Mathematics form 1&2 short simple notes By Kelvin 2H/2017

1. Prepared by, Kelvin, Class 2H, SMJK Heng Ee
Prepared date on 28/10/2017.
Mathematics Form 1 and 2 Short Simple Notes (Form 2/2017 – 2B EXAM)
Form 1
Chp 1 +, -, × and ÷
Chp 2 Number Sequence, Prime Number, Factor, Prime Factor,
common multiply, common factor, LCM and HCF
Chp 3, 4, 5 – Fraction, Decimal and Percentage.
Chp 6 Integer – Positive and Negative (+ and -)
Chp 7 Algebraic Expression –Unknown and Coefficient (2x, 3y)
- Like terms and Unlike terms
Chp 8 Measurement – Length – (mm,cm,m and km)
- Mass – (mg, g kg and ton)
– Area – (mm2,cm2,m2 and km2)
- Volume – (mm3,cm3,m3 and km3)
– Money – (mm2,cm2,m2 and km2)
– Time – (second, minute, hour, day, week, month, year and decade)
Length
1 km = 1000 m, 1 m = 100 cm, 1 cm = 10 mm
Mass
1 tonne = 1 000kg,1 kg = 1 000 g, 1 g = 1 000mg3.
Time
1 millennium = 1 000 years, 1 century = 100 years, 1 decade = 10 years,
1 year = 12 months = 52 weeks = 365 days, 1 leap year = 366 days,
1 week = 7 days, 1 day = 24 hours, 1 hour = 60 minutes, 1 minute = 60 seconds.
Time can be stated using the 12-hour system or the 24-hour system.
For example, the starting time of a show is 9.00 p.m. or 2100 hours.
Chp 9 Angle
Name of angle Value of angle Diagram of angle
Acute angle 0° < 90°
Right angle =90°
Obtuse angle 90° < 180°
Straight angle = 180°
Reflex angle 180° < 360°
Whole angle =360°
To determine properties of complementary angles
In diagram 10, shown a complementary angle.
Complementary angles are two angles whose sum is 90°.
a+b=90°.
To determine properties of supplementary angles
In diagram 10, shown a supplementary angles.
Supplementary angles are
two angles whose sum is 180°.
a+b=180°.
Chp 10
10.3 Triangles
A triangle is a polygon with 3 straight sides and 3 vertices.
Geometric properties and name of triangles
1. A triangle is a three-sided polygon.
2. A triangle is classified based on:
Table 1

2. Prepared by, Kelvin, Class 2H, SMJK Heng Ee
Prepared date on 28/10/2017.
a) The length of its sides (Figure 7)
b) The size of its angles (Figure 8&9)
To solve problems involving triangles
1. The sum of the angles of a triangle is 180°.
2. This is an example of following example 1.
10.4 Quadrilaterals
To determine and drawing the line(s) of symmetry of quadrilateral
1. A quadrilateral is a polygon with 4 straight sides and 4 vertices.
2. In figure 10, shown example of quadrilateral and their properties.
Quadrilaterals Symmetry of quadrilaterals (diagram) Properties
Square
Rectangle
Parallelogram
Rhombus
Trapezium
Kite
To solve problems involving quadrilaterals
1. The sum of the angles of a quadrilateral is 360°.
2. This is an example of following example 2. 
11.1 Perimeter and area
To identify the perimeter of a region
1. Perimeter is total distance around the outer edge of a figure.
2. The perimeter of figure 12 is the total length of all the sides.
3. The perimeter of figure 12 is (8+10+2+5+5+6) cm.
11.2 Area of rectangles
To find the area of a unit of square and rectangle.
1. The area of a unit square = 1 unit X 1 unit = 1 unit2
2. The standard units for measuring area are square centimeter (𝐜𝐦2
) and square meter ( 𝐦2
).
3. Area of a rectangle (cm2
) = length X breadth
11.3 Area of triangles, parallelograms and trapeziums
To identify the height and bases of triangles, parallelograms and trapeziums
1. The height of a shape is the perpendicular distance from the base to the highest point.
2. Height and base must meet at right angle (90°).
3. In figure 13, shown the height and bases of triangles, parallelograms and trapeziums.
To find the areas of triangles, parallelograms and trapeziums
1. Area of triangles = ½Xbase(b)Xheight(h) or base(b)Xheight(h)÷2
2. Area of parallelograms = base(b)Xheight(h)
3. Area of trapeziums = sum of parallel sidesX½Xheight(h)
Figure 10

