E Maths Formulas
Numbers
1. Summary
A rational number can be expressed as an exact fraction while an irrational number cannot be expressed as an exact fraction.
An integer is a number without decimals or fractional component. Integers can be further classified as:
- Whole numbers are positive integers and the integer $0$
- Natural numbers are positive integers
- Prime numbers are positive integers with only two factors, one and itself
- Composite numbers are positive integers with more than two factors
- Perfect square (or square number) is the square of an integer, i.e. $1^2, 2^2, 3^2, ... $
- Perfect cube (or cube number) is the cube of an integer, i.e. $1^3, 2^3, 3^3, ... $
Example
Consider the set of numbers below.
$$ \pi, -2.4, 9, \sqrt{4}, \sqrt{7} $$
The rational numbers are $ -2.4 = -{12 \over 5}, 9 = {9 \over 1}, \sqrt{4} = 2 = {2 \over 1} $.
The irrational numbers are $ \pi, \sqrt{7} $.
The integers are $ 9, \sqrt{4} = 2 $.
1. Express a number in terms of it's prime factors
Example
Express $1386$ as a product of it's prime factors, leaving your answer in index notation.
\begin{align*} 2 | & \underline{1386} \\ 3 | & \underline{\phantom{0} 693 } \\ 3 | & \underline{\phantom{0} 231 } \\ 7 | & \underline{\phantom{00} 77} \\ 11 | & \underline{\phantom{00} 11} \\ & \phantom{000} 1 \end{align*} \begin{align*} 1386 & = 2 \times 3 \times 3 \times 7 \times 11 \\ & = 2 \times 3^2 \times 7 \times 11 \end{align*}
2. Highest common factor (i.e. the greatest common divisor of two or more numbers)
Example
Find the highest common factor (HCF) of $1170$ and $1485$.
Solutions (by selecting common factors)
\begin{align*} 1170 & = 2 \times 3^2 \times 5 \times 13 \\ 1485 & = 3^3 \times 5 \times 11 \\ \\ \text{HCF} & = 3^2 \times 5 \\ & = 45 \end{align*}
Solutions (by dividing by prime numbers)
\begin{align*} 3 | & \underline{1170, 1485} \\ 3 | & \underline{\phantom{0} 390, \phantom{0} 495} \\ 5 | & \underline{\phantom{0} 130, \phantom{0} 165 } \\ & \phantom{00} 26, \phantom{00} 33 \phantom{000000} [\text{Stop - no more common factors}] \\ \\ \text{HCF} & = 3 \times 3 \times 5 \\ & = 45 \end{align*}
3. Lowest common multiple (i.e. the smallest number that is divisible by two or more numbers)
Example
Find the lowest common multiple of $36$, $45$ and $112$.
Solutions (by selecting highest power of common factors and uncommon factors)
\begin{align*} 36 & = 2^2 \times 3^2 \\ 45 & = 3^2 \times 5 \\ 112 & = 2^4 \times 7 \\ \\ \text{LCM} & = 2^4 \times 3^2 \times 5 \times 7 \\ & = 5040 \end{align*}
Solutions (by dividing by prime numbers)
\begin{align*} 2 | & \underline{36, 45, 112} \\ 2 | & \underline{18, 45, \phantom{0} 56} \\ 2 | & \underline{\phantom{1} 9, 45, \phantom{0} 28} \\ 2 | & \underline{\phantom{1} 9, 45, \phantom{0} 14} \\ 3 | & \underline{\phantom{1} 9, 45, \phantom{00} 7} \\ 3 | & \underline{\phantom{1} 3, 15, \phantom{00} 7} \\ 5 | & \underline{\phantom{1} 1, \phantom{4} 5, \phantom{00} 7} \\ 7 | & \underline{\phantom{1} 1, \phantom{4} 1, \phantom{00} 7} \\ & \phantom{1} 1, \phantom{4} 1, \phantom{00} 1 \\ \\ \text{LCM} & = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 \\ & = 5040 \end{align*}
4. Perfect square
For a perfect square, the powers of each factor is a multiple of $ 2 $.
Example
Find the smallest integer value of $h$ such that $336h$ is a perfect square.
\begin{align*} 336 & = 2^4 \times 3 \times 7 \\ \\ 336h & = 2^4 \times 3 \times 7 \times 3 \times 7 \\ & = 2^4 \times 3^2 \times 7^2 \phantom{000000} [\text{Perfect square}] \\ \\ \therefore h & = 3 \times 7 \\ & = 21 \end{align*}
5. Perfect cube
For a perfect cube, the powers of each factor is a multiple of $ 3 $.
Example
Find the smallest value of $h$ such that ${336 \over h}$ is a perfect cube.
\begin{align*} 336 & = 2^4 \times 3 \times 7 \\ \\ {336 \over h} & = {2^4 \times 3 \times 7 \over 2 \times 3 \times 7} \\ & = 2^3 \phantom{000000000} [\text{Perfect cube}] \\ \\ \therefore h & = 2 \times 3 \times 7 \\ & = 42 \end{align*}
1. Multiply terms with the same base
$ a^m \times a^n = (a^m)(a^n) = \phantom{.} $ $ a^{m + n} $
Example
Simplify $ x^{-2} y^6 \times x^4 y^{-4} $.
\begin{align*} x^{-2} y^6 \times x^4 y^{-4} & = x^{-2 + 4} y^{6 + (-4)} \\ & = x^2 y^2 \end{align*}
2. Divide terms with the same base
$ a^m \div a^n = {a^m \over a^n} = \phantom{.} $ $ a^{m - n} $
Example
Simplify $ {x^5 y^3 \over x^2 y^{-7} } $.
\begin{align*} {x^5 y^3 \over x^2 y^{-7} } & = \left(x^5 \over x^2\right) \left(y^3 \over y^{-7}\right) \\ & = x^{5 - 2} y^{3 - (-7)} \\ & = x^3 y^{10} \end{align*}
3. Power of $n$
$ (a^m)^n = \phantom{.} $ $ a^{mn} $
Example
Simplify $ {(2x^{10} y)^3 \over 4x^2 } $.
\begin{align*} { (2x^{10} y)^3 \over 4x^2 } & = { 2^3 x^{10(3)} y^3 \over 4x^2 } \\ & = {8 x^{30} y^3 \over 4x^2 } \\ & = \left(8 \over 4\right) \left(x^{30} \over x^2\right) y^3 \\ & = 2 x^{30 - 2} y^3 \\ & = 2 x^{28} y^3 \end{align*}
4. Power of $0$
$ a^0 = \phantom{.} $ $ 1 $
5. Multiply terms with the same power
$ a^m \times b^m = (a^m)(b^m) = \phantom{.} $ $ (ab)^m $
Example
Express $ 2^{2y} \times 3^y $ as power of $12$.
\begin{align*} 2^{2y} \times 3^y & = (2^2)^y \times 3^y \\ & = 4^y \times 3^y \\ & = 12^y \end{align*}
6. Divide terms with the same power
$ a^m \div b^m = {a^m \over b^m} = \phantom{.} $ $ \left(a \over b \right)^m $
Example
Express $ {3^{2y} \over 2^y} $ as power of $ 4.5 $.
\begin{align*} {3^{2y} \over 2^y} & = { (3^2)^y \over 2^y} \\ & = {9^y \over 2^y} \\ & = \left(9 \over 2\right)^y \\ & = 4.5^y \end{align*}
7. Negative indices
$ a^{-n} = \phantom{.} $ $ {1 \over a^n} $
$ {1 \over a^{-n}} = \phantom{.} $ $ a^{-(-n)} = a^n $
$ \left(a \over b\right)^{-n} = \phantom{.} $ $ \left(b \over a\right)^n $
Example 1
Simplify $ 1 \div 2 x^{-3} $, leaving your answer in positive index form.
\begin{align*} 1 \div 2x^{-3} & = {1 \over 1} \times {1 \over 2x^{-3}} \\ & = {1 \over 2x^{-3} } \\ & = {x^3 \over 2} \end{align*}
Example 2
Simplify $ \left( x^3 \over y^{-2} \right)^{-5} $, leaving your answer in positive index form.
\begin{align*} \left( {x^{3} \over y^{-2} } \right)^{-5} & = \left(y^{-2} \over x^{3}\right)^5 \\ & = {y^{-2(5)} \over x^{3(5)}} \\ & = {y^{-10} \over x^{15}} \\ & = {1 \over x^{15} y^{10} } \end{align*}
8. Fractional indices
$ a^{1 \over n} = \phantom{.} $ $ \sqrt[n]{a} $
$ a^{1 \over 2} = \phantom{.} $ $ \sqrt{a} $
$ a^{m \over n} = \phantom{.} $ $ \sqrt[n]{a^m} $
Example
Simplify $ \sqrt{ x^6 y } \div \sqrt[3]{y} $ .
\begin{align*} \sqrt{x^6 y} \div \sqrt[3]{y} & = { (x^6 y)^{1 \over 2} \over y^{1 \over 3} } \\ & = { x^{6 \left(1 \over 2\right)} y^{1 \over 2} \over y^{1 \over 3}} \\ & = { x^3 y^{1 \over 2} \over y^{1 \over 3}} \\ & = x^3 y^{ {1 \over 2} - {1 \over 3} } \\ & = x^3 y^{1 \over 6} \end{align*}
1. Express in standard form (positive power)
$ 52900 = \phantom{.} $ $ 5.29 \times 10^4 $
$ 6453.7 \times 10^4 = \phantom{.} $ $ (6.4537 \times 10^3) \times 10^4 = 6.4573 \times 10^7 $
2. Express in standard form (negative power)
$ 0.0034 = \phantom{.} $ $ 3.4 \times 10^{-3} $
$ 0.082 \times 10^{-4} = \phantom{.} $ $ (8.2 \times 10^{-2}) \times 10^{-4} = 8.2 \times 10^{-6} $
3. Million, billion and trillion
| Quantity | Value |
| Million | $ 10^6 $ |
| Billion | $ 10^9 $ |
| Trillion | $ 10^{12} $ |
Example
(i) Express $238$ million in standard form.
\begin{align*} 238 \text{ million} & = 238 \times 10^6 \\ & = (2.38 \times 10^2) \times 10^6 \\ & = 2.38 \times 10^8 \end{align*}
(ii) Express $4.2 \times 10^8$ in billion.
\begin{align*} 4.2 \times 10^8 & = {4.2 \times 10^8 \over 10^9} \text{ billion} \\ & = 0.42 \text{ billion} \end{align*}
1. Units
$ 1 \text{ cm} = \phantom{.} $ $ 10 $ $\text{ mm}$
$ 1 \text{ cm}^2 = \phantom{.} $ $ 10^2 = 100 $ $\text{ mm}^2$
$ 1 \text{ m} = \phantom{.} $ $ 100 $$ \text{ cm}$
$ 1 \text{ m}^2 = \phantom{.} $ $ 100^2 = 10 \phantom{.} 000 $$ \text{ cm}^2$
$ 1 \text{ km} = \phantom{.} $ $ 1000 $ $\text{ m}$
$ 1 \text{ km}^2 = \phantom{.} $ $ 1000^2 = 1 \phantom{.} 000 \phantom{.} 000$ $ \text{ m}^2 $
2. Changing scale from length to area
Example
A map is drawn to the scale of $1 : 40 \phantom{.} 000 $. On the map, a resort has an area of $20$ cm2. Calculate the actual area of the resort in km2.
\begin{align*} 1 & : 40 \phantom{.} 000 \\ 1 \text{ cm} & : 40 \phantom{.} 000 \text{ cm} \\ 1 \text{ cm} & : 400 \text{ m} \\ 1 \text{ cm} & : 0.4 \text{ km} \\ [\text{Square}] \phantom{0000} 1^2 \text{ cm}^2 & : (0.4)^2 \text{ km}^2 \phantom{0000} [\text{Square}] \\ 1 \text{ cm}^2 & : 0.16 \text{ km}^2 \\ [\times 20] \phantom{0000} 20 \text{ cm}^2 & : 3.2 \text{ km}^2 \phantom{0000000} [\times 20] \\ \\ \therefore \text{Actual area of resort} & = 3.2 \text{ km}^2 \end{align*}
3. Changing scale from area to length
Example
$4$ cm2 on a map represents $100$ m2 in real-life. Find the scale of the map in the form $1: n$, where $n$ is an integer.
