GCSE Mathematics AQA YEAR 11

NAME:_____________________________

TUTOR GROUP:______________________

Knowledge Organiser

MATHEMATICS GCSE Mathematics AQA

MA

THEM

ATI

CS

GC

SE M

ath

em

atic

s A

QA

YEAR 11

(Autumn) 2021-2023

Topic 11 1.1 Polygons and Circles

Topic/Skill Definition/Tips Example Your turn

1 Types of Angles

Acute angles are less than 90°.

Right angles are exactly 90°.

Obtuse angles are greater than 90° but less than

180°.

Reflex angles are greater than 180° but less than

360°.

What type of angle is this?

Measure the angle.

2 Angle Notation Can use one lower-case letters, e.g. 𝜃 or 𝑥

Can use three upper-case letters, e.g. 𝐵𝐴𝐶.

Write the angle x using three upper case

letters.

3 Angles at a Point Angles around a point add up to 360°.

Work out the size of angle x.

4 Angles on a

Straight Line Angles on a straight line add up to 180°.


5 Opposite Angles Vertically opposite angles are equal.

Work out the size of angle x and y..

Give reasons for your answers

Work out the size of angle FGD.

Give reasons for your answer.

6 Alternate Angles

Alternate angles are equal.

They look like Z angles, but never say this in the

exam.

7 Corresponding

Angles

Corresponding angles are equal.

They look like F angles, but never say this in the

exam.

8 Co-Interior Angles

Co-Interior angles add up to 180°.

They look like C angles, but never say this in the

exam.

9 Angles in a

Triangle Angles in a triangle add up to 180°.

Work out the size of the missing angle.

10 Types of Triangles

Right Angle: Triangles that have a 90° angle.

Isosceles: Triangles that have 2 equal sides and 2

equal base angles.

Equilateral: Triangles that have 3 equal sides and

3 equal angles (60°).

Scalene: Triangles that have different sides and

different angles.

Base angles in an isosceles triangle are equal.


11 Angles in a

Quadrilateral Angles in a quadrilateral add up to 360°.

Find x.

12 Polygon A 2D shape with only straight edges. Rectangle, Hexagon, Decagon, Kite etc.

13 Regular A shape is regular if all the sides and all the angles

are equal.

14 Names of

Polygons

3-sided = Triangle.

4-sided = Quadrilateral.

5-sided = Pentagon.

6-sided = Hexagon.

7-sided = Heptagon/Septagon.

8-sided = Octagon.

9-sided = Nonagon.

10-sided = Decagon.

15 Interior and

Exterior Angles

Interior angle + exterior angle = 180°.

Sum of Exterior Angles = 360°.

16 Sum of Interior

Angles

(𝒏 − 𝟐) × 𝟏𝟖𝟎

Where n is the number of sides.

Sum of Interior Angles in a Heptagon = (7 − 2) × 180 = 900°

96 + 105 + 161+ 123 + 155 + 114= 754

900 − 754 = 146

𝑥 = 146

Work out the sum of the interiors angle in

pentagon.

Work out x.

17 Size of Interior

Angle in a Regular

Polygon

(𝒏 − 𝟐) × 𝟏𝟖𝟎

𝒏

You can also use the formula: 𝟏𝟖𝟎 − 𝑺𝒊𝒛𝒆 𝒐𝒇 𝑬𝒙𝒕𝒆𝒓𝒊𝒐𝒓 𝑨𝒏𝒈𝒍𝒆

Size of Interior Angle in a Regular Pentagon = (5 − 2) × 180

5= 108°

Work out the size of each interior angle in a

regular octagon.

18 Size of Exterior

Angle in a Regular

Polygon

𝟑𝟔𝟎

𝒏

You can also use the formula: 𝟏𝟖𝟎 − 𝑺𝒊𝒛𝒆 𝒐𝒇 𝑰𝒏𝒕𝒆𝒓𝒊𝒐𝒓 𝑨𝒏𝒈𝒍𝒆

Size of Exterior Angle in a Regular Octagon = 360

8= 45°

Work out the size of each exterior angle in a

regular hexagon.

19

H Circle Theorem 1

Angles in a semi-circle have a right angle at the

circumference.

𝑦 = 90° 𝑥 = 180 − 90 − 38 = 52°

Find x.

20

H Circle Theorem 2

Opposite angles in a cyclic quadrilateral add up

to 180°.

𝑥 = 180 − 83 = 97° 𝑦 = 180 − 92 = 88°

Find x and y.

