General algebra and numbers

… A ‘term’ is a number or letter (or group of number and letters) that adds or subtracts. For example,

2y + 5zy is an expression with two terms.

… Solving equations with the unknown on both sides. For example:

2x + 3 = 9 – x

It can be useful to ‘bubblize’ the term that you wish to ‘zap’. Then write what you need to do to both sides in order to ‘zap’ that term. For example:

(-3) (-3)

2x = 6 – x

(+x) (+x)

3x = 6

(÷3) (÷3)

x = 2

… A fraction is part of a whole (think slices of cake!).

… You can scale a fraction up from its lowest form by multiplying the denominator and numerator by the same number. It will then become an equivalent fraction. For example

2/5 = 6/15 (by multiplying top and bottom by 3)

… Finding the fraction of a slice in a pie chart. For example, if a segment has an angle of 9°, then the fraction that this segment represents would be:

9°/360° = 3/120 = 1/40

The segment (slice) is 1/40 th of the pie.

We can then use this to calculate the frequency that this fraction represents = (1/40) x total frequency.

… To compare fractions (e.g. if you are asked to write them in ascending size order), you need to scale up each fraction so that their …?… are equal.

Which is the larger fraction? 3/4 or 4/5 ?

… Multiplying fractions by first cancelling (dividing) numerators with denominators and then multiplying the “top with the top” and the “bottom with the bottom”.

… Dividing fractions by flipping the divider fraction and turning the division into a multiplication? For example:

2/5 ÷ 1/3

2/5 × 3/1

= 6/5

… Adding and subtracting fractions. To do this, scale one or both of the fractions so that both denominators are the same. Then you can add the numerators together. For example:

2/5 + 1/3

6/15 + 5/15 (Now both denominators are the same, the ‘slices of the cake’ are the same size)

= 11/15

… Always check to see if you can cancel down a fraction into its lowest form.

… Forming equations from words. For example:

1) “The area of the triangle is double the area of the square”

The “is” can be thought of as an equals sign, and “double” means multiply by 2:

Area of triangle = 2 x area of square.

2) “Alex has some marbles

Barbara has twice as many marbles as Alex

Clive has 8 more than Barbara.

Together they all have 38 marbles.

How many does Alex have?”

… Comparing coefficients

If:

4x + c = 4ax + 8 + a

Then looking at the x coefficients of both sides:

4 = 4a, so

a = 1

Now comparing the coefficients with no x’s on both sides (I.e. just the numbers):

c = 8 + a

and as a = 1, then

c = 9

Algebraic Fractions

Simplifying algebraic fractions

… First multiply the denominator and numerator of one or both fractions so that you make their denominators the same. Then you can add the numerators together (and put it over the shared denominator).

… Take care when subtracting algebraic fractions, for example:

1/x – (2+x)/5 Now making the denominators the same:

5/5x – x(2+x)/5x Now putting the numerators over the shared denominator:

[ 5 – x(2+x) ] / 5x And simplify:

(5 – 2x – x²) / 5x Ta da!

It’s usually a good idea to keep the denominators nicely factorised once you have added up (or subtracted) two algebraic fractions.

… Take care when cancelling terms of the numerator with the denominator in an algebraic fraction. Think about what you are dividing the top and bottom of the fraction by and remember that you have to do this to every individual term.

… When solving algebraic equations, it is often useful to start by multiplying both sides by ‘awkward’ denominators, to remove them. For example:

4(x – 3)² / (4x + 1) = 2x – 1

Multiply both sides by the denominator (4x + 1) to remove it from the LHS. Take care though, as we need to introduce brackets on the RHS!

4(x – 3)² = (2x – 1)(4x + 1)

… BIDMAS tells us that we should always do brackets and indices (powers) first. For example, to expand:

4(x – 3)²

We cannot simplify anything further within the Bracket (B in BIDMAS), so we must apply the indice (I in BIDMAS), but keep the 4 outside the brackets:

4(x – 3)(x – 3)

4 -3x -3x + 9)

Now we can simplify within the bracket:

4(x² – 6x + 9)

and expand:

4x² – 24x + 36

… Sometimes it might be possible to make the denominators of two fractions the same in one single multiplication, for example:

y / (x-y) + x / (y-x)

Here we can match the denominators by multiplying the top and bottom of just ONE of the fractions by -1:

-y / (-x+y) + x / (y-x) Note that -x + y is the same denominator as y – x

so we get:

(-y + x) / (y-x)

= (x-y) / (y-x)

… Watch out when multiplying the ‘Last’ terms when expanding double brackets, for example, spot the mistake:

(x + 8)(x – 3)

= x^2 – 3x + 8x + 5

… Where possible, try using brackets instead of the × sign, as it makes working out much clearer to see (especially when ‘x’s are involved!). For example,

2.3 × 45 – 2x × 5

Can be written:

2.3(45) – 2x(5)

Changing the subject of a formula

… Usually you will need to add or subtract terms from both sides FIRST, then divide or multiply both sides.