\begin{align*} 4 \text{ cm}^2 & : 100 \text{ m}^2 \\ \sqrt{4} \text{ cm} & : \sqrt{100} \text{ m} \phantom{000000} [\text{Square root both sides}] \\ 2 \text{ cm} & : 10 \text{ m} \\ 1 \text{ cm} & : 5 \text{ m} \\ 1 \text{ cm} & : 500 \text{ cm} \\ 1 & : 500 \end{align*}
1. Direct proportion
If $y$ is directly proportional to $x$, then $y = $ $ kx $
2. Inverse proportion
If $y$ is inversely proportional to $x$, then $y = $ $ {k \over x} $
3. Graphs:
(a) Graph representing $y = kx$
As $x$ increases, $y$ increases at a constant rate
(b) Graph representing $y = kx^2$
As $x$ increases, $y$ increases at an increasing rate
(c) Graph representing $y = {k \over x} $
As $x$ increases, $y$ decreases
1. Percentage change
$ \text{Percentage change} = $ $ { \text{Final} - \text{Initial} \over \text{Final} } \times 100 $
Example
The price of a Tesla stock decreased from $ \$ 900 $ to $ \$ 400 $. Find the percentage decrease in the price.
\begin{align*} \text{Percentage change} & = {\text{Final} - \text{Initial} \over \text{Initial}} \times 100 \\ \\ & = {900 - 400 \over 900} \times 100 \\ \\ & = 55{5 \over 9} \% \end{align*}
2. Rate
$ \text{Rate of quantity 1 per quantity 2} = $ $ { \text{Quantity 1} \over \text{Quantity 2} } $
Example
A particular person views 20 Instagram profiles in 10 minutes. Find the average minutes spent on each profile.
\begin{align*} \text{Average minutes spent per profile} & = {\text{10 minutes} \over 20 \text{ profiles}} \\ & = 0.5 \text{ minutes per profile} \end{align*}
3. Speed - Units & formulas
$ 1 \text{ cm} = $ $ 10 \text{ mm} $
$ 1 \text{ m} = $ $ 100 \text{ cm} $
$ 1 \text{ km} = $ $ 1000 \text{ m} $
$ \text{Speed} = $ $ { \text{Distance} \over \text{Time} } $
$ \text{Average speed} = $ $ { \text{Total distance} \over \text{Total time} } $
4. Speed - conversion
Example
(i) Express $90$ km/h in m/s.
\begin{align*} 90 \text{ km/h} & = {90 \text{ km} \over 1 \text{ hour}} \\ & = {90 \times 1000 \text{ m} \over 1 \times 60 \times 60 \text{ seconds}} \\ & = 25 \text{ m/s} \end{align*}
(ii) Express $10$ m/s in km/h.
\begin{align*}
10 \text{ m/s} & = {10 \text{ m} \over 1 \text{ second}} \\
& = {10 \div 1000 \text{ km} \over 1 \div 60 \div 60 \text{ hour}} \\
& = 36 \text{ km/h}
\end{align*}
1. Terminology
$T_n$ is commonly used to denote the $n$th term.
$S_n$ is commonly used to denote the sum of the first $n$ terms. For example, $S_5 = T_1 + T_2 + T_3 + T_4 + T_5$.
Example
The sum of the first $n$ terms of a sequence is given by $S_n = n^2 + 4n$. Find the 4th term of the sequence, $T_4$.
\begin{align*} S_4 & = 4^2 + 4(4) \\ & = 32 \phantom{0000000} [T_1 + T_2 + T_3 + T_4] \\ \\ S_3 & = 3^2 + 4(3) \\ & = 21 \phantom{0000000} [T_1 + T_2 + T_3]\\ \\ T_4 & = 32 - 21 \\ & = 11 \end{align*}
2. Adding or subtracting the same term
$ T_n = $ $ a + (n - 1)(d) $
$a$ represents the first term, $T_1$.
$d$ represents the common difference between successive terms.
Example
Form the $n$th term for the following number pattern: $1, -0.5, -2, -3.5, ...$.
\begin{align*} T_n & = T_1 + (n - 1)(d) \\ & = (1) + (n - 1)(-1.5) \\ & = 1 - 1.5n + 1.5 \\ & = 2.5 - 1.5n \end{align*}
3. Perfect squares
$$ 1^2, 2^2, 3^2, 4^2, ... = 1, 4, 9, 16, ... $$
$ T_n = $ $ n^2 $
Example
Form the $n$th term for the following number pattern: $9, 16, 25, 36, ...$.
\begin{align*} T_1 & = 9 = 3^2 = (1 + 2)^2 \\ T_2 & = 16 = 4^2 = (2 + 2)^2 \\ T_3 & = 25 = 5^2 = (3 + 2)^2 \\ T_4 & = 36 = 6^2 = (4 + 2)^2 \\ \\ \therefore T_n & = (n + 2)^2 \end{align*}
4. Perfect cubes
$$ 1^3, 2^3, 3^3, 4^3, ... = 1, 8, 27, 64, ... $$
$ T_n = $ $ n^3 $
Example
Form the $n$th term for the following number pattern: $2, 9, 28, 65, ...$.
\begin{align*} T_1 & = 2 = 1 + 1 = 1^3 + 1 \\ T_2 & = 9 = 8 + 1 = 2^3 + 1 \\ T_3 & = 28 = 27 + 1 = 3^3 + 1 \\ T_4 & = 65 = 64 + 1 = 4^3 + 1 \\ \\ \therefore T_n & = n^3 + 1 \end{align*}
1. Order of a matrix
Order of a matrix is given by: Row $ \times $ Column
Example
State the order of the following matrices.
$$ \left( \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right) $$
Order: $ 2 \times 3 $
$$ \left( \begin{matrix} 7 & 8 & 9 & 10 \end{matrix} \right) $$
Order: $ 1 \times 4 $
$$ \left( \begin{matrix} 11 \\ 12 \\ 13 \\ 14 \\ 15 \end{matrix} \right) $$
Order: $ 5 \times 1 $
2. Addition and subtraction
For addition and subtraction, matrices must have the same order.
Example
Evaluate the following if possible:
(i) $ \left( \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right) + \left( \begin{matrix} 5 & 6 & 7 \\ 8 & 9 & 10 \end{matrix} \right) $
\begin{align*} \mathop{\left( \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right)}_{2 \times 2} & + \mathop{\left( \begin{matrix} 5 & 6 & 7 \\ 8 & 9 & 10 \end{matrix} \right)}_{2 \times 3} \\ \\ \text{Not possible, } & \text{since matrices have different order} \end{align*}
(ii) $ 2 \left( \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right) - 3\left( \begin{matrix} 5 & -6 \\ -7 & 8 \end{matrix} \right) $
\begin{align*} 2 \left( \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right) - 3\left( \begin{matrix} 5 & -6 \\ -7 & 8 \end{matrix} \right) & = \left( \begin{matrix} 2 & 4 \\ 6 & 8 \end{matrix} \right) - \left( \begin{matrix} 15 & - 18 \\ -21 & 24 \end{matrix} \right) \\ & = \left( \begin{matrix} -13 & 22 \\ 27 & -16 \end{matrix} \right) \end{align*}
3. Multiplication of two matrices
The column(s) of the first matrix must be equal to the row(s) of the second matrix.
Example
Evaluate the following if possible:
(i) $ \left( \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right)\left( \begin{matrix} 4 & 5 & 6 \\ 7 & 8 & 9 \\ 10 & 11 & 12 \end{matrix} \right) $
\begin{align*} \mathop{\left( \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right)}_{2 \times \underline{2}} & \mathop{\left( \begin{matrix} 4 & 5 & 6 \\ 7 & 8 & 9 \\ 10 & 11 & 12 \end{matrix} \right)}_{\underline{3} \times 3} \\ \\ \text{Not } & \text{possible} \end{align*}
(ii) $ \left( \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right)\left( \begin{matrix} 7 \\ 8 \\ 9 \end{matrix} \right) $
\begin{align*} \mathop{\left( \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix} \right)}_{2 \times \underline{3}} \mathop{\left( \begin{matrix} 7 \\ 8 \\ 9 \end{matrix} \right)}_{\underline{3} \times 1} & = \mathop{ \left( \begin{matrix}1 \times 7 + 2 \times 8 + 3 \times 9 \\ 4 \times 7 + 5 \times 8 + 6 \times 9 \end{matrix} \right)}_{2 \times 1} \\ & = \left( \begin{matrix} 50 \\ 122\end{matrix} \right) \end{align*}
4. Square of matrix
Example
Given that $ \textbf{A} = \left( \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right)$, find $ \textbf{A}^2$.
\begin{align*} \textbf{A}^2 & = \textbf{A} \textbf{A} \\ & = \mathop{\left( \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right)}_{2 \times \underline{2}} \mathop{\left( \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right)}_{\underline{2} \times 2} \\ & = \mathop{ \left( \begin{matrix} 1 \times 1 + 2 \times 3 & 1 \times 2 + 2 \times 4 \\ 3 \times 1 + 4 \times 3 & 3 \times 2 + 4 \times 4 \end{matrix} \right)}_{2 \times 2} \\ & = \left( \begin{matrix} 7 & 10 \\ 15 & 22 \end{matrix} \right) \end{align*}
Set language & notations (Only for O levels)
1. Notations
| Symbol | Meaning |
|---|---|
| $\xi$ | Universal set |
| $\{ \} $ | Empty set |
| $\emptyset$ | Null set |
| $ \in $ | "is an element of" |
| $ n(A) $ | Number of elements in set $A$ |
| $ \subset $ | Proper subset |
| $ \cap $ | Intersect |
| $ \cup $ | Union |
| $ B' $ | Complement of set $B$ |
Example 1
If $A = \{ \text{a, b, c} \}$, fill in the following blanks with appropriate symbols.
$ \text{c} $$ \in $ $ A $
$ \{ \text{c} \} $$ \subset $ $ A $
$ \text{d} $$ \notin $ $ A $
$ \{ \text{c, d} \} $$ \not\subset $ $ A $
Example 2
If $\xi = \{ 0, 1, 2, 3, 4, 5\}$, $B = \{ 0, 1, 2 \}$, $C = \{ 2, 3 \}$ and $D = \{ 3, 4 \}$, complete the following.
$ n(B) = $$ 3 $
$ B \cap C = $$ \{ 2 \} $
$ B \cap D = $$ \emptyset $
$ B \cup D = $$ \{ 0, 1, 2, 3, 4 \} $
$ B' = $$ \{ 3, 4, 5 \} $
2. Venn diagram to represent set notations:
$$ A \cap B \ne \emptyset $$
$$ A \cap B = \emptyset $$
$$ A \subset B $$
3. Describe shaded region in Venn diagram:
$ A \cup B $
$ B' $
$ (A \cap B)' $ or $ A' \cup B' $
$ (A \cup B)' $ or $ A' \cap B' $
$ A' \cap B $
$ A' \cup B $
(From O level 2018)
$ ( A \cap B') \cup (A' \cap B) $
Algebra
1. Identities
$ (a + b)^2 = \phantom{.} $$ a^2 + 2ab + b^2 $
$ (a - b)^2 = \phantom{.} $$ a^2 - 2ab + b^2 $
$ (a + b)(a - b) = \phantom{.} $$ a^2 - b^2 $
Example
Expand and simplify the following:
(i) $ (2x + 3y)^2 $
\begin{align*} (2x + 3y)^2 & = (2x)^2 + 2(2x)(3y) + (3y)^2 \\ & = 4x^2 + 12xy + 9y^2 \end{align*}
(ii) $ (a - 3b)^2 - (2a + 5b)(2a - 5b) $
\begin{align*} (a - 3b)^2 - (2a + 5b)(2a - 5b) & = (a)^2 - 2(a)(3b) + (3b)^2 - [ (2a)^2 - (5b)^2 ] \\ & = a^2 - 6ab + 9b^2 - (4a^2 - 25b^2) \\ & = a^2 - 6ab + 9b^2 - 4a^2 + 25b^2 \\ & = a^2 - 4a^2 - 6ab + 9b^2 + 25b^2 \\ & = - 3a^2 - 6ab + 34b^2 \end{align*}
2. Common misconceptions
Example
Expand and simplify the following:
(i) $ (1 + 2x)(3x - 4) $
(ii) $ 1 + 2x(3x - 4) $
(iii) $ (1 + 2x) - (3x - 4) $
1. Factorise common factor(s)
Example
Factorise the expression $ 2ab - 6a^2 b + 10 a b^2 $.
\begin{align*} 2ab - 6a^2b + 10ab^2 & = 2ab ( 1 - 3a + 5b) \end{align*}
2. Factorise by grouping
Example
Factorise the expression $ ab + 2bc - ac - 2b^2 $.
\begin{align*}
ab + 2bc - ac - 2b^2 & = ab - ac - 2b^2 + 2bc \phantom{00000} [\text{Rearrange}] \\
& = a(b - c) - 2b (b - c) \phantom{00000} [\text{Factorise}] \\
& = (b - c)(a - 2b) \phantom{000000000} [\text{Factorise } (b - c)]
\end{align*}
3. Factorise quadratic expressions
Methods:
- Cross method
- Box method
- By calculator (make $x^2$ positive first)
Example
Factorise the expression $ -2x^2 + 3x + 9 $.
