1B_Ch7(1). 7.2Surface Areas and Volumes of Prisms A What is a Prism? Index 1B_Ch7(2) B Total Surface Areas of Prisms C Volumes of Prisms

1B_Ch7(1)

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


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


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


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


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


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


(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


= (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


Back to Question

Index


(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


(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


(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


(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.


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.


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.


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


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?


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


=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


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.


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


= (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


= (2 × 48 + 31h) m2

= (96 + 31h) m2

Therefore 96 + 31h = 220

31h = 124

h = 4

∴ The height of the flat is 4 m.


Application questions. Key Concept 7.2.2

Volumes of Prisms

‧ Volume of prism = area of base × height


Index

1B_Ch7(26)

C)

Example

Index 7.2

Find the volumes of the rectangular prism and the cube as shown.

Index


(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.


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



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.


65 cm 65 cm

1.8 m

1.2 m

1 m

Index

Back to Question


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?


50 m

1.2 m

5 m

25 m

Index

Back to Question


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


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

Documents

1B_Ch7(1). 7.2Surface Areas and Volumes of Prisms A What is a Prism? Index 1B_Ch7(2) B Total Surface Areas of Prisms C Volumes of Prisms