… For example, make p the subject of:

q = (3 – p) / (2 + p)

… First multiply out any denominators on both sides.

q(2 + p) = 3 – p

… Expand both sides so that you have separate terms.

2q + pq = 3 – p

… Then move all the p terms to one side and everything else to the other side:

p + 2q + pq = 3

p + pq = 3 – 2q

… Now the trick… FACTORISE for p:

p(1 + q) = 3 – 2q

… and divide both sides:

p = (3 – q) / (1 + q)

Iteration and iteration formulae

Method 1: trial and improvement

… Use a table with column headings: the value you are trying ‘x’, the expression (the result you get using this value) and then a comment (too high or too low).

… To choose each value of x, aim at the middle of the range (or close to it). So if the equation has a solution of x between 4 and 5, then start at 4.5 .

… If 4.5 (in this example) gives a result that is too high, then try 4.3 (halfway-ish between 4.5 and 4)

… Once you’re close to the answer, try the “decider 5”. For example, if x= 4.4 gives a result which is close but too low, try x=4.45. This will help you decide if the value of x to 1d.p. is 4.4 or 4.5.

So, if x=4.45 gives a value that is too high, then your final answer (to 1 d.p.) would round down to x = 4.4

Iterative formulae – Method 2

Some equations can be rewritten in different ways so that x’s appear on both sides. For example:

“The solution (the value of x) to the equation x³ – 5x² – 1 = 0 lies somewhere in the interval 5 < x < 6. Use an iterative method with x[0] = 6 to find the solution to 3 decimal places.”

x[0] means the starting value of x that we will try (0 is written in subscript).

This equation can be rewritten

x³ = 5x² + 1

x = cubed root of (5x² + 1)

The iterative formula can now be written

x[n+1] = cubed root of (5x[n]² + 1)

This looks horrible!… but x[n+1] just means ‘the next (and hopefully better) estimate for x’ and x[n] means ‘the current value of x’

Start by entering the first value of x, x[0] = 6 into the formula:

x[1] = cubed root (5(6)² + 1)

and press the Equals button.

= 5.657

Now use the Replay key on your calculator to replace ‘6’ by ‘ANS’. This will feed the answer back into the formula each time you press the ENTER key. After several iterations the answer ‘converges’ to 5.657 (to 3 d.p.)

… If you are asked to show that a solution to an equation lies between two x values, for example x = 1 and x = 2, then sub in these x values and see what comes out of the equation. One answer should be a positive and the other should be a negative if the solution lies between the two x values.

Questionnaires

… Good Questionnaires require time-specific questions, e.g. How often do you go to the cinema PER MONTH? The examples you often get in a question are usually too vague and subjective (i.e. the answers are not measurable)

… Remember to use tick boxes in your questionnaire with values that do not overlap, for example:

0 times, 1-2 times, 3-4 times, 5+ times

Algebraic fractions

… Adding or subtracting algebraic fractions by multiplying the numerator and denominator of each fraction by the same term to obtain a common denominator.

… When simplifying algebraic fractions, it is usually best to keep denominators factorised, as as one of the bracketed terms may cancel with a term in the numerator (when it’s factorised).

… Take care with subtracting algebraic fractions. The minus sign will affect all terms in the numerator of the second fraction (best to use brackets!)

… Note that you can only multiply and divide the numerator and denominator by the same thing so that the fraction remains the same (an equivalent fraction). It is not possible to add or subtract, or square or square root. For example:

2x/3 is the same as 2x(x+1)/3(x+1) because the fractions are equivalent. However:

2x/3 is NOT equal to (2x+7)/(3+7) (Adding)

Also, 2x/3 is NOT equal to (4x^2)/9 (Squaring)

… You can always check that you have correctly solved an algebraic equation by substituting the values back into the original equation and seeing if it works.

Functions

… Functions such as “f(x)” can be thought of as ‘sausage machines’ with an input ‘x’ (the ingredients) and an output y (the sausage!).