21

H Circle Theorem 3

The angle at the centre is twice the angle at the

circumference.

𝑥 = 104 ÷ 2 = 52° Find x.

22

H Circle Theorem 4

Angles in the same segment are equal. 𝑥 = 42° 𝑦 = 31°

Find x and y .

23 Tangent

A straight line that touches a circle at exactly one

point, never entering the circle’s interior.

A radius is perpendicular to a tangent at the

point of contact.

Topic 11 1.2 Statistics 2


1 Mean from a Table

1) Find the midpoints (if necessary).

2) Multiply Frequency by values or midpoints.

3) Add up these values.

4) Divide this total by the Total Frequency.

If grouped data is used, the answer will be an

estimate.

Estimated mean height: 450 ÷ 24 = 18.75.

Calculate an estimate for the mean.

2 Median from a

Table

Use the formula (𝒏+𝟏)

𝟐 to find the position of the

median.

𝑛 is the total frequency.

The total frequency is 24, the median will be

the (24+1

2) = 12.5𝑡ℎ position.

The interval containing the median is 10 < ℎ ≤

30.

Which interval contains the median in the table

above?

3 Line Graph

A graph that uses points connected by straight

lines to show how data changes in values.

This can be used for time series data, which is a

series of data points spaced over uniform time

intervals in time order.

4

H

Histograms

A visual way to display frequency data using bars.

Bars can be unequal in width.

Histograms show frequency density on the y-axis,

not frequency.

𝑭𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 𝑫𝒆𝒏𝒔𝒊𝒕𝒚 = 𝑭𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚

𝑪𝒍𝒂𝒔𝒔 𝑾𝒊𝒅𝒕𝒉

Draw a histogram

5

H

Interpreting

Histograms

The area of the bar is proportional to the

frequency of that class interval.

𝑭𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 = 𝑭𝒓𝒆𝒒 𝑫𝒆𝒏𝒔𝒊𝒕𝒚 × 𝑪𝒍𝒂𝒔𝒔 𝑾𝒊𝒅𝒕𝒉

A histogram shows information about the

heights of a number of plants. 4 plants were

less than 5cm tall. Find the number of plants

more than 5cm tall.

Above 5cm:

1.2 x 10 + 2.4 x 15 = 12 + 36 = 48

30 pigs weigh between 50 and 65 kg.

Work out an estimate for the number of pigs

which weigh more than 80kg.

Topic 11 1.3 Algebra


1 Rearranging

Formulae

Use inverse operations on both sides of the

formula (balancing method) until you find the

expression for the letter.

Make x the subject of 𝑦 =2𝑥−1

𝑧

Multiply both sides by z 𝑦𝑧 = 2𝑥 − 1

Add 1 to both sides 𝑦𝑧 + 1 = 2𝑥

Divide by 2 on both sides 𝑦𝑧 + 1

2= 𝑥

We now have x as the subject.

Make t the subject of the formula. 𝑢 = 4𝑡 – 21

2 Simultaneous

Equations

A set of two or more equations, each involving

two or more variables (letters).

The solutions to simultaneous equations satisfy

both/all of the equations.

2𝑥 + 𝑦 = 7 3𝑥 − 𝑦 = 8 5𝑥 = 15

𝑥 = 3

Now substitute this value of x into one of the

equations. 2 × 3 + 𝑦 = 7

𝑦 = 1

Solve: 6𝑥 + 𝑦 = 18 4𝑥 + 𝑦 = 14

3 Variable A symbol, usually a letter, which represents a

number which is usually unknown. In the equation 𝑥 + 2 = 5, 𝑥 is the variable.

4 Coefficient

A number used to multiply a variable.

It is the number that comes before/in front of a

letter.

6z.

6 is the coefficient.

z is the variable.

+ + ÷ 5 ÷ 5

5

Solving

Simultaneous

Equations (by

Elimination)

1) Balance the coefficients of one of the variables.

2) Eliminate this variable by adding or subtracting

the equations (Same Sign Subtract, Different Sign

Add).

3) Solve the linear equation you get using the other

variable.

4) Substitute the value you found back into one of

the previous equations.

5) Solve the equation you get.

6) Check that the two values you get satisfy both of

the original equations.

5𝑥 + 2𝑦 = 9 10𝑥 + 3𝑦 = 16

Multiply the first equation by 2.

10𝑥 + 4𝑦 = 18 10𝑥 + 3𝑦 = 16

Same Sign Subtract (+10x on both) 𝑦 = 2

Substitute 𝑦 = 2 into equation.