3. Prepared by, Kelvin, Class 2H, SMJK Heng Ee
Prepared date on 28/10/2017.
To find the height or base of triangles, parallelograms and trapeziums
1. Height of triangle =
2Xarea of triangle
base
2. Base of triangle =
2Xarea of triangle
height
3. Height of parallelogram =
area of parallelogram
base
4. Base of parallelogram =
area of parallelogram
height
5. Height of trapezium =
2Xarea of trapezium
base
6. (Sum of two parallel sides) Base of trapezium =
2Xarea of trapezium
height
To find the areas of figures are made up of triangles, parallelograms or trapeziums
To find the areas of such figures, we must:
a) Separate or identify it into triangles, parallelograms or trapeziums.
b) Find the area of each shape.
c) Add up all the areas.
12.1 Geometric Solids
To identify geometric solids
1. A solid is a three-dimensional (3D) object that has length, width and height.
2. Every geometric solid has a fixed number of edges, vertices and surfaces. (Refer figure 14)
Name of solid Diagram of solid Edges Vertex Flat surface Curved surface
Cube 12 8 6 0
Cuboid 12 8 6 0
Pyramid 8 5 5 0
Cylinder 0 0 2 1
Cone 0 1 1 1
Sphere 0 0 0 1
To state the geometric properties of cubes and cuboids
1. A cube is a solid has six square faces are same size. (Refer figure 15)
2. A cuboid is a solid has six rectangular faces. (Refer figure 16)
3. The edge of a cube or a cuboid is the line where the faces meet.
4. The vertex of a cube or a cuboid is the point where the edges meet.
12.2 Volume of Cuboids
To find the volume of a solid
1. Volume is the space taken up by a solid.
2. The volume of a unit cube = 1 unit X 1 unit X 1 unit = 1 unit3
3. The standard units for measuring volume are cubic centimeter (𝐜𝐦3
)
and cubic meter ( 𝐦3
).
To find the volume of cubes and cuboids
Volume of cubes and cuboids (cm3
) = length X width X height
-------------------------------------- Mathematics Short Simple Notes (Form 1) --------------------------------------
Figure 14

4. Prepared by, Kelvin, Class 2H, SMJK Heng Ee
Prepared date on 28/10/2017.
Form 2
Chp 1 Directed Numbers – Form 1 Chp 1 and 6 – Positive and Negative (+, -, × and ÷)
Chp 2 – Square, Square root, Cube and Cube root
Chp 3, 4 – Algabric Expresion II and Linear Equation (x=5) (2+x=4+3)
Chp 5 Ratio - A:B A:B:C
5:4 5:6
4:3
___________
Chp 6 – Pythogras Theorem (Triangle 90o) – Hypotheses long2=short2+short2
short2=long2 - short2
Chp 7, 9 – Composes – Form 1 (Protractor) [draw angles]
- Angles 90o, 45o, 60o, 120o and others.
Perpendicular Line, Perpendicular Bisector, Angle Bisector and Circle (chp 9)
Chp 8 – Coordinates (x,y) – eg: (3,-5)
- Formula: Distance Between =√(𝐱 𝟐 − 𝐱 𝟏)𝟐
+ (𝐲𝟐 − 𝐲𝟏)𝟐
and Midpoint =
Chp 10 - Circle – Parts of Circle –
Radius, Diameter, Centre,Chord,
Segment, Arc,Sector and
Circumference.
-Formula: Circumference = 2𝜋r,
Arc = 2πrX
𝜃
360 𝑜
, Area = 𝜋r2
and
Area of Sector = 𝜋r2
X
𝜃
360 𝑜
Chp 11 – Transformation
-Translation (move same shape)
-Reflection (reflect opposite shape)
-Rotation (rotate/turn shape)
– clockwise and anti-clockwise
isometry-same, congruency-not same
& Properties of Quadrilateral
(Form 1 - 10.4) -Polygons
Chp 13 – Statistics Data
-Bar graph,
-Line graph,
-Chart/pie chart
-Pictogram
Frequency, Tally
KBAT, fill up the data,
*(total number)- (data value)
Chp 12 – Solid Geometry II – Nets and Properties (form 1-12.1) of 6 types solid geometry
Surface Area (Formula):-
A= [2 X (a X b)] + [2 X (a
X c)] + [2 X (b X c)]
A= 6 X (a X a) = 6a2
A= (area of triangular
faces) X (base area)
A= 2𝜋r2
+ 2𝜋rh
A= (2Xcongruent faces) X
(totalarea of rectangular
faces)
A= 𝜋r2
+ πrs
*s = slant