… The set of allowed input x values is called the ‘domain’, for example -2 < x < 4

… The set of possible output y values of the function is called the ‘range’, for example f(x) < 8

… f(x) = 2x – 1 means “put in a value of x, double it and then subtract 1”. So f(3) would mean our input value of x is 3. The output, y, would be 2(3) – 1 = 5

… fg(x) is called a composite function. This is where we feed g(x) into f(x). For example,

if f(x) = 2x – 1 and g(x) = 3 – x, then:

fg(x) = 2(3 – x) – 1

= 6 – 2x – 1

= 5 – 2x

So fg(2) = 5 – 2(2) = 1

… For the above example, fg(2) could also be found by first finding g(2) = 3 – 2 = 1 and then feed this result into function f:

f(1) = 2(1) – 1 = 1

… Finding the inverse function, for example f-1(x), (where the ‘-1’ is in superscript) by making x the subject of the function and then swapping x and y around.

So if f(x) = 2x – 1

y = 2x – 1

y + 1 = 2x

x = (y+1)/2

y = (x+1)/2

So the inverse function would be:

f-1(x) = (x+1)/2

Coordinate Geometry

Straight line graphs

… Straight line graphs can be rearranged into the form y = mx + c, where m is the …?… and c is the …?…

… If possible, you may be able to find the y-axis intercept, c from the graph or given coordinates.

… To calculate the gradient (m) of a line joining two points, draw a large …?… under or over the line and measure the change in …?… and the change in …?… from one point to the other.

The formula for calculating the gradient is…?

Always check: if your line slopes downwards to the right, then it will have a …?.. gradient.

… If two lines are parallel, then they have the same …?…

… Once you have found the gradient (let’s say that m = 3), substitute the x and y coordinates of a point on the line into y = 3x + c. You can then solve this equation to find the value of c, the y-axis intercept.

… Alternatively, once you have found the value of the y-axis intercept, c, substitute the x and y coordinates of a point on the line into y = mx + c. You can then solve this equation to find the value of m, the gradient.

… You may need to rearrange equations to get them into the form “y = mx + c”.

… What is the gradient of a line which is perpendicular to, say, y = 2x – 3 ? (hint: remember “flip ‘N’ minus!”)

… If two lines are perpendicular, then the product of their gradients = -1

m(line a) x m(line b) = -1

… How would you check if the point (2,-3) lies on the line y = 2x – 7 ?

Midpoints

… To find the midpoint between two points, find the average of the x coordinates and the average of the y coordinates.

… Calculate the midpoint coordinates between points A (6 , 2) and B (-4 , 8)

… The midpoint M (3 , 4) lies between points P and Q.

If Q has coordinates (2 , -6), what are the coordinates of P?

General Equation of a Circle

… What is are the coordinates of the centre of the circle x² + y² = 4 ?

What is the radius of this circle?

… If you need to find the equation of a tangent to a circle at a point, P, then the stages of working are:

1) find the …?… of the line joining the centre of the circle to the point.

2) use the rule…?… …to find the g…?… of the tangent line.

3) use y = mx + c to find the value of c by…?

Value for money

… Divide cost by volume, mass etc. and then use the units to guide you.

For example if 300g costs £1.50, then first convert to pence (to make it easier):

150p / 300g = 0.5p/g “pence per gram”

So we can see that the best value for money in this case will have the lowest value possible for “pence per gram”

Direct proportionality

Step 1. Form the equation by translating the words into maths:

“… is directly proportional to”

translates to

“= k x…”.

Note that y ∝ x is shorthand for saying “y is directly proportional to x”.

Step 2: Find the constant of proportionality, k (the ‘key’) by putting in the data

Step 3: Write down the formula (sometimes called an ‘expression for x in terms of y’)

Step 4: Use the formula to work stuff out.

… Note that ‘Varies as’ means the same thing as ‘is directly proportional to’.

… The graph of a directly proportional relationship is a straight line going through the origin (with either a positive or negative gradient).

Inverse proportionality

… This is the same method as for direct proportion except that

“…is inversely proportional to…”

translates to

“= k /…”

Note that y ∝ 1/x is shorthand for saying “y is inversely proportional to x”.

… The graph of an inverse proportional relationship is a reciprocal y = 1/x type of curve (see https://goo.gl/images/auDtHv)

… Sometimes a question might not mention “directly proportional” or “inversely proportional” so you will need to recognise that it is one or the other.

For example:

“Ishmael is making some T-shirts. It takes 5 m² of cotton to make 8 T-shirts. How much cotton will Ishmael need to make 85 T-shirts?”

Ask yourself, “if I double X does it make sense if Y also doubles?”

If it does make sense, then this relationship must be …?… proportional.

… Direct proportion involving three or more types of quantities.

It can be useful to use ‘ratio tables’ for these often very ‘wordy’ questions, for example:

“Ed runs a go-cart track. It takes 12 litres of petrol to race 8 go-carts for 20 minutes.

6 go-carts used 18 litres of petrol. How many minutes did they race for?”