5𝑥 + 2 × 2 = 9

5𝑥 + 4 = 9 5𝑥 = 5 𝑥 = 1

Solution: 𝑥 = 1, 𝑦 = 2

Solve: 4𝑥 + 9𝑦 = 10 2𝑥 + 3𝑦 = 2

6

Solving

Simultaneous

Equations (by

Substitution)

1) Rearrange one of the equations into the form

𝑦 = ⋯ or 𝑥 = ⋯.

2) Substitute the right-hand side of the rearranged

equation into the other equation.

3) Expand and solve this equation.

4) Substitute the value into the 𝑦 = ⋯ or 𝑥 = ⋯

equation.

5) Check that the two values you get satisfy both

of the original equations.

𝑦 − 2𝑥 = 3 3𝑥 + 4𝑦 = 1

Rearrange: 𝑦 − 2𝑥 = 3 → 𝑦 = 2𝑥 + 3

Substitute: 3𝑥 + 4(2𝑥 + 3) = 1

Solve: 3𝑥 + 8𝑥 + 12 = 1 11𝑥 = −11

𝑥 = −1

Substitute: 𝑦 = 2 × −1 + 3 𝑦 = 1

Solution: 𝑥 = −1, 𝑦 = 1

Solve: 𝑥 = 10 − 𝑦

2𝑥 + 𝑦 = 17

7

Solving

Simultaneous

Equations

(Graphically)

Draw the graphs of the two equations.

The solutions will be where the lines meet.

The solution can be written as a coordinate.

Solve 2𝑥 − 1 = 5 − 𝑥

𝑦 = 5 − 𝑥 and 𝑦 = 2𝑥 − 1.

They meet at the point with coordinates (2,3)

so the answer is:

𝑥 = 2 and 𝑦 = 3

Use the graph to solve 𝑥2 − 𝑥 + 1 = 7

We draw the line y = 7 on our graph and look

at where it crosses the curve.

The lines meet at the points with coordinates

(-2,7) and (3,7), so the answers are:

𝑥 = −2, 𝑦 = 7 and 𝑥 = 3, 𝑦 = 7

a) Use the graph to solve −𝑥 + 4 = 𝑥 − 2

b) Use the graph to solve 2𝑥2 − 5𝑥 + 1 = 1

8

H

Solving Linear and

Quadratic

Simultaneous

Equations

Method 1: If both equations are in the same form

(e.g. Both 𝑦 =…):

1) Set the equations equal to each other.

2) Rearrange to make the equation equal to zero.

3) Solve the quadratic equation.

4) Substitute the values back in to one of the

equations.

Method 2: If the equations are not in the same

form:

1) Rearrange the linear equation into the form 𝑦 =

⋯ or 𝑥 = ⋯

2) Substitute into the quadratic equation.

3) Rearrange to make the equation equal to zero.

4) Solve the quadratic equation.

5) Substitute the values back in to one of the

equations.

You should get two pairs of solutions (two values

for 𝑥, two values for 𝑦).

Graphically, you should have two points of

intersection.

Example 1

Solve :

𝑦 = 𝑥2 − 2𝑥 − 5 and 𝑦 = 𝑥 − 1

𝑥2 − 2𝑥 − 5 = 𝑥 − 1 𝑥2 − 3𝑥 − 4 = 0

(𝑥 − 4)(𝑥 + 1) = 0

𝑥 = 4 and 𝑥 = −1

𝑦 = 4 − 1 = 3 and 𝑦 = −1 − 1 = −2

Answers: (4,3) and (-1,-2)

Example 2

Solve:

𝑥2 + 𝑦2 = 5 and 𝑥 + 𝑦 = 3

𝑥 = 3 − 𝑦

(3 − 𝑦)2 + 𝑦2 = 5 9 − 6𝑦 + 𝑦2 + 𝑦2 = 5

2𝑦2 − 6𝑦 + 4 = 0 𝑦2 − 3𝑦 + 2 = 0

(𝑦 − 1)(𝑦 − 2) = 0

𝑦 = 1 and 𝑦 = 2

𝑥 = 3 − 1 = 2 and 𝑥 = 3 − 2 = 1

Answers: (2,1) and (1,2)

Solve: 𝑦 = 𝑥 + 3

𝑦 = 𝑥2 + 5𝑥 − 2

9

H

Completing the

Square (when 𝑎 =1)

A quadratic in the form 𝑥2 + 𝑏𝑥 + 𝑐 can be written

in the form (𝒙 + 𝒑)𝟐 + 𝒒.

