1. 9 Area and Volume [I] Case Study 9.1 Areas of Polygons
9.2 Volumes and Total Surface Areas of Prisms
Chapter Summary

2. Case Study ∵ Perimeter of rectangle  (Length  Width)  2
Here is a wire of length 20 cm.
Can you bend it into a rectangle
with the largest area?
Let me try.
∵ Perimeter of rectangle
 (Length  Width)  2
(Length  Width)  2  20
∴ Length  Width  10
Area (cm2)
Width (cm)
Length (cm) 1 9 25 24 21 16 9
Length  Width 2 8 3 7 4 6 5
∴ If the wire is bent into a square of side length 5 cm, it will attain the largest area of 25 cm2.

3. 9.1 Areas of Polygons
In primary level, we learnt how to find the area of different polygons by using the area-dissecting algorithm and the area-filling algorithm.
Area of the figure
 Area of square I  Area of triangle II
Area of the figure
 Area of rectangle  Area of trapezium IV

4. Example 9.1T 9.1 Areas of Polygons
In the figure, find the area of the polygon ABCDEF.
Solution: (area-dissecting algorithm)
From E, construct a line EG such that EG  BC.
There is more than one way to dissect the area. For example:
Area of the polygon
 Area of rectangle ABGF  Area of rectangle EGCD
 [6  3  4  (10  3)] cm2
 (18  28) cm2

5. Example 9.1T 9.1 Areas of Polygons Solution: (area-filling algorithm)
In the figure, find the area of the polygon ABCDEF.
Solution: (area-filling algorithm)
Extend the lines AF and CD to meet at K such that FK  DK.
Area of the polygon
 Area of rectangle ABCK  Area of rectangle FEDK
 [6  10  (6  4)  (10  3)] cm2
 (60 – 14) cm2

6. Example 9.2T 9.1 Areas of Polygons Solution:
In the figure, AB  6 cm, BC  8 cm and AC  10 cm.
Find the value of the unknown a. D
Solution:
Area of DABC with base BC
  BC  AB
Area of DABC with base AC
  AC  BD
We can first find the area of the triangle with base BC and height AB, and then use the result to find the value of a.
The area calculated by the two methods must be the same,
i.e., 5a  24

7. Example 9.3T 9.1 Areas of Polygons Solution:
In the figure, find the area of the polygon
ABCDEGH.
Solution:
Area of the polygon
 Area of rectangle ABCH  Area of parallelogram CDGH
 Area of triangle DEG
Figure CDGH is a parallelogram.
In this example, it is better for us to use the area-dissecting algorithm.
If we use the area-filling algorithm instead, the calculation will become more complicated and some information may not be given as well.

8. Example 9.4T 9.1 Areas of Polygons Solution:
The figure shows a metal plate.
(a) Find the area of the metal plate.
The cost of polishing the metal plate is 0.2/cm2.
Find the total cost of polishing the metal plate. X Y
Solution:
(a) Area of the metal plate
 Area of square  Area of triangle X  Area of triangle Y
(b) Total cost  $(0.2  112.5)

9. 9.2 Volumes and Total Surface Areas of Prisms
Prisms are 3-D solids with uniform cross-sections which are in form of a polygon.
For a prism,
 bases — 2 end faces
 height — distance between the 2 bases
 lateral faces — faces adjacent to the bases

10. 9.2 Volumes and Total Surface Areas of Prisms
We name a prism according to the shape of its base. For example:
Name
Shape
Base
Triangular prism
Triangle
Rectangular prism
Rectangle
Pentagonal prism
Pentagon
Hexagonal prism
Hexagon
Cubes and cuboids are kinds of rectangular prisms.

11. 9.2 Volumes and Total Surface Areas of Prisms
A. Volume of Prisms
In primary level, we learnt the formulas for finding the volumes of cubes and cuboids.
Volume of cube
 Length  Length  Length
 Base area  Height
Volume of cuboid
 Length  Width  Height
 Base area  Height
The volume of a prism equals to the product of its base area and its height.
Volume of a prism  Base area  Height
Note:
The unit of volume is cubic unit, that is mm3, cm3, m3, etc.

12. Example 9.5T 9.2 Volumes and Total Surface Areas of Prisms Solution:
A. Volume of Prisms
Example 9.5T
In the figure, find the volume of the prism.
Solution:
The base of the prism is formed by a square and a trapezium.
Base area cm2
Volume of the prism

13. Example 9.6T 9.2 Volumes and Total Surface Areas of Prisms Solution:
A. Volume of Prisms
Example 9.6T
In the figure, the volume of the block is 44 cm3. Find the value of x.
Solution:
The base of the block is formed by a square and a trapezium.
Volume of the block  Base area  Height

14. Example 9.7T 9.2 Volumes and Total Surface Areas of Prisms Solution:
A. Volume of Prisms
Example 9.7T
The figure shows a cup. Daniel fills the cup with water such that 1 cm of height is left unfilled.
(a) Find the volume of the water.
(b) Daniel adds 4 metal balls of volume 4 cm3 each
into the water. Find the rise in the water level.
Solution:
The base of the cup is formed by a rectangle and a trapezium.
Volume of the water

15. Example 9.7T 9.2 Volumes and Total Surface Areas of Prisms Solution:
A. Volume of Prisms
Example 9.7T
The figure shows a cup. Daniel fills the cup with water such that 1 cm of height is left unfilled.
(a) Find the volume of the water.
(b) Daniel adds 4 metal balls of volume 4 cm3 each
into the water. Find the rise in the water level.
Solution:
Let h cm be the rise in the water level.
Volume of metal balls  Volume of water rise
∴ The rise in the water level is 0.25 cm.

16. Example 9.8T 9.2 Volumes and Total Surface Areas of Prisms Solution:
A. Volume of Prisms
Example 9.8T
A piece of metal as shown in the figure is melted and recast into another rectangular metal.
(a) What is the volume of the metal?
(b) Find the value of a.
Solution:
(a) Volume of the metal
(b) Volume of the rectangular metal  (5  6  a) cm3

17. 9.2 Volumes and Total Surface Areas of Prisms
B. Total Surface Areas of Prisms
The total surface area of a prism is the sum of the areas of all its faces.
That is, the total surface area of a prism equals the total area of the 2 bases and the lateral faces.
Total surface area of a prism
 2  Base area  Total area of the lateral faces

18. Example 9.9T 9.2 Volumes and Total Surface Areas of Prisms Solution:
B. Total Surface Areas of Prisms
Example 9.9T
Find the total surface area of the prism.
Solution:
Area of the base of the prism  (5  3) cm2
left, right top, bottom
 15 cm2
Total area of the lateral faces  (3.5  8  2  5  8  2) cm2
 136 cm2
front, back
Total surface area of the prism  (15  2  136) cm2

19. Example 9.10T 9.2 Volumes and Total Surface Areas of Prisms Solution:
B. Total Surface Areas of Prisms
Example 9.10T
The figure shows an open box that was made from paper. Find the total area of the paper used to make it.
Solution:
Area of the trapezium
 36 cm2
Total area of 3 lateral faces  (5  12  6  12  5  12) cm2
 192 cm2
Total area of paper used  (36  2  192) cm2

20. Chapter Summary 9.1 Areas of Polygons
Areas of polygons can be calculated by the following 2 methods.
1. Area-dissecting algorithm
2. Area-filling algorithm
9.2 Volumes and Total Surface Areas of Prisms
1. Volume of a prism  Base area  Height
Total surface area of a prism
 2  Base area  Total area of the lateral faces