Litres Go-carts Minutes

12 8 20

Try to scale down to 1 unit, in this case, go-carts. Notice that the minutes column stays the same – that’s because we can only scale two columns at a time!

12/8=1.5 1 20

1.5×6=9 6 20

Now scale up the litres and minutes columns (leave the go-carts column the same)

18 6 40

Take care when two columns are inversely proportional to each other. For example, Go-cards and Minutes are inversely proportional, so if we were to keep Litres the same but halve the number of Go-carts, then we would have to DOUBLE the number of Minutes!

Litres Go-carts Minutes

12 8 20

12 4 40

Changing the subject of a formula

… Work a step at a time, showing what you are doing to both sides (e.g. multiplying, dividing, subtracting, adding, squaring or square rooting)

… If you are making x the subject, then you will often need to bring any x terms to one side of the equation. Put everything else on the other side and then FACTORISE using x as the common factor.

E.g.

2yx – 3 = 4x + 2y

2yx = 4x + 2y + 3

2yx – 4x = 2y + 3

x(2y – 4) = 2y + 3

x = (2y + 3)/(2y – 4)

Surds

… A Surd is a square root of a non-square numbers, for example √5. This means they are irrational numbers: non-terminating (never ending) and non-repeating (no pattern).

… Simplifying surds by ‘spotting the square factor’. For example:

√27 = √(3×9) = 3√3

… Multiplying and dividing surds: Surds go with surds and numbers go with numbers. For example:

(12√3) / 2√6

= 6 / √2

… Multiplying out surd brackets – well done here, just watch out for those pesky minus signs 🙂

… You can add and subtract surds if they are the same ‘animal’. For example:

2√15 + 4√15

= 6√15

But

2√15 + 4√7

cannot be added because they are different surd ‘animals’!

In the same way, 3 + 4√7 cannot be added either.

… When a surd is multiplied by itself: for example √3 x √3 = √9 = 3

… Watch out for ‘hidden’ double bracket multiplications, for example: (√2 – 4)²

… Rationalising the denominator by multiplying top and bottom of the fraction by the denominator surd.

… Rationalising the denominator when the denominator is, for example, (2 – √3). Here you would need to multiply the top and bottom of the fraction by…?

… When you get to a final surd answer, always ask yourself, “can I simplify this any further?”

… Sometimes you may need to equate coefficients in a surd equation to work out the value of unknowns. This method uses the idea that numbers and surds are totally different ‘animals’. For example:

3√2 + 2a = b√2 – 6

Equating the coefficients of the √2 surds, we get:

3 = b

Equating the numbers, we get:

2a = -6

a = -3

Indices (Powers)

… Negative powers mean ‘reciprocal’ (1/…). It can be useful to deal with these first, then what’s left goes underneath as a denominator.

A negative power applied to a fraction means flip the fraction:

(2/3)^(-2) = (3/2)^2

… Fractional indices are roots. E.g. The power of 1/2 means square root.

… A fractional power of 2/3 means first do the cubed root, then square the result (easiest to reduce the size of the number first).

… Solving indice equations by matching the base, for example:

3^x = 9^(x+1)

3^x = (3^2)^(x+1)

3^x = 3^(2x + 2)

And so…

x = 2x + 2

-x = 2

x = -2

… Simplifying powers and roots, such as ∛(8 ×10^6)

This can be rewritten:

∛(8) × ∛(10^6)

∛(8) × (10^6)^(1/3) because the cubed root is the same as to the power of 1/3

2 × 10²

=200

… Combinations of powers, for example:

(8a³)^(⅔)

= (8^⅔)(a²)

= (2²)(a²)

= 4a²

… Estimating square roots by finding the nearest square numbers. For example, estimate √70:

√81 = 9

and

64 = 8

So our estimate lies between 8 and 9 (and closer to 8), so 8.4 would be a reasonable estimate.

Algebraic Proof

… If n can be any whole number, then 2n will always be an …?… number because it will give multiples of …?..

… If n can be any whole number, then an expression for an odd number can be 2n + …?…

… Consecutive whole numbers can be represented by n , n+…?.. , n+…?.. etc.

… Consecutive EVEN numbers can be written…?

… Consecutive ODD numbers can be written …?

… Most proof type questions involve three steps:

- Translate into maths (e.g. ‘product’ means multiply)
- Expand and Simplify
- Factorise, using a common factor that matches the question.

For example:

“Prove algebraically that the difference between the squares of any two consecutive even numbers is always a multiple of 4”:

(2n+2)² – (2n)²

(2n+2)(2n+2) – 4n²

4n² + 4n + 4n + 4 – 4n²

8n + 4

4(2n + 1) which is always a multiple of 4.

