1. AREA AND VOLUME
Form 1 Mathematics
Chapter 7

2. Area of plane figures (p.2)
1. Area of triangle
=  base  height 1 2
2. Area of square
= length  length
3. Area of rectangle
= length  width

3. Area of plane figures (p.2)
4. Area of parallelogram
= base  height
5. Area of trapezium
=  sum of lengths of parallel sides  height 1 2

4. Calculating areas (p.3) Splitting Method Filling Method
Area of the figure 1 2
=  (3 + 6)  2 cm2
= cm2
Area of the figure
=  4 –  2  2 cm2 1 2
= 38 cm2

5. Volume of Simple Solids (p.3)
1. Volume of cube
= length  length  length
2. Volume of cuboid
= length  width  height

6. PRISMS (柱體)
A solid with uniform cross- section in the shape of a polygon is called a prism.
A prism is named and classified according to the shape of its base.
The perpendicular distance between the two parallel bases of a prism is called its height (or length).
The faces (other than the two bases) of a prism are called lateral faces.
base
height
lateral faces
base
Triangular prism

7. Volume of Prisms Volume = area of base  height
e.g. Volume of the solid
= 260  20 cm3
= 5200 cm3

8. Surface Areas of Prisms
Total surface area
= areas of the two bases
+ total area of all lateral faces
e.g. Total surface area of the solid
= [2   (20  10)] cm2
= 1720 cm2

9. Volume of Prisms Volume of the triangular prism = ?
Area of the base of the prism
= ( 7  4  2 ) cm2
= 14 cm2
So, the volume
= ( 14  10 ) cm3
= 140 cm3

10. Volume of Prisms Volume of the rectangular prism (cuboid) = ?
The volume
= [ ( 5  3 )  6 ] cm3
= 90 cm3
= [ ( 3  6 )  5 ] cm3
= [ ( 5  6 )  3 ] cm3

11. Volume of Prisms Find the volume of the prism. Area of base
= (12  7 – 8  5) cm2
= 44 cm2
= 44  11 cm3
= 484 cm3

12. Volume & Surface Area of Prisms
The figure shows a gold ingot in the shape of a prism. Its base is a trapezium, and the other faces are rectangles. If the volume of the gold ingot is 540 cm3, find (a) the value of d,
(b) the total surface area of the gold ingot. 1 2
= (6 + 12)  4 cm2
(a) Area of trapezium
∴ The volume of the gold ingot
i.e. 36d = 540
d = 15
(b) Total surface area of the gold ingot
= [2   (15  5) + 15   6] cm2
= 492 cm2
= 36 cm2
= 36d cm3

13. WORKBOOK

14. WORKBOOK 8 8 8
512 5 10 4
200
cm3
200
12d
512 4 3 d
12d = 312
12d
cm3 26

15. WORKBOOK
2.5 1 6 15
cm2 19 15
4 cm2 4
2 cm2 2
2.5 5
cm3 1 2 3
6 cm3
6 cm3
5 cm3 B

16. Enjoy the world of Mathematics!
Ronald HUI