Solutions (cross method or box method)
$$ -2x^2 + 3x + 9 = (- x + 3)(2x + 3) $$
\begin{align*} -2x^2 + 3x + 9 & = -(2x^2 - 3x - 9) \\ & = -(x - 3)(2x + 3) \phantom{000000} [x = 3, x = -{3 \over 2}] \end{align*}
4. Factorise using $a^2 - b^2$
$ a^2 - b^2 = \phantom{.} $$ (a + b)(a - b) $
Example
Factorise the expression $ 5x^2 - 45 $.
\begin{align*} 5x^2 - 45 & = 5(x^2 - 9) \\ & = 5 ( x^2 - 3^2 ) \\ & = 5 ( x + 3)(x - 3) \end{align*}
Operations on algebraic fractions
1. Add or subtract algebraic fractions
To combine fractions into a single fraction, make sure all denominators have the same denominator.
Example
Express ${x - 4 \over x^2 - 4} + {x \over 2 - x}$ as a single fraction.
\begin{align*} {x - 4 \over x^2 - 4} + {x \over 2 - x} & = {x - 4 \over x^2 - 2^2} + {x \over -(x - 2)} \\ & = {x - 4 \over (x + 2)(x - 2)} - {x \over x - 2} \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \\ & = {x - 4 \over (x + 2)(x - 2)} - {x(x + 2) \over (x + 2)(x - 2)} \\ & = {x - 4 - x(x + 2) \over (x + 2)(x - 2)} \\ & = {x - 4 - x^2 - 2x \over (x + 2)(x - 2)} \\ & = {-x^2 - x - 4 \over (x + 2)(x - 2)} \end{align*}
2. Multiply or divide algebraic fractions
To multiply two algebraic fractions, multiply the numerator with the numerator and the denominator with the denominator.
Example
Simplify $ {2x \over x + 2} \div {x \over x^2 + x - 2} $.
\begin{align*} \require{cancel} {2x \over x + 2} \div {x \over x^2 + x - 2} & = {2x \over x + 2} \div {x \over (x + 2)(x - 1)} \\ & = {2x \over x + 2} \times {(x + 2)(x - 1) \over x} \\ & = {2\cancel{x}\cancel{(x + 2)}(x - 1) \over \cancel{x}\cancel{(x + 2)}} \\ & = {2(x - 1) \over 1} \\ & = 2(x - 1) \end{align*}
3. Equation with fractions
Example
Solve the equation $ 1 + {x \over 2} = {x \over 5} $.
\begin{align*} 1 + {x \over 2} & = {x \over 5} \\ {2 \over 2} + {x \over 2} & = {x \over 5} \\ {2 + x \over 2} & = {x \over 5} \\ 5(2 + x) & = 2x \phantom{00000} [\text{Cross-multiply}] \\ 10 + 5x & = 2x \\ 5x - 2x & = -10 \\ 3x & = -10 \\ x & = -{10 \over 3} \end{align*}
1. Reorganise and factorise
Example
Given that $1 + {x \over 3} = {xy \over z} $, make $x$ the subject of the equation.
\begin{align*} 1 + {x \over 3} & = {xy \over z} \\ {3 \over 3} + {x \over 3} & = {xy \over z} \\ {3 + x \over 3} & = {xy \over z} \\ z(3 + x) & = 3xy \phantom{000000} [\text{Cross-multiply}] \\ 3z + xz & = 3xy \\ xz - 3xy & = - 3z \\ x(z - 3y) & = - 3z \phantom{000000} [\text{Factorise } x] \\ x & = -{3z \over z - 3y} \end{align*}
2. Square root $\sqrt{ \phantom{0} }$ or $\sqrt[n]{\phantom{0}}$
Remember to include $\pm$ when taking square root $\sqrt{ \phantom{0} }$ or for $\sqrt[n]{\phantom{0}}$, where $n$ is an even integer.
Example
Given that $y = x^4 + z$, make $x$ the subject of the equation.
\begin{align*} y & = x^4 + z \\ -x^4 & = z - y \\ x^4 & = y - z \\ x & = \pm \sqrt[4]{y - z} \end{align*}
3. Square/cube both sides of equation
Example
Given that $2y = \sqrt[3]{x + 3z}$, make $x$ the subject of the equation.
\begin{align*} 2y & = \sqrt[3]{x + 3z} \\ (2y)^3 & = \left( \sqrt[3]{x + 3z} \right)^3 \\ (2)^3 (y)^3 & = x + 3z \\ 8 y^3 & = x + 3z \\ - x & = 3z - 8y^3 \\ x & = 8y^3 - 3z \end{align*}
1. Solve by elimination method
Example
Solve the following pair of simultaneous equation:
$$ 2x + 3y = 5 $$
$$ x - 2y = 6 $$
\begin{align*} 2x + 3y & = 5 \phantom{0} \text{--- (1)} \\ \\ x - 2y & = 6 \phantom{0} \text{--- (2)} \\ \\ (2) & \times 2, \\ 2x - 4y & = 12 \phantom{0} \text{--- (3)} \\ \\ (1)& - (3), \phantom{000000} [\text{Eliminate } x ] \\ 2x + 3y - (2x - 4y) & = 5 - 12 \\ 2x + 3y - 2x + 4y & = -7 \\ 7y & = -7 \\ y & = -1 \\ \\ \text{Substitute } & y = -1 \text{ into (1),} \\ 2x + 3(-1) & = 5 \\ 2x - 3 & = 5 \\ 2x & = 5 + 3 \\ 2x & = 8 \\ x & = 4 \\ \\ \therefore x & = 4, y = -1 \end{align*}
2. Solve by substitution method
Example
Solve the following pair of simultaneous equation:
$$ 2x + 3y = 5 $$
$$ x - 2y = 6 $$
\begin{align*} 2x + 3y & = 5 \phantom{0} \text{--- (1)} \\ \\ x - 2y & = 6 \\ x & = 6 + 2y \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 2(6 + 2y) + 3y & = 5 \\ 12 + 4y + 3y & = 5 \\ 7y & = 5 - 12 \\ 7y & = -7 \\ y & = -1 \\ \\ \text{Substitute } & y = -1 \text{ into (2)}, \\ x & = 6 + 2(-1) \\ x & = 4 \\ \\ \therefore x & = 4, y = -1 \end{align*}
1. Solve by factorising common factor
Example
Solve the equation $ 2x^2 = 3x $.
\begin{align*} 2x^2 & = 3x \\ 2x^2 - 3x & = 0 \\ x(2x - 3) & = 0 \end{align*} \begin{align*} x & = 0 && \text{ or } & 2x - 3 & = 0 \\ & &&& 2x & = 3 \\ & &&& x & = {3 \over 2} \end{align*}
2. Solve by factorising quadratic expression (cross method or box method or by calculator)
Example
Solve the equation $2x^3 - 3x + 1 = 0$ by factorisation.
\begin{align*} 2x^2 - 3x + 1 & = 0 \\ (2x - 1)(x - 1) & = 0 \end{align*} \begin{align*} 2x - 1 & = 0 && \text {or } & x - 1 & = 0 \\ 2x & = 1 &&& x & = 1 \\ x & = {1 \over 2} \end{align*}
3. Solve by quadratic formula
$ x = $$ {-b \pm \sqrt{b^2 - 4ac} \over 2a} $
Example
Solve the equation $ 2x^2 - 3x = 4 $, leaving your answer correct to two decimal places.
\begin{align*} 2x^2 - 3x & = 4 \\ 2x^2 - 3x - 4 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-3) \pm \sqrt{(-3)^2 - 4(2)(-4)} \over 2(2)} \\ & = {3 \pm \sqrt{41} \over 4} \\ & = 2.3507 \text{ or } -0.8507 \\ & \approx 2.35 \text{ or } -0.85 \text{ (2 d.p.)} \end{align*}
4. Solve by taking square root
Example
Solve the equation $ 2(x - 3)^2 = 8 $.
\begin{align*} 2(x - 3)^2 & = 8 \\ (x - 3)^2 & = {8 \over 2} \\ (x - 3)^2 & = 4 \\ \sqrt{(x - 3)^2} & = \pm \sqrt{4} \\ x - 3 & = \pm 2 \end{align*} \begin{align*} x - 3 & = 2 && \text{ or } & x - 3 & = - 2 \\ x & = 2 + 3 &&& x & = -2 + 3 \\ x & = 5 &&& x & = 1 \end{align*}
5. Solve by completing the square, then taking square root
$ x^2 \pm bx + c = $$ \left(x \pm {b \over 2}\right)^2 - \left(b \over 2\right)^2 - c $
Make the coefficient of $x^2$ is $1$ and $x^2$ is positive before completing the square!
Example
Solve the equation $ 2x^2 - 3x - 4= 0 $ by completing the square, leaving your answer correct to two decimal places.
\begin{align*} 2x^2 - 3x - 4 & = 0 \\ x^2 - {3 \over 2}x - 2 & = 0 \phantom{000000} [\text{Divide every term by } 2] \\ \left(x - {3 \over 4}\right)^2 - \left(3 \over 4\right)^2 - 2 & = 0 \phantom{000000} \left[ {3 \over 2} \div 2 = {3 \over 4} \right] \\ \left(x - {3 \over 4}\right)^2 - {9 \over 16} - 2 & = 0 \\ \left(x - {3 \over 4}\right)^2 - {41 \over 16} & = 0 \\ \left(x - {3 \over 4}\right)^2 & = {41 \over 16} \\ \sqrt{ \left(x - {3 \over 4}\right)^2 } & = \pm \sqrt{41 \over 16} \\ x - {3 \over 4} & = \pm \sqrt{41 \over 16} \\ x & = \pm \sqrt{41 \over 16} + {3 \over 4} \end{align*} \begin{align*} x & = \sqrt{41 \over 16} + {3 \over 4} && \text{ or } & x & = - \sqrt{41 \over 16} + {3 \over 4} \\ & = 2.3507 &&& & = -0.8507 \\ & \approx 2.35 \text{ (2 d.p.)} &&& & = -0.85 \text{ (2 d.p.)} \end{align*}
1. Inequality signs
| Signs | Meaning |
|---|---|
| $ \ge $ | Greater than or equals to |
| $ > $ | Greater than |
| $ \le $ | Less than or equals to |
| $ < $ | Less than |
2. Represent inequality on number line
3. Tips on solving inequality
1. Change the direction of the inequality when you multiply or divide both sides by a negative number:
\begin{align*} -2x & > 10 \\ {-2x \over -2} & < {10 \over -2} \phantom{00000} [\text{Change from > to <}] \\ x & < -5 \end{align*}
2. For inequality with fractions, make all denominators the same before simplifying.
3. For two sided inequalities, split into two inequalities and solve each inequality (see example below).
Example
Solve the inequality $ -x \le {3x - 2 \over 4} \le 13 $.
\begin{align*}
-x & \le {3x - 2 \over 4} &&& {3x - 2 \over 4} & \le 13 \\
{-4x \over 4} & \le {3x - 2 \over 4} &&& {3x - 2 \over 4} & \le {52 \over 4} \\
-4x & \le 3x - 2 &&& 3x - 2 & \le 52 \\
-4x - 3x & \le -2 &&& 3x & \le 52 + 2 \\
-7x & \le -2 &&& 3x & \le 54 \\
[ \text{Flip sign}] \phantom{0000} x & \ge {-2 \over -7} &&& x & \le {54 \over 3} \\
x & \ge {2 \over 7} &&& x & \le 18
\end{align*}
$$ {2 \over 7} \le x \le 18 $$
Graphs
Linear functions (i.e. straight lines)
1. Gradient of line
$ \text{Gradient} = \phantom{.} $$ {\text{Rise} \over \text{Run}} = {y_2 - y_1 \over x_2 - x_1}$
If gradient $> 0$, then the line is an upward sloping line.