… But what if you had to prove that an expression is NOT a multiple of (for example) 4? Let’s say you simplified the algebra down to:

8n – 10

Now factorise using a common factor which matches the question. But first we have to make an adjustment so that the factorisation can work:

8n – 8 – 2

4(2n – 2) – 2

Here 4(2n – 2) is always a multiple of 4, but because we then subtract 2, the result can never be a multiple of 4.

…. You may also be asked to prove other types of mathematical statements. For example, “if a < b < c < d then is a/b < c/d always true?”

In this case, try to find a counter example where the statement is NOT true. For example, if a = 1, b = 2, c = 3 and d = 100 then:

1/2 < 3/100 is not true and so you have disproved the statement.

… Proving that a given number is NOT prime by finding factors which are not equal to the number or 1, for example:

2^64 – 1

(2^32)² – 1 … This can now be factorised using Differ ence of Two Squares:

(2^32 + 1)(2^32 – 1)

So both (2^32 + 1) and (2^32 – 1) are factors of 2^64 – 1, and are not equal to 1 or 2^64 – 1.

Therefore 2^64 – 1 is NOT prime.

… Sneaky algebra terms which can be divisible by a number. For example:

2n(n-1) doesn’t look like it can be divided by 4. However…

n(n-1) is the product of an odd number and an even number = even number.

Then 2 x an even number must be divisible by 4.

Ta da! So 2n(n-1) is divisible by 4.

… Prove that the difference between a whole number and its cube is always a multiple of 6.

n³ – n

n(n²-1)

n(n-1)(n+1)

This is the product of three consecutive integers and can also be written as

n(n+1)(n+2)

One of the terms n, n+1 and n+2 must be even, and so divisible by 2.

One of them must be divisible by 3.

So their product must be divisible by 2× 3 = 6

… A 3-digit number “abc” can be written as:

100a + 10b + 1c

Because a is the hundreds column, b is the tens column and c is the units column.

Vectors:

… Vectors tell us how to get from one place to another. They have both s…?… (magnitude) and d…?…

… Finding alternative routes using vector notation.

… Midpoints and finding the vector expressions.

… Rather than write a vector as “-2b + a”, it can make your working easier if you write “a – 2b”

… First rule of vectors: as soon as you work out some vector information, put it on your vector diagram (if there is space) as it will help you to answer the next part. Remember the arrows!.

When a line is split up into sections, it can help to draw a parallel line alongside it in order to show the vector.

… Second rule of vectors: always try to SIMPLIFY and then FACTORISE your answer (take a common factor outside a pair of brackets). This can help to prove that two vectors are in the same direction.

… If you are given a ratio, for example AP : PB is 2 : 3

A———>———-P——————–>——————-B

2 3

First write these ratio numbers on the corresponding vector sections.

Let’s say you worked out vector AB = (a+b).

Then vector AP would be 2/5 of AB

AP = (2/5) (a+ b)

If vector AP = c, then can you find vector PB be in terms of c?

… Ratios may also be given in the form (for example):

3TW = 2WV

To better ‘see’ this ratio (and annotate our diagram), first rearrange this equation:

TW/WV = 2/3

We can now see that if TW is 2 parts long, then WV is 3 parts in length.

T———>———-W——————–>——————-V

2 3

… Proving that two vectors are parallel: find the vector of each part of the straight line and compare them (remember to factorise – sometimes by taking a fraction out of the brackets).

You should find that one vector is a multiple of the other, i.e. they have a ‘common vector factor’, which means that they have the same direction.

… If two vectors are in the same direction (parallel) and if they join up (i.e. go through the same point), then they must lie on a straight line.

Drawing pie charts for a frequency table

… constructing each sector of the chart by finding a fraction of the whole and x360 for each category.

… Drawing a pie chart by finding out how many degrees each item is worth. For example, if there are 72 different types of fish in a lake, then each fish can be represented by 360/72 = 5 degrees on a pie chart.

Ratio and scales

… Using scales, for e.g. A map scale is 1:10000. The numbers in a scale will always in the same units. So in this case 1cm on the map represents 10,000cm in real life.

… To find a scale, use column headings to organise the information in the question. For example:

Model : Real life

6m : 120km

Then make sure that the units are the same, so here we would have to convert 120km into metres:

Model : Real life

6m : 120,000m

Now simplify this ratio into its lowest form by dividing by 6

1m : 2,000m

The scale is then 1:2000.

… Part-to-total type ratios, e.g. “Share £180 in the ratio 1:3:5”

First add up the number of parts = 9

Then find out how much each part is worth = 180/9 = £20

Then draw the columns (with headings) and multiply by what each part is worth:

1 : 3 : 5 = 9

20 : 60 : 100 = 180

… Part-to-part type ratios, e.g. “150ml cream and 300g of chicken are used in a receipt for making chicken soup for 4 people. How much cream and chicken will you need to make the recipe for 6 people?”