1) Write a set of brackets with 𝑥 in and half the

value of 𝑏.

2) Square the bracket.

3) Subtract (𝑏

2)

2and add 𝑐.

4) Simplify the expression.

You can use the completing the square form to

help find the maximum or minimum of quadratic

graph.

Complete the square of :

𝑦 = 𝑥2 − 6𝑥 + 2

Answer:

(𝑥 − 3)2 − 32 + 2 = (𝑥 − 3)2 − 7

The minimum value of this expression occurs

when (𝑥 − 3)2 = 0, which occurs when 𝑥 = 3

When 𝑥 = 3, 𝑦 = 0 − 7 = −7.

Minimum point = (3, −7)

Complete the square and find the minimum

value of:

𝑥2 + 8𝑥 + 1

10

H

Completing the

Square (when 𝑎 ≠1)

A quadratic in the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 can be written

in the form p(𝒙 + 𝒒)𝟐 + 𝒓.

Use the same method as above but factorise out 𝑎

at the start.

When a question asks for the answer in the form

(𝑥 + 𝑎)2 + 𝑏 you need to complete the square.

Complete the square of:

4𝑥2 + 8𝑥 − 3

Answer:

4[𝑥2 + 2𝑥] − 3 = 4[(𝑥 + 1)2 − 12] − 3

= 4(𝑥 + 1)2 − 4 − 3 = 4(𝑥 + 1)2 − 7

Complete the square of:

2𝑥2 + 8𝑥 + 2

11

H

Solving Quadratics

by Completing the

Square

Complete the square in the usual way and use

inverse operations to solve.

Solve:

𝑥2 + 8𝑥 + 1 = 0

Answer: (𝑥 + 4)2 − 42 + 1 = 0

(𝑥 + 4)2 − 15 = 0 (𝑥 + 4)2 = 15

(𝑥 + 4) = ±√15

𝑥 = −4 ± √15

Solve by completing the square:

𝑥2 − 8𝑥 + 3 = 0

2𝑥2 − 12𝑥 + 7 = 0

12

H

Turning Point of a

Quadratic

A turning point is the point where a quadratic

turns.

When the equation of the curve is in the form:

𝒚 = 𝒑(𝒙 + 𝒒)𝟐 + 𝒓

or 𝒚 = −𝒑(𝒙 + 𝒒)𝟐 + 𝒓

the turning point is (−𝒒, 𝒓).

Find the turning points of:

𝑥2 + 6𝑥 + 1 = 0

Complete the square. (𝑥 − 3)2 − 8 = 0

So the turning point is (3,-8)

Find the turning points of:

𝑥2 − 4𝑥 + 3 = 0

Maximum Minimum

13

H

Quadratic

Inequalities

Sketch the quadratic graph of the inequality.

If the expression is > 𝒐𝒓 ≥ then the answer will be

above the x-axis.

If the expression is < 𝒐𝒓 ≤ then the answer will be

below the x-axis.

Look carefully at the inequality symbol in the

question.

Look carefully if the quadratic is a positive or

negative parabola.

Solve the inequality 𝑥2 − 𝑥 − 12 < 0

Sketch the quadratic:

The required region is below the x-axis, so the

final answer is: −3 < 𝑥 < 4

If the question had been > 0, the answer

would have been: 𝑥 < −3 𝑜𝑟 𝑥 > 4

Solve the inequality:

𝑥2 − 3𝑥 − 10 ≤ 0

Write the solution using set notation.

14 Set Notation

A set is a collection of things, usually numbers,

denoted with brackets { }.

{𝑥 | 𝑥 ≥ 7} means ‘the set of all x’s, such that x is

greater than or equal to 7’.

Some people use ‘:’ instead of ‘|’

The ‘x’ can be replaced by any letter.

{3, 6, 9} is a set.

{𝑥 ∶ −3 ≤ 𝑥 < 4}

15

Function notation

𝑓(𝑥)

𝒙 is the input value

𝒇(𝒙) is the output value.

𝑓(𝑥) = 3𝑥 + 11

Suppose the input value is 𝑥 = 5.

The output value is 𝑓(5) = 3 × 5 + 11 = 26

𝑓(𝑥) = 2𝑥 − 5

Find:

a) 𝑓(3) =

b) 𝑓(−1) =

16

H Inverse function

𝑓−1(𝑥)

A function that performs the opposite process of

the original function.