If gradient $< 0$, then the line is a downward sloping line.
2. Equation of a non-vertical straight line
$ y = \phantom{.} $$ mx + c $
$m$ represents the gradient of the line
$c$ represents the $y$-intercept of the line
3. Horizontal line
General equation: $ y = k $
$ \text{Gradient} = \phantom{.} $$ \phantom{.} 0 \phantom{.} $
Example
Equation of line: $ y = 2.5 $
4. Vertical line
General equation: $ x = k $
Gradient of vertical line is undefined
Example
Equation of line: $ x = 2 $
1. Shape and features for $y = ax^2 + bx + c$, where $a > 0$
2. Shape and features for $y = ax^2 + bx + c$, where $a < 0$
Sketch quadratic functions (Only for O levels)
1. Sketch graph of quadratic functions in factorised form, $y = \pm (x - a)(x - b)$
Find the following features before sketching:
- Shape: Minimum curve ($ \cup $) or maximum curve ($ \cap $)
- Intercepts: $y$ -intercept (let $x = 0$) and $x$-intercepts (let $y = 0$)
- Line of symmetry: $ x = {a + b \over 2}$, where $a$ and $b$ are the $x$-intercepts
- Turning point: $x$-coordinate is the line of symmetry. Substitute $x$-coordinate into equation of curve to find the $y$-coordinate
Example
Sketch the graph of $y = (3 - x)(x + 1)$, indicating the axial intercepts, line of symmetry and coordinates of the turning point.
\begin{align*} y & = (3 - x)(x + 1) \\ y & = 3x + 3 - x^2 - 2x \\ y & = - x^2 + x + 3 \phantom{000000} [\text{Maximum curve, } \cap] \\ \\ \text{Let } & x = 0, \\ y & = (3 - 0)(0 + 1) \\ y & = 3 \phantom{0000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = (3 - x)(x + 1) \end{align*} \begin{align*} 3 - x & = 0 && \text{ or } & x + 1 & = 0 \\ -x & = -3 &&& x & = -1 \\ x & = 3 &&&& \phantom{0000000} [x \text{-intercepts}] \end{align*} \begin{align*} x & = {a + b \over 2} \\ x & = {3 + (-1) \over 2} \\ x & = 1 \phantom{00000000} [\text{Line of symmetry}] \\ \\ \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = (3 - 1)(1 + 1) \\ y & = 4 \\ \\ \text{Turning } & \text{point: } (1, 4) \end{align*}
2. Sketch graph of quadratic functions after completing the square, $y = \pm (x - h)^2 + k $
Complete the square formula:
$ x^2 \pm bx + c = $$ \left(x \pm {b \over 2}\right)^2 - \left(b \over 2\right)^2 - c $
Make the coefficient of $x^2$ is $1$ and $x^2$ is positive before completing the square!
$$ y = \pm (x - h)^2 + k $$
Find the following features before sketching:
- Shape: Minimum curve ($ \cup $) or maximum curve ($ \cap $)
- Intercepts: $y$ -intercept (let $x = 0$)
- Turning point: $(h, k)$
Example
(i) Express $y = x^2 - 2x + 2 $ in the form $(x - h)^2 + k$.
\begin{align*} x^2 - 2x + 2 & = x^2 - 2x + \left(2 \over 2\right)^2 - \left(2 \over 2\right)^2 + 2 \\ & = \underbrace{x^2 - 2x + 1^2}_\text{Factorise} - 1 + 2 \\ & = (x - 1)(x - 1) + 1 \\ & = (x - 1)^2 + 1 \end{align*}
(ii) Hence, sketch the graph of $y = x^2 - 2x + 2 $, indicating the $y$-intercept and coordinates of the turning point.
\begin{align*} y & = x^2 - 2x + 2 \phantom{000000} [\text{Minimum curve, } \cup] \\ y & = (x - 1)^2 + 1 \phantom{00000.} [\text{From part i}] \\ \\ \text{Let } & x = 0, \\ y & = (0 - 1)^2 + 1 \\ y & = 2 \phantom{000000000000000} [y \text{-intercept}] \\ \\ \text{Turning } & \text{point: } (1, 1) \end{align*}
1. $y = ax^0 = a(1) = a$
2. $y = ax^1 = ax $
3. $y = ax^{-1} = {a \over x} $
4. $y = ax^{2} $
5. $y = ax^{-2} = {a \over x^2} $
6. $y = ax^3 $
Problems in real-world context
1. Formula
$ I =$$ {PRT \over 100} $
$ I $ represents interest
$P$ represents principal amount of money
$R$ represents interest rate per annum (without percentage)
$T$ represents time (in years)
1. Formula provided
$$ \text{Total amount} = P \left(1 + {r \over 100} \right)^n $$
$P$ represents principal amount of money
$r$ represents interest rate per compounding period (without percentage)
$n$ represents number of compounding periods
2. Compounding frequency:
| $P$ | Interest rate per annum | Duration | Compounding frequency | $r$ | $n$ | Total amount |
|---|---|---|---|---|---|---|
| $ \$10 \phantom{.} 000 $ | $ 12 \% $ | $ 3 $ years | Annually | $12$ | $3$ | $ \$ 14 \phantom{.} 049.28 $ |
| $ \$10 \phantom{.} 000 $ | $ 12 \% $ | $ 3 $ years | Half-yearly | $12 \div 2 = 6$ | $3 \times 2 = 6$ | $ \$ 14 \phantom{.} 185.19 $ |
| $ \$10 \phantom{.} 000 $ | $ 12 \% $ | $ 3 $ years | Quarterly (i.e. 4 times a year) | $12 \div 4 = 3$ | $3 \times 4 = 12$ | $ \$ 14 \phantom{.} 257.61 $ |
| $ \$10 \phantom{.} 000 $ | $ 12 \% $ | $ 3 $ years | Monthly | $12 \div 12 = 1$ | $3 \times 12 = 36$ | $ \$ 14 \phantom{.} 307.69 $ |
3. Simple interest vs. compound interest
Example
(i) Ben invested $ \$ 1000 $ at an annual rate of $ 5 \% $ simple interest for two years. Find the interest Ben earned.
\begin{align*} I & = {PRT \over 100} \\ & = {(1000)(5)(2) \over 100} \\ & = \$ 100 \end{align*}
(ii) Andrew invested $ \$1000 $ at an annual rate of $ 5\% $ interest compounded annually for two years. Find the interest Andrew earned.
\begin{align*} \text{Total amount} & = P \left(1 + {r \over 100} \right)^n \\ & = (1000) \left(1 + {5 \over 100} \right)^2 \\ & = \$ 1 \phantom{.} 102.50 \\ \\ \text{Interest} & = 1 \phantom{.} 102.50 - 1000 \\ & = \$ 102. 50 \end{align*}
(iii) Explain why Andrew received more interest than Ben.
\begin{align*} 1000 \times {5 \over 100} & = \$ 50 \\ \\ \text{Ben earned } \$ 50 \text{ int} & \text{erest each year.} \\ \\ \text{For the first year, Andrew} & \text{ earned } \$ 50 \text{ int} \text{erest.} \\ \\ 50 \times {5 \over 100} & = \$ 2.50 \\ \\ \text{For the second year, Andrew} & \text{ earned } \$ 50 \text{ interest from the initial} \\ \$ 1000 \text{ and } \$ 2.50 \text{ interest from} & \text{ the } \$ 50 \text{ interest earned in the first year.} \end{align*}
1. Calculating income tax
$$ \text{Chargeable income} = \text{Gross annual income} - \text{Total tax reliefs} $$
Note: Income tax payable will be calculated based on Chargeable income!.
Example
The table shows the calculation of income tax payable based on different tiers of chargeable income for the residents in Singapore.
| Chargeable income | Income tax rate ($ \% $) | Gross tax payable ( $ \$ $) |
|---|---|---|
| First $ \$20 \phantom{.} 000 $ Next $ \$ 10 \phantom{.} 000 $ |
$0$ $0$ |
$0$ $200$ |
| First $ \$30 \phantom{.} 000 $ Next $ \$ 10 \phantom{.} 000 $ |
- $3.5$ |
$200$ $350$ |
| First $ \$40 \phantom{.} 000 $ Next $ \$ 40 \phantom{.} 000 $ |
- $7$ |
$550$ $2800$ |
| First $ \$80 \phantom{.} 000 $ Next $ \$ 40 \phantom{.} 000 $ |
- $11.5$ |
$3350$ $4600$ |
The tax reliefs given to assist taxpayers in reducing their chargeable income are as follows:
- Earned income relief - $ \$ 1000 $
- Spouse relief - $ \$ 2000 $
- Parent relief (per parent) - $ \$ 3000 $
- Qualifying child relief (per child) - $ \$ 4000 $
Brian is single and does not have any children. He lives with and supports both of his retired parents. He has a full-time job and his gross annual income for year 2023 is $ \$ 85 \phantom{.} 000 $. Find the amount of income tax he has to pay for 2023.
\begin{align*} \text{Total tax reliefs} & = 1000 + \underbrace{2(3000)}_\text{Parent reliefs} \\ & = \$ 7000 \\ \\ \text{Chargeable income} & = 85 \phantom{.} 000 - 7000 \\ & = \$ 75 \phantom{.} 000 \phantom{000000} [\text{Falls under tier 3 - \$ 40k to \$80k}] \\ \\ 75 \phantom{.} 000 - 40 \phantom{.} 000 & = 35 \phantom{.} 000 \\ \\ \text{Income tax payable} & = 550 + \left( 35 \phantom{.} 000 \times {7 \over 100} \right) \\ & = \$ 3000 \end{align*}
1. Features
Gradient of graph = Speed of object
2. Interpret distance against time graph
 Object is stationary (or at rest) |
 Object is moving with constant speed |
 Object is moving with increasing speed (i.e. accelerating) |
 Object is moving with decreasing speed (i.e. decelerating) |
1. Features:
Gradient of graph = Acceleration of object
Area under graph = Distance travelled by object
Area of trapezium = $ {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} $
2. Interpret speed against time graph:
 Object is stationary (or at rest) |
 Object is moving with constant speed |
Object undergoes acceleration (i.e. speed is increasing):
 Object undergoes uniform acceleration |
 Object undergoes increasing acceleration |
 Object undergoes decreasing acceleration |
Object undergoes deceleration (i.e. speed is decreasing):
 Object undergoes uniform deceleration |
 Object undergoes increasing deceleration |
 Object undergoes decreasing deceleration |
Geometry
1. Terms:
The size of an acute angle is $ 0^\circ < \theta < 90^\circ $.
The size of an obtuse angle is $ 90^\circ < \theta < 180^\circ $.
The size of a reflex angle is $ 180^\circ < \theta < 360^\circ $.
Complementary angles add up to $ 90^\circ $.
Supplementary angles add up to $ 180^\circ $.
2. Conversion between degree and radians:
$ \theta_{\text{radian}} = \theta_{\text{degree}} \times $$ {\pi \over 180} $
$ \theta_{\text{degree}} = \theta_{\text{radian}} \times $$ {180 \over \pi} $
3. Important angles in radians:
$ {\pi \over 2} \text{ radians} = $$ 90^\circ $
$ \pi \text{ radians} = $$ 180^\circ $
$ 2\pi \text{ radians} = $$ 360^\circ $
4. Angle properties:
 $ \angle a = \angle c, \angle d = \angle b $ (Vertically opposite angles) |
 $ \angle a + \angle b + \angle c = 360^\circ $ (Angles at a point) |
 $ \angle a + \angle b = 180^\circ $ (Adjacent angles on a straight line) |
5. Angle properties involving parallel lines:
$ \angle a = \angle b $ (Alternate angles)
$ \angle a = \angle d $ (Corresponding angles)
$ \angle a + \angle c = 180^\circ $ (Interior angles)
1. Properties of angle bisector:
If line $BD$ is the angle bisector of angle $ABC$, then
- Angle $ABD$ = Angle $CBD$
- Any point on line $BD$ is equidistant from lines $AB$ and $CB$
2. Steps to construct angle bisector:
 Before |
 After |
- Place metal tip of compass at $B$ and draw an arc that cuts line $AB$ (at $E$) and line $CB$ (at $F$).