Start by writing the ratio down using two columns

People 4 : 6

Chicken 300g : ?

Cream 150ml : ?

To go from 4 to 6 you need to ÷ 4 and then × 6. This is becomes your conversion for the other quantities.

So chicken = 300 ÷ 4 × 6 = 450g

Cream = 150 ÷ 4 × 6 = 225g

Graph Transformations

Enlargements

… To enlarge a shape using a scale factor and centre of enlargement:

locate the centre of enlargement (CoE)

Draw lines from the CoE through each corner of the shape and extend beyond

Count squares from the CoE across and up/down to each corner, e.g. “2 across and 3 down”

Multiply your count by the scale factor. So if our scale factor is 2, the new count from the CoE would be 4 across and 6 down to the new position of the corner.

Repeat for the other corners

Check that your enlarged shape has the same proportions as the original, just larger or smaller!

You may sometimes need to locate the CoE by joining and extending lines from the matching corners of each shape. Where the lines cross is the CoE.

A negative scale factor means count in the opposite direction from the CoE (but also check that the resulting point lies on the line joining the CoE to the corner of the shape).

Rotations

… Use tracing paper and mark on it the centre of rotation and the corners of the shape.

… Rotate the tracing paper through the required angle and direction, usually 90° or 180° clockwise or anti-clockwise.

Reflections

… For harder reflections, when the mirror line is at an angle, do use tracing paper. Draw on the mirror line and the corners of the shape – remember to mark an ‘X’ at a known point on the mirror line. Flip the paper around to find the reflected shape.

Translations

… All you need to describe a translation (movement of a shape) is a translation v…?… This is usually written a vertical column with the x movement at the top and the y movement at the bottom.

Congruent triangles

… Congruent means same …?… and same …?…

… First draw both triangles in the same orientation, then you can compare them easily

… Triangles are congruent if they have the same (in order):

- Side – Angle – Side (SAS or SSA)
- Sides (SSS)
- Angle – Angle – Side (AAS or ASA)

… Note that if you have a ‘gap’ where there is no angle and no length, then you cannot move in that direction… you’ll have to go the other way.

Note: AAA doesn’t mean that two triangles will be congruent. Why not?

Circle theorems

… Circle theorems come in two flavours: ‘centre’ and ‘non-centre’ types.

… Sometimes you will have to ‘angle chase’ – find as many angles as you can until you get to the angle you want!

… Ask yourself, “what do I see?” Look out for keywords such as diameter, triangle, tangent, radius, chord, segment, or sector. Then ask yourself, “what do I know about [keyword]?”

… You’ll need to learn the ‘wordy’ descriptions so that you can give a ‘reason’ for finding a particular angle.

… Sometimes you may have to add lines to a circle diagram in order to spot which theorem to use – a kind of creative ‘stepping stone’.

… Also try turning the paper sideways or upside down, or covering up parts of the diagram – this can also help your brain to spot different features!

… If you get stuck, go on to the next question and come back after that. It is surprising how often you will suddenly ‘see’ what you need to do (your brain will work on the problem in the background!)

Circle theorems which have lines going through the centre of the circle:

… The angle in a semicircle (from a diameter) is always …?…

… A …?… or a …?… always meets a tangent at a right angle.

… Sneaky …?… triangles are often made by radii.

… The angle at the …?… is twice the angle at the …?… (Star Trek)

… The angle between a radius and a tangent is always …?… degrees.

Circle theorems for lines which do NOT go through the centre of the circle:

… Angles (subtended from a chord) in the same …?… are equal.

… Opposite angles in a …?… quadrilateral add up to …?… degrees

… The angle between a tangent and a chord is equal to the angle in the …?… …?…

… The two triangles formed when two tangents meet two radii are c…?… (same size, same shape).

… Sometimes you may need to prove a circle theorem algebraically. This can be tricky! However, to do this:

- break up the geometry into parts by drawing extra lines if needed.
- Then introduce some variables (e.g. x, y) to represent the important angles in the geometry.
- Using basic principles of isosceles triangles, angles on a straight line, opposite angles, angles in a triangle and a quadrilateral, see if you can create equations to relate the variables together.
- It is then possible to combine the equations to eliminate the variables you don’t need.

… How to prove circle theorems. This looks complicated at first sight, but is actually just a lot of little steps using familiar geometry ideas such as angles in a triangle add up to 180.

1) split up the shape in the circle into isosceles triangles using lines drawn from the centre of the circle.