1) Write the function as 𝑦 = 𝑓(𝑥)

2) Rearrange to make 𝑥 the subject.

3) Replace the 𝒚 with 𝒙 and the 𝒙 with 𝒇−𝟏(𝒙)

𝑓(𝑥) = (1 − 2𝑥)5. Find the inverse function.

𝑦 = (1 − 2𝑥)5

√𝑦5 = 1 − 2𝑥

1 − √𝑦5 = 2𝑥

1 − √𝑦5

2= 𝑥

𝑓−1(𝑥) =1 − √𝑥

5

2

𝑓(𝑥) = (𝑥 + 3)3

Find 𝑓−1(𝑥)

17

H

Composite

function

A combination of two or more functions to create

a new function.

𝒇𝒈(𝒙) is the composite function that substitutes

the function 𝒈(𝒙) into the function 𝒇(𝒙).

𝒇𝒈(𝒙) means ‘do g first, then f’.

𝒈𝒇(𝒙) means ‘do f first, then g’.

𝑓(𝑥) = 5𝑥 − 3 𝑔(𝑥) =1

2𝑥 + 1

What is 𝑓𝑔(4)?

𝑔(4) =1

2× 4 + 1 = 3

𝑓(3) = 5 × 3 − 3 = 12 = 𝑓𝑔(4)

What is 𝑓𝑔(𝑥)?

𝑓𝑔(𝑥) = 5 (1

2𝑥 + 1) − 3 =

5

2𝑥 + 2

𝑓(𝑥) = 2𝑥 + 3, 𝑔(𝑥) = 𝑥2

Find 𝑓𝑔(𝑥)

Find 𝑔𝑓(𝑥)

18

H

Equation of a

Circle

The equation of a circle, centre (0,0), radius r, is:

𝑥2 + 𝑦2 = 𝑟2

𝑥2 + 𝑦2 = 25

Centre (0,0) radius 5. What is the centre and radius of a circle with

equation:

𝑥2 + 𝑦2 = 49

Write the equation of a circle with the centre

(0,0) and radius 4:

Topic 11 1.4 Congruence and Similarity


1 Congruent Shapes

Shapes are congruent if they are identical - same

shape and same size.

Shapes can be rotated or reflected but still be

congruent.

These are all congruent to each other. Are these shapes congruent?

a)

b)

2 Congruent

Triangles

4 ways of proving that two triangles are congruent:

1) SSS (Side, Side, Side)

2) RHS (Right angle, Hypotenuse, Side)

3) SAS (Side, Angle, Side)

4) ASA (Angle, Side, Angle) or AAS

ASS does not prove congruency.

𝐵𝐶 = 𝐷𝐹 ( = 8𝑐𝑚)

𝐴𝐵𝐶 = 𝐸𝐷𝐹 ( = 73°) 𝐴𝐶𝐵 = 𝐸𝐹𝐷 ( = 61°)

So, the triangles are congruent by ASA.

Show these two triangles are congruent:

3 Similar Shapes

Shapes are similar if they are the same shape but

different sizes.

The proportion of the matching sides must be the

same, meaning the ratios of corresponding sides

are all equal.

These shapes are all similar. The two on the

right are congruent to each other, but not to

the one on the left.

4 Scale Factor

The ratio of corresponding sides of two similar

shapes.

To find a scale factor, divide a length on one

shape by the corresponding length on a similar

shape.

Scale Factor = 15 ÷ 10 = 1.5

5 Finding missing

lengths in similar

shapes

1) Find the scale factor.

2) Multiply or divide the corresponding side to

find a missing length.

If you are finding a missing length on the larger

shape you will need to multiply by the scale factor.

If you are finding a missing length on the smaller

shape you will need to divide by the scale factor.

Scale Factor = 3 ÷ 2 = 1.5 𝑥 = 4.5 × 1.5 = 6.75𝑐𝑚

Find the missing length.

6 Similar Triangles

To show that two triangles are similar, show that:

1) The three sides are in the same proportion.

2) Two sides are in the same proportion, and their

included angle is the same.

3) The three angles are equal.

a) Find the angle JKL.

b) What is the length of GI?

Topic 11 1.5 Trigonometry


1 See 10 1.4 for revision of Pythagoras and Trigonometry.

Find the value of x:

a)

b)

c)

2 Exact Values for

Angles in

Trigonometry

0° 30° 45° 60° 90°

sin 0

𝟏

𝟐

√𝟐

𝟐

√𝟑

𝟐 1

cos 1

√𝟑

𝟐

√𝟐

𝟐

𝟏

𝟐 0

tan 0

𝟏

√𝟑 1 √𝟑 ---

Topic: 11 1.5 Trigonometry

3

H Sine Rule

Use with non right angle triangles.