- Place metal tip of compass at $E$ and draw an arc (labelled as arc 1). Do not adjust compass after this.
- Place metal tip of compass at $F$ and draw an arc (labelled as arc 2) that meets the previous arc (at $D$).
- Join points $B$ and $D$ with a straight line
1. Properties of perpendicular bisector:
If line $CD$ is the perpendicular bisector of line $AB$, then
- Line $CD$ is perpendicular to line $AB$
- Line $CD$ passes through the mid-point of line $AB$
- Any point on line $CD$ is equidistant from points $A$ and $B$
2. Steps to construct perpendicular bisector:
 Before |
 After |
- Set compass to more than half the length of $AB$ and do not adjust compass anymore.
- Place metal tip of compass at $A$ and draw an arc (labelled as arc 1)
- Place metal tip of compass at $B$ and draw an arc (labelled as arc 2) that meets the previous arc (at $C$ and $D$)
- Join points $C$ and $D$ with a straight line
1. Angle properties:
At a vertex (or corner), interior angle + exterior angle = $ 180^\circ $ .
Sum of interior angles = $ (n - 2) \times 180^\circ $ , where $n$ is the number of sides in the polygon
Sum of exterior angles = $ 360^\circ $
2. Names of polygons:
| Name of polygon | Number of sides |
|---|---|
| Pentagon | 5 sides |
| Hexagon | 6 sides |
| Octagon | 8 sides |
| Decagon | 10 sides |
Note: These are the only ones specified in the syllabus!
3. Regular polygon vs. irregular polygon
A regular polygon has equal lengths on all sides. This means that interior angles are equal and exterior angles are equal.
On the other hand, an irregular polygon has sides that are not equal and interior/exterior angles that are not equal.
Example: Regular polygon
The interior angle of a regular polygon is $156^\circ$. Find the number of sides in this polygon.
Solutions (by interior angles)
\begin{align*} \text{Let no. of sides} & = n \\ \\ \text{Sum of interior angles} & = (n - 2) \times 180 \\ \underbrace{ 156n }_\text{Sum of n sides} & = 180n - 360 \\ 156n - 180n & = -360 \\ -24n & = -360 \\ n & = {-360 \over -24} \\ n & = 15 \end{align*}
Solutions (by exterior angles)
\begin{align*} \text{Exterior angle} & = 180^\circ - 156^\circ \\ & = 24^\circ \\ \\ \text{Sum of exterior angles} & = 360^\circ \\ \\ \text{Number of sides} & = {360^\circ \over 24^\circ} \\ & = 15 \end{align*}
Example: Irregular polygon
Four interior angles of an octagon are $100^\circ$, $110^\circ$, $120^\circ$ and $130^\circ$ respectively. The rest of the interior angles are $x^\circ$ each, where $x$ is an integer.
Find the value of $x$.
\begin{align*} \text{Sum of interior angles} & = (8 - 2) \times 180^\circ \phantom{000000} [\text{Octagon} \implies 8 \text{ sides}] \\ & = 1080^\circ \\ \\ 1080 & = 100 + 110 + 120 + 130 + 4x \\ 1080 & = 460 + 4x \\ 1080 - 460 & = 4x \\ 620 & = 4x \\ {620 \over 4} & = x \\ 155 & = x \end{align*}
1. Congruent figures
If two figures are congruent, then they have the same exact shape (same corresponding angles) and size (same corresponding length).
2. Calculations involving congruent figures
Example
In the diagram, triangle ABC is congruent to triangle PQR.
Given that AB $ = 7 $ cm, AC $ = 5 $ cm, angle BAC $= 50^\circ$ and angle PQR $ = 44^\circ$, find the length of PR and the size of angle ACB.
\begin{align*} \text{Triangles } & ABC \text{ and } PQR \text{ are congruent} \\ \\ PR & = AC = 5 \text{ cm} \\ \\ \angle ABC & = \angle PQR = 44^\circ \\ \\ \angle ACB & = 180^\circ - 50^\circ - 44^\circ \phantom{0} \text{ (Angle sum of triangle}) \\ & = 86^\circ \end{align*}
3. Similar figures
If two figures are similar, then they have the same exact shape (same corresponding angles) but different size (different length). Corresponding sides have the same scale factor.
For the two similar regular pentagons above, the scale factor is $ 2:5 $ or ${2 \over 5}$.
4. Calculations involving similar figures
Example
In the diagram, triangle PQR is similar to triangle SQT.
Given that PS $ = 6 $cm, SQ $= 3$ cm and QT $= 2$ cm, find the length of TR.
\begin{align*} \text{Triangles } PQR & \text{ and } SQT \text{ are similar} \\ \\ {SQ \over PQ} & = {3 \over 9} \\ & = {1 \over 3} \\ \\ \implies {QT \over QR} & = {1 \over 3} \\ {2 \over QR} & = {1 \over 3} \\ 2(3) & = QR \phantom{000000} [\text{Cross-multiply}] \\ 6 & = QR \\ \\ TR & = 6 - 2 \\ & = 4 \text{ cm} \end{align*}
Tests for congruent triangles (Only for O levels)
1. Tests
SSS (Side-Side-Side)
AAS (Angle-Angle-Side) or ASA (Angle-Side-Angle)
RHS (Right angle-Hypotenuse-Side)
SAS (Side-Angle-Side)
2. Examples
Example 1
In the diagram, triangle PQR is isosceles with PQ = PR.
X is a point on QR such that angle PXQ and angle PXR are right angles. Prove that triangle PQX is congruent to triangle PRX.
\begin{align*} \angle PXQ & = \angle PXR = 90^\circ \phantom{0} \text{ (Given) [R]} \\ \\ PQ & = PR \phantom{0} \text{ (Given) [Hypotenuse]} \\ \\ PX & \text{ is a common side [S]} \\ \\ \therefore \text{Triangles } & PQX \text{ and } PRX \text{ are congruent (RHS)} \end{align*}
\begin{align*} \angle PXQ & = \angle PXR = 90^\circ \phantom{0} \text{ (Given) [A]} \\ \\ \angle PQX & = \angle PRX \phantom{0} \text{ (Base angles of isosceles triangle) [A]} \\ \\ PQ & = PR \phantom{0} \text{ (Given) [S]} \\ \\ \therefore \text{Triangles } & PQX \text{ and } PRX \text{ are congruent (AAS)} \end{align*}
Example 2
In the diagram, lines AB and CD are parallel and have the same length. AXD and BXC are straight lines.
Prove that triangle ABX is congruent to DCX.
\begin{align*} \angle AXB & = \angle DXC \phantom{0} \text{ (Vertically opposite angles) [A]} \\ \\ \angle XAB & = \angle XDC \phantom{0} \text{ (Alternate angles, AB // CD) [A]} \\ \\ AB & = DC \phantom{0} \text{ (Given) [S]} \\ \\ \therefore \text{Triangles } & ABX \text{ and } DCX \text{ are congruent (AAS)} \end{align*}
\begin{align*} \angle XAB & = \angle XDC \phantom{0} \text{ (Alternate angles, AB // CD) [A]} \\ \\ AB & = DC \phantom{0} \text{ (Given) [S]} \\ \\ \angle XBA & = \angle XCD \phantom{0} \text{ (Alternate angles, AB // CD) [A]} \\ \\ \therefore \text{Triangles } & ABX \text{ and } DCX \text{ are congruent (ASA)} \end{align*}
Tests for similar triangles (Only for O levels)
1. Tests
SSS (Side-Side-Side)
AA (Angle-Angle)
SAS (Side-Angle-Side)
2. Examples
Example 1
In the diagram, D and E are points on triangle ABC such that DE is parallel to BC.
Prove that triangle ADE is similar to ABC.
\begin{align*} \angle DAE & = \angle BAC \phantom{0} \text{ (Common angle) [A]} \\ \\ \angle ADE & = \angle ABC \phantom{0} \text{ (Corresponding angles, DE // BC) [A] } \\ \\ \therefore \text{Triangles } & ADE \text{ and } ABC \text{ are similar (AA)} \end{align*}
Example 2
In the diagram, S and T are points on triangle PQR such that PS = 3 cm, SQ = 6 cm, PT = 2 cm and T = 4 cm. In addition, ST = 4 cm and QR = 12 cm.
Prove that triangle PST is similar to triangle PQR.
\begin{align*}
{PS \over PQ} & = {3 \over 9} = {1 \over 3} \\
\\
{PT \over PR} & = {2 \over 6} = {1 \over 3} \\
\\
{ST \over QR} & = {4 \over 12} = {1 \over 3} \\
\\
\text{Since } {PS \over PQ} = {PT \over PR} = {ST \over QR}, &
\text{ triangles } PST \text{ and } PQR \text{ are similar (SSS)}
\end{align*}
\begin{align*}
{PS \over PQ} & = {3 \over 9} = {1 \over 3} \\
\\
\angle SPT & = \angle QPR \phantom{0} \text{ (Common angle)} \\
\\
{PT \over PR} & = {2 \over 6} = {1 \over 3} \\
\\
\text{Since } {PS \over PQ} = {PT \over PR} \text{ and } \angle SPT & = \angle QPR,
\text{ triangles } PST \text{ and } PQR \text{ are similar (SAS)}
\end{align*}
Geometrically similar figures/solids (Only for O levels)
1. Formulas for two geometrically similar figures/solids
Comparing area, $ {A_1 \over A_2} = $ $ \left( l_1 \over l_2 \right)^2 $
Comparing volume, $ {V_1 \over V_2} = $ $ \left( l_1 \over l_2 \right)^3 $
Comparing mass, $ {m_1 \over m_2} = $ $ \left( l_1 \over l_2 \right)^3 $
Example
The curved surface area of two geometrically similar cylinders are $54$ cm2 and $486$ cm2 respectively.
If the capacity of the smaller cylinder is $50$ millilitres (ml), find the capacity of the larger cylinder in ml.
\begin{align*}
{V_1 \over V_2} & = \left(l_1 \over l_2\right)^3 \phantom{000000} [ \text{Need to find } {l_1 \over l_2} \text{ first}] \\
\\
{A_1 \over A_2} & = {54 \over 486} \\
\left(l_1 \over l_2\right)^2 & = {1 \over 9} \\
{l_1 \over l_2} & = \sqrt{ 1 \over 9} \\
{l_1 \over l_2} & = {1 \over 3} \\
\\
{V_1 \over V_2} & = \left(1 \over 3\right)^3 \\
{50 \over V_2} & = {1 \over 27} \\
27(50) & = V_2 \phantom{00000000000} [\text{Cross-multiply}] \\
1350 & = V_2 \\
\\
\therefore \text{Capacity of larger} & \text{ cylinder} = 1350 \text{ ml}
\end{align*}
2. Compare area of two triangles
1. If triangles are similar, use the formula $ {A_1 \over A_2} = \left(l_1 \over l_2\right)^2 $ .
2. If triangles have a common height, use the following method:
\begin{align*} \require{cancel} { \text{Area of triangle } ABD \over \text{Area of triangle } ADC} & = { \cancel{1 \over 2} \times BD \times \cancel{h} \over \cancel{1 \over 2} \times DC \times \cancel{h} } \\ & = {BD \over DC} \end{align*}
Example
The diagram shows a trapezium PQRS, with parallel sides PQ and SR.
The diagonals PR and QS meet at point T and PQ : SR = 2 : 3.