2) write in ‘x’ and ‘y’ angles at angle locations which make sense depending on the theorem you are trying to prove.

3) Find out other angles in the triangles in terms of x and y using geometry rules such as ‘angles in a triangle add up to 180’.

Once you have found all of the angles in terms of x and y, you should be able to see a relationship between the angles to prove the theorem.

Regular polygons

… Exterior angle of a regular polygon = 360 / number of sides.

… I recommend always starting with the exterior angle for regular polygon questions. Then, if needed, use interior angle = 180° – exterior angle

Sectors and Arcs

… The formulae for the area and circumference of a circle are…?

… How to calculate the area of a circle sector by using the angle as a fraction of 360 degrees.

Area of sector = fraction of the circle x area of the circle

… This method can also be used to find the arc length of a sector:

Arc length = fraction of the circle x circumference of the circle

Triangles

… The basic formula for calculating the area of a triangle is 0.5 × b…?… × perpendicular …?…

… If you know a triangle’s Side – Angle – Side (SAS in that order), then you can calculate the area of a triangle by using the formula…?

Bearings

… ‘Three Figure Bearings’ (e.g. “065”) are always measured in the …?…wise direction from the …?… line.

… Three figure bearings should only be written as three figure bearings (so round to the nearest degree)

… Make sure there is always a North line on your diagram, and a starting point (where you are ‘standing’ looking at things from)

… “OF” means the object you’re looking at, and “FROM” means the place where you are standing. For example, “what is the bearing of L from H?” This means you are standing at …?… and looking at …?…

… Always remember to draw a straight line from the point where you are standing to the point you are looking at.

… Take care that you use the correct protractor scale when measuring bearings – you did this well.

… When working out distances from a map scale, start by writing down what you know using a ratio format.

For example, if a map has a scale of 4cm represents 1km, then 6cm would represent a distance of…

4cm : 1km < write down what you know first

(÷2) (÷2)

(×3) (×3)

6cm : 1.5km

Formulae, units and converting

… You can often work out a formula by looking at the units. For example, m/s is the unit for speed, so we have “metres ÷ seconds” and the formula is speed = distance ÷ time.

… Converting compound units, for example what is 20m/s in km/h? Go a step at a time and write down the units you have found each time. Check if the number at each step makes sense.

For example,

20m/s

÷ 1000

0.020km/s (this makes sense because 20m is the same as 0.020km)

x 3600

72km/h (this makes sense because we’re now travelling for 1 hour, not 1 second as before, so we must cover more distance.

… Prefixes can help with conversions. “km” stands for kilometre and the prefix “k”, or “kilo” means x1000

… Converting areas. First start by drawing a square which has an area of 1 metre-squared. The sides must be 1m each. Convert the sides into (for example) cm, so each side is 100cm.

Then we can see that 1 m2 = 10000 cm2, which is your conversion factor for converting m2 into cm2.

… Converting volumes. First start by drawing a cube which has a volume of 1 metre-cubed. The sides must be 1m each. Convert the sides into (for example) cm, so each side is 100cm.

Then we can see that 1 m3 = 1000000 cm3, which is your conversion factor for converting m3 into cm3.

Pythagoras’s theorem (finding lengths of right angled triangles)

… When you know two lengths of a right angled triangle, you can use Pythagoras’s theorem to calculate the other length.

… Start by labelling the three sides of the triangle ‘a’, ‘b’ and ‘c’ (the hypotenuse).

… Then write down the formula: a² + b²= c²

… Then put the numbers in and rearrange to solve for the length you need.

… As a check, ask yourself if your answer ‘feels right’ in the context of the question (compared to the other lengths of the triangle)

… You can split isosceles or equilateral triangles down the middle to create two right angled triangles to use Pythagoras.

… Sometimes you will need to ‘split up’ a complicated shape so that you make simpler shapes, for example, to make rectangles and right angled triangles.

Set notation, Set Theory and Venn diagrams

… Venn diagrams help us to visualise how members of groups are related.

… The Universal Set (squiggly ‘E’ symbol) is the square box containing the Venn Disgram. It represents the set of all the elements being considered.

… n(P ∩ Q) means “the number of members in the set P ∩ Q”. Take care with this because numbers in a Venn diagram could represent:

the number of people or items in a particular set, or

Items which are numbers (e.g. the set of prime numbers)

… Where possible, always draw a Venn diagram from the information given in the question. Then it’s a lot easier to visualise the situation.

… If set B (multiples of 8) is a subset of set A (multiples of 4), what would the Venn diagram look like? Note that A ⊂ B means “A is a subset of B”

… Intersection (an ‘∩’ shaped symbol) means ‘overlap’. For example, if A={0,3,5,6} and B={1,3,4,6,9}, then

A ∩ B = {3,6}

In this example, when you have hatched the area representing A and the area representing B, then A ∩ B is where there is an OVERLAP of hatching.