Use when the question involves 2 sides and 2

angles.

For missing side: 𝒂

𝐬𝐢𝐧 𝑨=

𝒃

𝐬𝐢𝐧 𝑩

For missing angle: 𝐬𝐢𝐧 𝑨

𝒂=

𝐬𝐢𝐧 𝑩

𝒃

There is an ambiguous case (where there are two

potential answers).

To find the two angles, use sine to find one, and

then subtract your answer from 180 to find the

other answer.

𝑥

sin 85=

5.2

sin 46

𝑥 =5.2 × sin 85

sin 46= 7.20𝑐𝑚

sin 𝜃

1.9=

sin 85

2.4

sin 𝜃 =1.9 × sin 85

2.4= 0.789

𝜃 = 𝑠𝑖𝑛−1(0.789) = 52.1°

a) Work out the value of 𝑥.

Give your answer to 1 decimal place.

b) Work out the size of angle 𝑥.

Give your answer to 3 significant figures.

4

H Cosine Rule

Use with non right angle triangles.

Use when the question involves 3 sides and 1

angle.

For missing side:

𝒂𝟐 = 𝒃𝟐 + 𝒄𝟐 − 𝟐𝒃𝒄𝒄𝒐𝒔𝑨

For missing angle:

𝐜𝐨𝐬 𝑨 =𝒃𝟐 + 𝒄𝟐 − 𝒂𝟐

𝟐𝒃𝒄

𝑥2 = 9.62 + 7.82 − (2 × 9.6 × 7.8 × cos 85) 𝑥 = 11.8

cos 𝜃 =7.22 + 8.12 − 6.62

2 × 7.2 × 8.1

𝜃 = 50.7°

a) Work out the value of 𝑥.

Give your answer to 1 decimal place.

b) Work out the value of 𝑥.

Give your answer to 3 significant figures.

Answers

11 1.1 Polygons and Circles

1) Acute, 65° 2) ABD or DBA 3) 125° 4) 59°

5, 6, 7, 8) a) 𝑥 = 53°, opposite angles are equal 𝑦 = 53° alternate angles are equal.

b) 𝑥 = 70°

9) 84° 10) 50° 11) 150° 16) 540°, 𝑥 = 67° 17) 135° 18) 60° 19) 18°

20) 𝑥 = 130°, 𝑦 = 95° 21) 82.5° 22) 𝑥 = 28°, 𝑦 = 60°

11 1.2 Statistics 2

1) Mean estimate = 167.9 (1dp)

2) 160 < ℎ ≤ 170

4)

5) 35

11 1.3 Algebra 2

1) 𝑡 =𝑢+21

4 2) 𝑥 = 2, 𝑦 = 6 5) 𝑥 = −2, 𝑦 = 2 6) 𝑥 = 7, 𝑦 = 3 7) a) 𝑥 = 3, 𝑦 = 1 b) 𝑥 = 0, 𝑦 = 1 and 𝑥 = 2.5, 𝑦 = 1

8) 𝑥 = 1, 𝑦 = 4 𝑎𝑛𝑑 𝑥 = −5, 𝑦 = −2 9) (𝑥 + 4)2 − 15 , Minimum at (−4, −15) 10) 2(𝑥 + 2)2 − 6 11) a) 𝑥 = 4 ± √13, b) 𝑥 = 3 ± √11

2

12) (2, −1) 13) −2 ≤ 𝑥 ≤ 5 14) {𝑥: −2 ≤ 𝑥 ≤ 5} 15) a) 1 b) -7 16) 𝑓−1(𝑥) = √𝑥3

− 3

17) a) 𝑓𝑔(𝑥) = 2𝑥2 + 3, 𝑔𝑓(𝑥) = (2𝑥 + 3)2 18) (0,0) radius 7, 𝑥2 + 𝑦2 = 2

11 1.4 Congruence and Similarity

1) a) Yes b) No 2) Yes, by SAS 5) 45cm 6) a) 65° b) 11cm

11 1.5 Trigonometry

1) a) 7cm b) 34.87cm c) 53.7° 3) a) 8.13cm b) 35.3° 4) a) 17.7 cm b) 80.3°

Knowledge Organiser

MATHEMATICS GCSE Mathematics AQA

Documents

GCSE Mathematics AQA YEAR 11