(a) Find the value of $ { \text{Area of triangle PQT} \over \text{Area of triangle RST} } $.
\begin{align*} \text{Triangles PQT } & \text{and RST are similar} \\ \\ { \text{Area of triangle PQT} \over \text{Area of triangle RST} } & = \left(l_1 \over l_2\right)^2 \\ & = \left(PQ \over SR\right)^2 \\ & = \left(2 \over 3\right)^2 \\ & = {4 \over 9} \end{align*}
(b) Find the value of $ { \text{Area of triangle PQT} \over \text{Area of triangle PST} } $.
\begin{align*} \require{cancel} { \text{Area of triangle PQT} \over \text{Area of triangle PST} } & = { \cancel{1 \over 2} \times QT \times \cancel{h} \over \cancel{1 \over 2} \times ST \times \cancel{h} } \\ & = {QT \over ST} \\ & = {PQ \over RS} \phantom{000000} [\text{Triangles PQT and RST are similar}] \\ & = {2 \over 3} \end{align*}
$ \angle ACB = \angle ADB, \angle CAD = \angle CBD \text{ (}$ $ \text{Angles in the same segment} $ $ \text{)} $
If O is the centre of the circle,
$ \angle AOB = 2 \times \angle ACB, \text{Reflex } \angle AOB = 2 \times \angle ADB \text{ (}$ $ \text{Angle at centre} = 2 \times \text{Angle at circumference} $ $ \text{)} $
$ \angle ACB + \angle ADB = 180^\circ, \angle CAD + \angle CBD = 180^\circ \text{ (}$ $ \text{Angles in opposite segments} $ $ \text{)} $
Note: All four points ($A$, $B$, $C$ and $D$) must lie on the circumference of the circle.
If AOB is the diameter of the circle,
$ \angle ACB = 90^\circ \text{ (}$ $ \text{Right-angle in semi-circle} $ $ \text{)} $
If O is the centre of the circle and the line AC is tangent to the circle at B,
$ \angle OBA = \angle OBC = 90^\circ \text{ (}$ $ \text{Tangent perpendicular to radius} $ $ \text{)} $
If O is the centre of the circle and tangents to the circle at A and at B meet at an external point T, then
$ AT = BT $ $ \text{ and } $ $ \angle OTA = \angle OTB $
The perpendicular bisector of chord $AB$ passes through the centre of the circle, $O$.
If chords $AB$ and $CD$ have the same length, then $ OM = ON $.
1. Pythagoras theorem
Formula: $a^2 + b^2 = c^2 $
Example
In triangle $ABC$, $AB = 5 $ cm, $BC = 12 $ cm and $AC = 13 $ cm. Prove that $ABC$ is a right-angle triangle.
\begin{align*} AC^2 & = 13^2 \\ & = 169 \\ \\ AB^2 + BC^2 & = 5^2 + 12^2 \\ & = 169 \\ \\ \text{Since } AC^2 = AB^2 + BC^2, \text{ by } & \text{the converse of Pythagoras theorem,} \\ \text{ triangle } ABC \text{ is a right-ang} & \text{led triangle (with angle } ABC = 90^\circ) \end{align*}
2. TOA CAH SOH (for right-angle triangle only)
Example: Find length
In triangle $ABC$, $AB = 4$ cm and angle $CAB = 50^\circ$. Find the length of $AC$.
\begin{align*} \cos 50^\circ & = {AB \over AC} \phantom{000000} \left[ {Adj \over Hyp} \right] \\ { \cos 50^\circ \over 1} & = {4 \over AC} \\ AC \cos 50^\circ & = 4 \phantom{0000000000} [\text{Cross-multiply}] \\ AC & = {4 \over \cos 50^\circ} \\ & = 6.2228 \\ & \approx 6.22 \text{ cm} \end{align*}
Example: Find angle
In triangle $PQR$, $PQ = 5$ cm, $QR = 12$ cm and $PRQ = \theta$ radians. Find the value of $ \theta $.
\begin{align*} \tan \angle PRQ & = {PQ \over QR} \phantom{000000000000} \left[ {Opp \over Adj} \right] \\ \tan \theta & = {5 \over 12} \\ \theta & = \tan^{-1} \left(5 \over 12\right) \phantom{000000} [\text{Radian mode! Remember to switch back after}] \\ & = 0.39479 \\ & \approx 0.395 \end{align*}
3. Trigonometric ratio of obtuse angles
Recap: The size of an obtuse angle is $ 90^\circ < \theta < 180^\circ $.
If $\theta$ is an acute angle, then
$ \sin (180^\circ - \theta) = $ $ \sin \theta $
$ \cos (180^\circ - \theta) = $ $ - \cos \theta $
$ \tan (180^\circ - \theta) = $ $ - \tan \theta $
Example
In triangle $PQR$, $PQ = 4$ cm, $QR = 3$ cm, $PR = 5$ cm and angle $PQR$ is a right angle. If $QRT$ is a straight line, find the value of $ \sin PRT $ and $ \cos PRT $.
\begin{align*} \sin PRT & = \sin PRQ \\ & = { PQ \over PR} \phantom{0000000} \left[ {Opp \over Hyp} \right] \\ & = {4 \over 5} \\ \\ \cos PRT & = - \cos PRQ \\ & = - {QR \over PR} \phantom{000000} \left[ {Adj \over Hyp} \right] \\ & = - {3 \over 5} \end{align*}
4. Find two possibles values of $\theta$ given $ \sin \theta $
Example
Given that $ \sin \theta = 0.5 $, find the two possible values of $\theta$ in degrees.
$$ \sin \theta = \sin (180^\circ - \theta) = 0.5 $$
\begin{align*}
\theta & = \sin^{-1} (0.5) && \text{ or } & 180^\circ - \theta & = \sin^{-1} (0.5) \\
\theta & = 30^\circ &&& 180^\circ - \theta & = 30^\circ \\
& &&& -\theta & = 30^\circ - 180^\circ \\
& &&& -\theta & = -150^\circ \\
& &&& \theta & = 150^\circ
\end{align*}
5. Area of triangle
Area of triangle $ABC = {1 \over 2} a b \sin C$ (provided)
Note: Angle $C$ is the angle between the two sides $a$ and $b$.
Example
In triangle $PQR$, $PQ = 8$ cm, $QR = 10$ cm and angle $PRQ = 30^\circ$.
(a) Find the area of triangle $PQR$.
\begin{align*} \text{Area of triangle } PQR & = {1 \over 2} (QR)(PR) \sin \angle PRQ \\ & = {1 \over 2}(10)(8)\sin 30^\circ \\ & = 20 \text{ cm}^2 \end{align*}
(b) Hence, find the shortest distance from $P$ to $QR$.
\begin{align*} \text{Area of triangle } PQR & = {1 \over 2} \times QR \times PF \\ 20 & = {1 \over 2} \times 10 \times PF \\ 20 & = 5PF \\ {20 \over 5} & = PF \\ 4 \text{ cm} & = PF \\ \\ \therefore \text{Shortest distance from } P \text{ to } QR & = 4 \text{ cm} \end{align*}
6. Sine rule
${a \over \sin A} = {b \over \sin B} = {c \over \sin C}$ (provided)
Example: Find length
In triangle $PQR$, $PQ = 6 $ cm, angle $PQR = 0.908$ radians and angle $PRQ = 0.484$ radians. Find the length of $PR$.
\begin{align*} \text{By Sine rule, } \phantom{.} {PR \over \sin \angle PQR} & = {PQ \over \sin \angle PRQ} \\ {PR \over \sin 0.908} & = {6 \over \sin 0.484} \\ PR \sin 0.484 & = 6 \sin 0.908 \phantom{000000} [\text{Cross-multiply}] \\ PR & = {6 \sin 0.908 \over \sin 0.484} \phantom{00000} [\text{Radian mode! Remember to switch back.}] \\ & = 10.164 \\ & \approx 10.2 \text{ cm} \end{align*}
Example: Find angle
In triangle $ABC$, $AB = 8$ cm, $BC = 9$ cm, $AC = 10$ cm and angle $BCA = 50^\circ$. Find the size of obtuse angle $ABC$.
\begin{align*} \text{By Sine rule, } \phantom{.} {AC \over \sin \angle ABC} & = {AB \over \sin \angle ACB} \\ {10 \over \sin \angle ABC} & = {8 \over \sin 50^\circ} \\ 10 \sin 50^\circ & = 8 \sin \angle ABC \phantom{00000} [\text{Cross-multiply}] \\ {10 \sin 50^\circ \over 8} & = \sin \angle ABC \\ \\ \therefore \angle ABC & = 180^\circ - \sin^{-1} \left( {10 \sin 50^\circ \over 8} \right) \\ & = 106.754^\circ \\ & \approx 106.8^\circ \end{align*}
7. Cosine rule
$ a^2 = b^2 + c^2 - 2bc \cos A $ (provided)
Example: Find length
In triangle $ABC$, $AB = 4$ cm, $AC = 10$ cm and angle $BAC = 60^\circ$. Find the length of $BC$.
\begin{align*} \text{By Cosine rule, } \phantom{.} BC^2 & = AB^2 + AC^2 - 2(AB)(AC)\cos \angle BAC \\ & = 4^2 + 10^2 - 2(4)(10) \cos 60^\circ \\ & = 76 \\ \\ BC & = \sqrt{76} \\ & = 8.7177 \\ & \approx 8.72 \text{ cm} \end{align*}
Example: Find angle
In triangle $PQR$, $PQ = 4$ cm, $QR = 9$ cm and $PR = 10$ cm. Find the size of angle $PQR$ in degrees.
\begin{align*} \text{By Cosine rule, } \phantom{.} PR^2 & = PQ^2 + QR^2 - 2 (PQ)(PR) \cos \angle PQR \\ 10^2 & = 4^2 + 9^2 - 2(4)(9) \cos \angle PQR \\ 100 & = 16 + 81 - 72\cos \angle PQR \\ 72 \cos \angle PQR & = 16 + 81 - 100 \\ 72 \cos \angle PQR & = -3 \\ \cos \angle PQR & = -{3 \over 72} \\ \angle PQR & = \cos^{-1} \left(-{3 \over 72}\right) \\ & = 92.388^\circ \\ & \approx 92.4^\circ \end{align*}
1. Bearings
Bearings must be:
- Calculated from the North and in the clockwise direction
- Presented in 3 digits (not including decimals, i.e. $123.4^\circ$, $023^\circ$)
Example
In triangle $ABC$, $C$ is due east of $A$ and angle $BAC = 15^\circ$.
(a) Find the bearing of $B$ from $A$.
\begin{align*} \angle NAC & = 90^\circ \phantom{000000} [C \text{ is east of } A] \\ \\ \text{Bearing of } B \text{ from } A & = 90^\circ - 15^\circ \\ & = 075^\circ \phantom{00000} [\text{3 digits}] \end{align*}
(b) Find the bearing of $A$ from $B$.
\begin{align*} \angle N_1 B A & = 180^\circ - 75^\circ \phantom{0} \text{ (Interior angles)} \\ & = 105^\circ \\ \\ \text{Bearing of } A \text{ from } B & = 360^\circ - 105^\circ \phantom{0} \text{ (Angles at a point)} \\ & = 255^\circ \end{align*}
2. Angle of elevation and angle of depression
$ \theta$ is the angle of elevation of object $A$ from $O$.
$ \alpha$ is the angle of depression of object $B$ from $O$.
Example
In the diagram, points $P$, $Q$ and $R$ are points on horizontal ground. $S$ is a point on $PQ$ such that $SR$ is perpendicular to $PQ$. $PR = 10$ m, $SR = 8 $ m and $QR = 17$ m.
A tree is located at $R$ and the top of the tree is denoted by $T$. The angle of depression of $P$ from $T$ is $45^\circ$.
(a) Find the height of the tree.
\begin{align*} \tan \angle TPR & = {TR \over PR} \phantom{000000} \left[ {Opp \over Adj} \right] \\ \tan 45^\circ & = {TR \over 10} \\ 1 & = {TR \over 10} \\ 10 & = TR \\ \\ \text{Height of the tree} & = 10 \text{ m} \end{align*}
(b) Find the greatest possible angle of elevation of $T$ from $PQ$.
The greatest possible angle of elevation of $T$ from $PQ$ occurs at the point that is closest to $R$, which is $S$.