… Union (a ‘U’ shaped symbol) means ‘together with’. For example, if A={0,3,5,6} and B={1,3,4,6,9}, then

A U B = {0,1,3,4,5,6,9}

Notice that you don’t count duplicates

In this example, when you have hatched the area representing A and the area representing B, then the area A U B is EVERYWHERE that is hatched.

… P(A ∩ B) means what is the probability of event A and event B happening? This can also mean what is the probability of selecting an item in the overlap between sets A and B?

… The ‘complement’ (an apostrophe symbol) means ‘Not’. For example,

A’ means everything outside of set A.

So, A’ means Not A or the Complement of A. Imagine there is a force field that doesn’t let your pencil enter the A circle. Hatch everywhere OUTSIDE of A.

… For complex set notation, work out the brackets first (starting with the most complicated term).

… Shading a couple of stages at a time can help visualise the situation, for example:

(A ∩ B’) U C

Here, first imagine everything OUTSIDE of circle B and how this area overlaps with circle A. Shade that region.

Then ADD that region TOGETHER WITH (union) everything in circle C.

… Venn diagrams can also be used with probabilities written as decimals or fractions in the circles. When dealing with probabilities, the universal set = …?…

… Conditional Probability: for example, the probability of event L happening GIVEN THAT event G has already happened:

P(L|G)

Think of ‘given that’ conditional probability situations by imagining that your universe ‘shrinks’… in this case to the bubble called event G (which has already happened).

Linear programming

… Formulating inequalities from ‘wordy’ questions

… Plotting the inequalities on a graph and finding the feasible region

… Finding the optimum values for a certain condition (e.g. Minimum price) by substituting-in the vertex coordinates of the feasible region. If the variables can only be whole numbers, then find the nearest integers within the feasible region.

… Plotting a line for profit by choosing an arbitrary profit amount. The equation is then called the ‘objective function’ and moving the line upward until it reaches the ‘highest’ vertex of the feasible region will identify the maximum profit conditions.

Quadratic sequences

Step 1. Find the first difference between each term

Step 2. Then find the SECOND difference, which should be a constant value. For example, if the sequence is:

3, 9, 19, 33, 51…

Then the first differences are

+6, +10, +14, +18

And the second difference is

+4

Step 3. Halve the second difference to find the coefficient (multiplier) on the n squared term. In the above case, the coefficient would be 4/2 = 2. So we would have the nth them as being 2x^2

Step 4. Test your nth term expression using n = 1. You may need to modify it by adding or subtracting a number. In the above case, n = 1 gives the value 2(1)^2 = 2, but the first term of the sequence is 3… So +1 to the expression:

The nth term = 2n^2 + 1

Density

… The density of a liquid is 1.3g/cm³. Using the units, what is the FORMULA for density in terms of volume and mass?

Proofs

Proofs can usually be approached using the “Expand – Simplify – Factorise” steps. For example:

a) “Prove that (x + 4)^2 – (3x + 4) = (n + 1)(n + 4) + 8”

Here, we would need to expand the LHS and then factorise the result (keeping out an 8 from the factorisation). We should be able to produce the RHS.

b) “Prove that (n + 1)^2 – (n – 1)^2 + 1 is always odd for all positive integer values of n.”

Here we would do the same “Expand – Simplify – Factorise” process. For the last step, try to factorise so that you get an even number plus one, e.g. 4n + 1, which will always be odd (because 4n is always even).

If n is an integer, then

… 2n is always even (it is a multiple of 2)

… Two consecutive odd numbers can be written as:

(2n + 1), and (2n + 3)

… Two consecutive even numbers can be written as:

(2n), and (2n + 2)

… To prove that an expression is a multiple of 4 (for example), try to factorise it so that 4 is the common factor outside the front of the brackets:

8n + 2

4(2n + 1) … Must be a multiple of 4

c) “Prove algebraically that the difference between the squares of any two

consecutive even numbers is always a multiple of 4.”

Nth term – harder quadratic sequences!

These can be tricky:

1) If the SECOND difference is, for example, 4 then you know that the nth term formula will start with:

(4 ÷2)n^2

= 2n^2

2) Then write down the numbers this gives for n=1, 2, 3… above each of the values of the sequence. Do they match? If yes, you are finished. If they don’t match go on to 3)…

3) If the differences between what you GET and what you WANT go up by 3 each time for each term, then the next term in the formula would be “3n”.

4) Add a + or – number modifier to complete your formula.