(but not that part of the wall covered by the wardrobe).
Index
7.1 Areas of Simple Polygons 1B_Ch7(12)
= (14 – 2) m2
= 12 m2
(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
(b) Volume of paint required = litres212
= 6 litresFulfill Exercise Objective
Application questions.
Back to Question
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’.
0.3 m
0.3 m
0.3 m
0.3 m
0.3 m
1.2 m
0.4 m
0.4 m
1.5 m
Soln
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.
( Note: If the area of coloured paper required is less than 1
m2, the cost is still $15. )
Soln
Index
7.1 Areas of Simple Polygons 1B_Ch7(15)
(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
Back to Question
Index
7.1 Areas of Simple Polygons 1B_Ch7(16)
(b) Area of the Chinese character formed
= 2 × 1.56 m2
= 3.12 m2
i.e. The area is (3 + 0.12) m2.
∴ Total cost = $(3 × 15 + 15)
= $60
3 m2 of coloured paper will cost $3 × 15 and 0.12 m2 of
coloured paper will still cost $15.
Fulfill Exercise Objective
Application questions. Key Concept 7.1.1
Back to Question
What is a Prism?
1. A solid with uniform cross-sections (which are polygons)
is called a prism.
7.2 Surface Areas and Volumes of Prisms
Index
1B_Ch7(17)
A)
2. The perpendicular distance between
the two parallel bases is called the
height of the prism. The faces of a
prism other than the two parallel
bases are called lateral faces.
base
base
height
lateralfaces
Example
Index 7.2
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 pris
m.
Index
Solid A Solid B Solid C
Solid A Key Concept 7.2.1
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(18)
Total Surface Areas of Prisms
‧ For any prism,
total surface area
= areas of the two bases + total area of all lateral faces.
7.2 Surface Areas and Volumes of Prisms
Index
1B_Ch7(19)
B)
Example
Index 7.2
Find the total surface area of a
square prism (cube).
Index
Total surface area
8 cm8 cm
8 cm
【 A cube is a prism with 6 equal faces. 】
= 8 × 8 × 6 cm2
= 384 cm2
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(20)
Index
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?
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(21)
6 cm
8 cm
10 cm
15 cmA journey of a thousand
leagues begins w
ith a
single ste
p.- L
ao Tzu
Index
Total surface area of one paperweight
Back to Question
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(22)
=2cm 1561581510
286
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 020Fulfill Exercise Objective
Application questions.
Index
The figures below show the floor plan, the three-
dimensional 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.
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(23)
9.5 m
4.2 m
4.5 m
6 m
floor plan three-dimensional plan
9.5 m
4.2 m
4.5 m
6 mh m
window
door
Index
Area of the floor
Back to Question
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(24)
= (6 × 4.5 + 5 × 4.2) m2
= 48 m2
Area of the ceiling = area of the floor
= 48 m2
9.5 m
6 m4.2 m
4.5 m
5 m
1.8 m
Total area of the walls = (6 + 4.5 + 1.8 + 5 + 4.2 + 9.5) × h m2
= 31h m2
Index
Total surface area of the flat
Back to Question
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(25)
= (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
Volumes of Prisms
‧ Volume of prism = area of base × height
7.2 Surface Areas and Volumes of Prisms
Index
1B_Ch7(26)
C)
Example
Index 7.2
Find the volumes of the rectangular prism and the cube as shown.
Index
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(27)
(a)
4 m2.5 m
3 m
(b)
3.1 cm3.1 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
The figure shows a victory-
stand in the shape of a prism.
Find its volume.
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(28)
2.4 m
0.8 m
0.5 m0.4 m
0.4 m
Area of base = (0.8 × 0.4 + 2.4 × 0.4) m2
= (0.32 + 0.96) m2
= 1.28 m2
0.8 m
0.4 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
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.
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(29)
65 cm 65 cm
1.8 m
1.2 m
1 m
Index
Back to Question
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(30)
Fulfill Exercise Objective Application questions.
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 = 1.35 × 1.8 m3
= 2.43 m3
Index
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?
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(31)
50 m
1.2 m
5 m
25 m
Index
Back to Question
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(32)
Original depth of water at the
shallow end
= (1.2 – 0.4) m
= 0.8 m
top
1.2 m
0.4 m
bottom
5 m
New depth of water at the shallow end = 0.8 × (1 + 25%) m
= 1 m
New depth of water at the deep end = [5 – (1.2 – 1)] m
= 4.8 m
Index
Back to Question
7.2 Surface Areas and Volumes of Prisms 1B_Ch7(33)
The water in the pool takes up the
shape of a prism. The cross-sections
of the water and the pool are shown
on the right.1 m
50 m
Area of base =2m 50)8.41(
2
1
= 145 m2
∴ Volume of water in the pool = 145 × 25 m3
= 3 625 m3
Fulfill Exercise Objective Application questions.
Key Concept 7.2.3