(Imagine looking at a clock on the wall - the closer you walk towards the wall, the more you have to tilt your head up to look at the clock!)
\begin{align*} \tan \angle TSR & = {TR \over SR} \phantom{000000} \left[ {Opp \over Adj} \right] \\ & = {10 \over 8} \\ \angle TSR & = \tan^{-1} \left(10 \over 8\right) \\ & = 51.34^\circ \\ & \approx 51.3^\circ \end{align*}
Area and perimeter of plane figures
1. Units
$ 1 \text{ cm} = \phantom{.} $ $ 10 $ $\text{ mm}$
$ 1 \text{ cm}^2 = \phantom{.} $ $ 10^2 = 100 $ $\text{ mm}^2$
$ 1 \text{ m} = \phantom{.} $ $ 100 $$ \text{ cm}$
$ 1 \text{ m}^2 = \phantom{.} $ $ 100^2 = 10 \phantom{.} 000 $$ \text{ cm}^2$
$ 1 \text{ km} = \phantom{.} $ $ 1000 $ $\text{ m}$
$ 1 \text{ km}^2 = \phantom{.} $ $ 1000^2 = 1 \phantom{.} 000 \phantom{.} 000$ $ \text{ m}^2 $
2. Formulas
1. Parallelogram or rhombus
$ \text{Area of parallelogram/rhombus} = $$ \phantom{.} \text{Base} \times \text{Height} $
2. Trapezium
$ \text{Area of trapezium} = $$ \phantom{.} {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} $
3. Triangle
$ \text{Area of triangle} = $$ \phantom{.} {1 \over 2} \times \text{Base} \times \text{Height} $
Note: The other formula, ${1 \over 2} ab \sin C$, is provided.
4. Circle
$ \text{Area of circle} = $$ \phantom{.} \pi r^2 $
$ \text{Area of triangle} = $$ \phantom{.} 2\pi r = \pi d $
Arc length, sector area and segment of a circle
1. Major sector vs. minor sector
The shaded region is the minor sector $AOB$ while the unshaded region is the major sector $OACB$.
$ \text{Sector area (degrees)} = $$ \phantom{.} {\theta \over 360} \times \pi r^2 $
$ \text{Sector area (radians)} = {1 \over 2} r^2 \theta $ (provided)
$AB$ is the minor arc while $ACB$ is the major arc.
$ \text{Arc length (degrees)} = $$ \phantom{.} {\theta \over 360} \times 2 \pi r $
$ \text{Arc length (radians)} = r \theta $ (provided)
Example
The diagram shows major sector $OPQ$ with centre $O$ and radius of $4$ cm. If angle $POQ = 1.75$ radians, find the length of major arc $PQ$ and the area of major sector $OPQ$.
\begin{align*} \text{Reflex } \angle POQ & = 2\pi - 1.75 \phantom{000000} [360^\circ = 2\pi \text{ radians}] \\ \\ \text{Arc length} & = r \theta \\ & = 4(2\pi - 1.75) \\ & = 8\pi - 7 \\ & \approx 18.1 \text{ cm} \\ \\ \text{Sector area} & = {1 \over 2} r^2 \theta \\ & = {1 \over 2} (4)^2 (2\pi - 1.75) \\ & = {1 \over 2} (16) (2\pi - 1.75) \\ & = 8(2\pi - 1.75) \\ & = 16\pi - 14 \\ & \approx 36.3 \text{ cm}^2 \end{align*}
\begin{align*} \pi \text{ radians} & = 180^\circ \\ 1 \text{ radian} & = \left(180 \over \pi\right)^\circ \\ 1.75 \text{ radians} & = {180 \over \pi} \times 1.75 \\ & = 100.27^\circ \\ \\ \text{Reflex } \angle POQ & = 360^\circ - 100.27^\circ \\ & = 259.73^\circ \\ \\ \text{Arc length} & = {\theta \over 360} \times 2 \pi r \\ & = {259.73 \over 360} \times 2 \pi (4) \\ & = 18.13 \\ & \approx 18.1 \text{ cm} \\ \\ \text{Sector area} & = {\theta \over 360} \times \pi r^2 \\ & = {259.73 \over 360} \times \pi (4)^2 \\ & = 36.27 \\ & \approx 36.3 \text{ cm}^2 \end{align*}
2. Segment of circle
A segment of a circle is the area enclosed by an arc and a chord.
In the diagram, the minor segment (shaded region) is enclosed by arc $ACB$ and the chord $AB$.
$ \text{Area of minor segment} = $$ \phantom{.} \text{Area of minor sector } OACB - \text{Area of triangle } OAB $
$ \text{Perimeter of minor segment} = $$ \phantom{.} \text{Minor arc length } ACB + AB $
In the same diagram, the major segment (unshaded region) is enclosed by arc $ADB$ and the chord $AB$.
$ \text{Area of major segment} = $$ \phantom{.} \text{Area of major sector } OADB + \text{Area of triangle } OAB $
$ \text{Perimeter of major segment} = $$ \phantom{.} \text{Major arc length } ADB + AB $
Volume and surface area of solids
1. Units
$ 1 \text{ m}^3 = \phantom{.} $ $ 100^3 = 1 \phantom{.} 000 \phantom{.} 000 $$ \text{ cm}^3$
$ 1 \text{ litre} = \phantom{.} $ $ 1000 $$ \text{ cm}^3$
2. Cuboid and cube
$ \text{Volume of cuboid} = \phantom{.} $ $ l \times b \times h $
$ \text{Surface area of closed cuboid} = \phantom{.} $ $ 2 (l \times b) + 2(l \times h) + 2(b \times h) $
A cube is a special case of a cuboid where all sides have the same length
$ \text{Volume of cube} = \phantom{.} $ $ l \times l \times l = l^3 $
$ \text{Surface area of closed cube} = \phantom{.} $ $ 6 (l \times l) = 6l^2 $
3. Cylinder
$ \text{Volume of cylinder} = \phantom{.} $ $ \pi r^2 h $
$ \text{Curved surface area} = \phantom{.} $ $ 2 \pi r h $
4. Prism
$ \text{Volume of prism} = \phantom{.} $ $ \text{Cross-sectional area} \times \text{Height/Length *} $
*: It depends on the orientation of the prism
Example
The diagram shows a prism where the cross-section is an equilateral triangle with length of $6$ cm. The length of the prism is $12$ cm.
Find the volume of the prism.
\begin{align*} \text{Interior angle} & = {180^\circ \over 3} \\ & = 60^\circ \\ \\ \text{Area of triangle} & = {1 \over 2} ab \sin C \phantom{000000} [\text{Provided}] \\ & = {1 \over 2}(6)(6) \sin 60^\circ \\ & = 15.588 \text{ cm}^2 \\ \\ \text{Volume of prism} & = \text{Cross-sectional area} \times \text{Length} \\ & = 15.588 \times 12 \\ & = 187.056 \\ & \approx 187 \text{ cm}^2 \end{align*}
5. Pyramid
$ \text{Volume of pyramid} = \phantom{.} $ $ {1 \over 3} \times \text{Base area} \times \text{Height} $
Note: There is no specific formula to find the surface area of a pyramid. The key is to find the height of the slanted triangle (see Example below).
Example
The diagram shows a pyramid with a square base $ABCD$, where $AB = 5$ cm and $VO$ is the height of the pyramid.
Given that the volume of the pyramid is $75$ cm3, find the surface area of the pyramid.
Find the volume of the prism.
\begin{align*} \text{Volume of pyramid} & = {1 \over 3} \times \text{Base area} \times \text{Height} \\ 75 & = {1 \over 3} \times (5 \times 5) \times \text{Height} \\ {75 \over {1 \over 3} \times (5 \times 5) } & = \text{Height} \\ 9 & = \text{Height} \end{align*}
\begin{align*} OE & = {5 \over 2} = 2.5 \text{ cm} \\ \\ \text{By Py} & \text{thagoras theorem,} \\ VE^2 & = 9^2 + 2.5^2 \\ VE & = \sqrt{9^2 + 2.5^2} \\ & = 9.3408 \text{ cm} \\ \\ \text{Area of triangle } VBC & = {1 \over 2} \times 5 \times 9.3408 \\ & = 23.352 \text{ cm}^2 \\ \\ \text{Surface area of pyramid} & = 4(23.352) + (5 \times 5) \\ & = 118.408 \\ & \approx 118 \text{ cm}^2 \end{align*}
6. Cone
$ \text{Volume of cone} = {1 \over 3} \pi r^2 h $ (provided)
$ \text{Curved surface area of cone} = \pi r l $ (provided)
The relationship between the radius ($r$), height ($h$) and slant length ($l$) is given by $ l^2 = r^2 + h^2 $
7. Sphere & hemisphere
$ \text{Volume of sphere} = {4 \over 3} \pi r^3 $ (provided)
$ \text{Curved surface area of sphere} = 4 \pi r^2 $ (provided)
$ \text{Volume of hemisphere} = \phantom{.} $ $ {2 \over 3} \pi r^3 $
$ \text{Curved surface area of hemisphere} = \phantom{.} $ $ 2 \pi r^2 $
1. Formulas
For the points $A(x_1, y_1)$ and $B(x_2, y_2)$,
$ \text{Length of } AB = \phantom{.} $ $ \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 }$
$ \text{Gradient of } AB = \phantom{.} $ $ {y_2 - y_1 \over x_2 - x_1} $
2. Equation of straight line
$$ y = mx + c $$
$ m $ represents the gradient of the line and $c$ represents the $y$-intercept.
Example
Find the gradient and $y$-intercept of the straight line $2y + 3x = 6$.
\begin{align*} 2y + 3x & = 6 \\ 2y & = -3x + 6 \\ y & = -{3 \over 2}x + 3 \phantom{00000} [\text{Divide each term by 2} ] \\ \\ \therefore \text{Gradient, } m & = -{3 \over 2} \\ \therefore y \text{-intercept, } c & = 3 \end{align*}
3. Form equation of straight line
To form the equation of a straight line, we need the gradient of the line and the coordinates of one point on the line.
Example
Form the equation of the line that passes through the points $P(-2, -5)$ and $Q(2, 3)$.
\begin{align*} \text{Gradient} & = {y_2 - y_1 \over x_2 - x_1} \\ & = {3 - (-5) \over 2 - (-2)} \\ & = 2 \\ \\ y & = mx + c \\ y & = 2x + c \\ \\ \text{Using } & P(-2, -5), \\ -5 & = 2(-2) + c \\ -5 & = -4 + c \\ -5 + 4 & = c \\ -1 & = c \\ \\ \text{Eqn of } PQ: & \phantom{.} y = 2x - 1 \end{align*}
4. Finding coordinates of points on the line
Any point that lie on the line must satisfy the equation of the line (see part ai).
To find the $x$-intercept of the line, we let $y = 0$.
To find the $y$-intercept of the line, we let $x = 0$.
Example
The equation of line $l$ is $2y + 3x = 6$. Line $l$ cuts the $y$-axis and the $x$-axis at points $A$ and $B$ respectively.
(a) Show that the point $C(4, -3)$ lies on $l$.
\begin{align*} 2y + 3x & = 6 \\ \\ \text{Let } & x = 4, \\ 2y + 3(4) & = 6 \\ 2y + 12 & = 6 \\ 2y & = 6 - 12 \\ 2y & = -6 \\ y & = {-6 \over 2} \\ y & = -3 \\ \\ \therefore C(4, -3) & \text{ lies on line } l \end{align*}
(b) Find the coordinates of points $A$ and $B$.
\begin{align*} 2y + 3x & = 6 \\ \\ \text{Let } & x = 0, \\ 2y + 3(0) & = 6 \\ 2y + 0 & = 6 \\ 2y & = 6 \\ y & = {6 \over 2} \\ y & = 3 \\ \\ \therefore & \phantom{.} A(0, 3) \\ \\ \text{Let } & y = 0, \\ 2(0) + 3x & = 6 \\ 0 + 3x & = 6 \\ 3x & = 6 \\ x & = {6 \over 3} \\ x & = 2 \\ \\ \therefore & \phantom{.} B(2, 0) \end{align*}
Point $D$ is a point on line $l$ such that the $x$-coordinate of $D$ is equal to the $y$-coordinate of $D$.
(c) Find the coordinates of the point $D$.
\begin{align*} \text{Let } x \text{-coordinate of } D & = d \\ \\ \therefore & \phantom{.} D(d, d) \\ \\ 2y + 3x & = 6 \\ \\ \text{Using } & D(d, d), \\ 2d + 3d & = 6 \\ 5d & = 6 \\ d & = {6 \over 5} \\ d & = 1.2 \\ \\ \therefore & \phantom{.} D(1.2, 1.2) \end{align*}
5. Parallel lines
If two lines are parallel, they have the same gradient. In addition, if both lines have different $y$-intercepts (as seen in the diagram above), they would not intersect.
On the other hand, two lines that are not parallel will meet at a point. We can solve for the coordinates of the point by solving simultaneous equations using the equations of the two lines.
Example
Find the coordinates of the point of intersection of the lines $y = x + 1$ and $y - 2x = -1$.
\begin{align*} y & = x + 1 \phantom{0} \text{--- (1)} \\ \\ y - 2x & = -1 \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x + 1 - 2x & = -1 \\ x - 2x & = -1 - 1 \\ -x & = -2 \\ x & = 2 \\ \\ \text{Substitute } & x = 2 \text{ into (1),} \\ y & = 2 + 1 \\ y & = 3 \\ \\ \therefore \text{Point of} & \text{ intersection: } (2, 3) \end{align*}
1. Column vectors
$$ {x \choose y} $$
The top value is the $x$-component and the bottom value is the $y$-component
Example
$ \overrightarrow{AB} = \phantom{.}$ $ {1 \choose -8} $
$ \overrightarrow{CD} = \phantom{.}$ $ {6 \choose 0} $
$ \textbf{p} = \phantom{.}$ $ {3 \choose 2} $
$ \textbf{q} = \phantom{.}$ $ {-3 \choose -2} $
2. Position vectors
The position vector of a point is the vector that represents the location of the point with respect to the origin $O(0, 0)$.
Example
$ \overrightarrow{OA} = \phantom{.}$ $ {4 \choose 3} $
$ \overrightarrow{OB} = \phantom{.}$ $ {-2 \choose -2} $
3. Magnitude (or length of vector)
$ \text{Magnitude of vector } {x \choose y} = \phantom{.} $ $ \sqrt{x^2 + y^2} $
4. Vector addition
$ \overrightarrow{OA} = \phantom{.} $ $ \overrightarrow{OB} + \overrightarrow{BA} $
$ \overrightarrow{OB} = \phantom{.} $ $ \overrightarrow{OA} + \overrightarrow{AB} $
$ \overrightarrow{AB} = \phantom{.} $ $ \overrightarrow{AO} + \overrightarrow{OB} $
5. Parallel vectors
If vectors $\textbf{a}$ and $\textbf{b}$ are parallel, then $ \textbf{a} = k \textbf{b} $, where $k$ is a real number.
6. Collinear points
Collinear points lie on the same straight line.
To prove that three points $A$, $B$ and $C$ are collinear, we can show that $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are parallel.
Example
Given that $ \overrightarrow{PQ} = {4 \choose -10} $ and $ \overrightarrow{QR} = {6 \choose -15} $, show that the points $P$, $Q$ and $R$ are collinear.
\begin{align*} \text{Let } \overrightarrow{PQ} & = k \overrightarrow{QR}, \text{ when } k \text{ is a real value} \\ {4 \choose -10} & = k {6 \choose -15} \\ {4 \choose -10} & = {6k \choose -15k} \\ \\ 4 = 6k \phantom{000.} & \text{ or } \phantom{00} -10 = -15k \\ {4 \over 6} = k \phantom{0000.} & \phantom{00000} {-10 \over -15} = -k \\ {2 \over 3} = k \phantom{0000.} & \phantom{0000.000} {2 \over 3} = k \\ \\ \text{Since } \overrightarrow{PQ} = {2 \over 3} \overrightarrow{QR} \text{, both vectors}& \text{ are parallel and points } P, Q \text{ and } R \text{ are collinear} \end{align*}
Data analysis & Probability
1. Mean, mode and median
Mean is the average of the data. The formula is provided: $ \text{Mean} = { \sum fx \over \sum f} $.
Mode is the data that occurs most frequently.
Median is the data that occurs at the centre. We can use the formula $ {n + 1 \over 2}$, where $n$ is the number of data to locate the centre.
Example
$$ 15, 20, 9, 4, 9, 11, 15, 20, 20 $$
Find the mean, mode and median of the data set above.
\begin{align*} 4 \phantom{0} 9 & \phantom{0} 9 \phantom{0} 11 \phantom{0} 15 \phantom{0} 15 \phantom{0} 20 \phantom{0} 20 \phantom{0} 20 \phantom{000000} [\text{Rearrange - important for median}] \\ \\ \text{Mean} & = {\sum fx \over \sum f} \\ & = {1 \times 4 + 2 \times 9 + 1 \times 11 + 2 \times 15 + 3 \times 20 \over 9} \\ & = 13{2 \over 3} \\ \\ \text{Mode} & = 20 \\ \\ \text{Median position} & = {9 + 1 \over 2} \\ & = 5 \\ \\ \text{Median} & = 15 \end{align*}
2. Range and interquartile range
$ \text{Range} = \phantom{.} $ $\text{Maximum} - \text{Minimum}$
$ \text{Interquartile range} = \phantom{.} $ $\text{Upper quartile } (Q_3) - \text{Lower quartile } (Q_1) $
Example
$$ 4, 9, 9, 11, 15, 15, 20, 20, 20 $$
Find the range and the interquartile range of the data set above.
\begin{align*} \text{Range} & = 20 - 4 \\ & = 16 \\ \\ \\ \boxed{4 \phantom{0} 9 \phantom{0} 9 \phantom{0} 11} & \phantom{0} \underline{15} \phantom{0} \boxed{15 \phantom{0} 20 \phantom{0} 20 \phantom{0} 20} \\ \\ \text{Median position} & = {9 + 1 \over 2} \\ & = 5 \\ \\ \text{Lower quartile} & = {9 + 9 \over 2} \phantom{00000000} [\text{Median of 1st 4}] \\ & = 9 \\ \\ \text{Upper quartile} & = {20 + 20 \over 2} \phantom{000000} [\text{Median of last 4}] \\ & = 20 \\ \\ \text{Interquartile range} & = 20 - 9 \\ & = 11 \end{align*}
3. Standard deviation
$ \text{Standard deviation} = \sqrt{ { \sum fx^2 \over \sum f } - \left( \sum fx \over \sum f\right)^2 } $ (provided)
$ \sum fx^2 $ is the sum of the square of each data.
$ \sum fx $ is the sum of the data.
$ \sum f$ is the total frequency (or total number of data).
Example
| Data ($x$) | $ 20 \le x < 22 $ | $ 22 \le x < 24 $ | $ 24 \le x < 26 $ | $ 26 \le x \le 30 $ |
|---|---|---|---|---|
| Frequency | 2 | 4 | 6 | 8 |
Find the mean and standard deviation of the data set above.
Solutions (by manual calculation)
\begin{align*} [\text{Use mid-values } & 21, 23, 25, 28 \text{ for calculations}] \\ \\ \sum fx^2 & = 2 \times 21^2 + 4 \times 23^2 + 6 \times 25^2 + 8 \times 28^2 \\ & = 13020 \\ \\ \sum fx & = 2 \times 21 + 4 \times 23 + 6 \times 25 + 8 \times 28 \\ & = 508 \\ \\ \sum f & = 2 + 4 + 6 + 8 \\ & = 20 \\ \\ \text{Mean} & = { \sum fx \over \sum f} \\ & = {508 \over 20} \\ & = 25.4 \\ \\ \text{Standard deviation} & = \sqrt{ { \sum fx^2 \over \sum f } - \left( \sum fx \over \sum f\right)^2 } \\ & = \sqrt{ { 13020 \over 20} - \left(508 \over 20\right)^2 } \\ & = 2.4166 \\ & \approx 2.42 \end{align*}
Solutions (by Casio fx-97SG X)
Steps:
- Press SHIFT, then MENU, then DOWN and select 2: Statistics. Select 1: On to turn on Frequency
- Press MENU, the select 3: Statistics, then select 1: 1-Variable
- Enter data: For $x$, use the mid-values 21, 23, 25, 28. For Freq, enter as per the table. Once done, press AC
- Press OPTN (below SHIFT), then select 2: 1-Variable Calc
- Note down the values for $\sum x$, $\sum x^2$ and $n$ (scroll down - the value of $n$ is equal to $ \sum f$). Use these values to calculate mean and standard deviation.
- To reset the calculator, press AC, then MENU and select 1: Calculate
\begin{align*} \text{Mean} & = { \sum fx \over \sum f} \\ & = {508 \over 20} \\ & = 25.4 \\ \\ \text{Standard deviation} & = \sqrt{ { \sum fx^2 \over \sum f } - \left( \sum fx \over \sum f\right)^2 } \\ & = \sqrt{ { 13020 \over 20} - \left(508 \over 20\right)^2 } \\ & = 2.4166 \\ & \approx 2.42 \end{align*}
4. Comparison question
Mean, mode or median is used to compare the average of two data sets.
- If there are outliers (i.e. extreme values) in the data set, the mean may be skewed. Thus, it is better to compare the median of the two data sets.
Range, interquartile range or standard deviation is used to compare the consistency of two data sets.
- The lower the range/interquartile range/standard deviation, the more consistent the data
- If there are outliers in the data set, the range and standard deviation may be skewed. Thus, it is better to compare the interquartile range of the two data sets.
- If there are changes (i.e. increase/decrease) that applies to all the data in the data set, the range/interquartile range/standard deviation will remain the same since consistency remains the same.
Example
Students in class 3A and 3B were given the same mathematics test. The total marks for the test is 50. The mean mark and standard deviation for each class are shown below.
| Class | Mean mark | Standard deviation |
|---|---|---|
| 3A | 28.5 | 9.2 |
| 3B | 25.1 | 8.4 |
Compare and comment briefly on the results of the two classes.
Class 3A performed better as a whole since the mean mark of 3A is higher than that of 3B.
The scores obtained by students in class 3B are more consistent as the standard deviation of scores in 3B is lower than the standard deviation of scores in 3A.
1. Formulas
$ \text{Total probability} = \phantom{.} $ $ 1 $
$ \text{P(Event A does not occur}) = \phantom{.} $ $ 1 - \text{P(Event A occurs)} $
$ \text{P(Impossible event)} = \phantom{.} $ $ 0 $
Example
A bag is said to contain red marbles, green marbles and blue marbles. A marble is picked at random from the bag.
If the probability that the marble is red is 0.31 and the probability that the marble is green is 0.69, find the probability that the marble is blue. Suggest what does your answer implies about the number of blue marbles in the bag.
\begin{align*} \text{P(Blue marble picked up)} & = 1 - \text{P(Blue marble not picked up)} \\ & = 1 - \text{P(Marble is red or green)} \\ & = 1 - (0.31 + 0.69) \\ & = 0 \\ \\ \implies \text{There are } & \text{no blue marbles in the bag} \end{align*}
2. Possibility diagram
The possibility diagram lists out the possible outcomes in a table form.
Example
A bag contains 4 identical cards numbered 1, 2, 3 and 4. Brian draws two cards at random, one after the other, without replacement.
By first constructing a possibility diagram, find the probability that the sum of the two numbers is a prime number.
\begin{align*} [\text{Prime numbers: } & 2, 3, 5, 7, ...] \\ \\ \text{P(Sum is a prime number)} & = { 2 + 2 + 2 + 2 \over 3(4)} \phantom{000000} \left[ {\text{No. of outcomes} \over \text{Total outcomes}}\right] \\ & = {2 \over 3} \end{align*}
3. Tree diagram
Example
A bag contains 16 red marbles and 9 blue marbles.
One marble is taken from the bag at random.
- If the marble is red, it is replaced in the bag. Then, a second marble is drawn.
- If the marble is blue, it is not replaced in the bag and a second marble is drawn.
By first constructing a tree diagram to show the possible outcomes and their probabilities, find the probability both marbles are drawn are of the same colour.
\begin{align*} \text{Case 1: P(Red, Red)} & = {16 \over 25} \times {16 \over 25} \\ & = {256 \over 625} \\ \\ \text{Case 2: P(Blue, Blue)} & = {9 \over 25} \times {8 \over 24} \\ & = {3 \over 25} \\ \\ \text{Required probability} & = {256 \over 625} + {3 \over 25} \\ & = {331 \over 625} \end{align*}