Download Revision Posters - Beaumont Leys School

The angle in a
Angles at the
0
B
circumference
semi-circle is 90
They must come
from the same arc.
are equal.
A
Look out for a
2x
diameter.
A
Cyclic
Quadrilateral
They must come
from the same arc.
x
B
is twice the angle at the
Circle Theorems
Opposite angles
The angle at the centre
circumference.
From any point you can
only draw two tangents...
equal
add up to 1800
The angle between
a tangent and a
equal
0
radius is 90
Alternate Segment
Theorem.
Look out
for radii.
@lhmaths
... and they’ll be equal.
Lesley Hall @ Soar Valley College
0
180
straight lines
round a point
360
triangles
allied angles
0
quadrilaterals
parallel
opposite
lines
alternate
angle sum =
(n – 2) x 1800
polygons
corresponding
exterior angle
Equal
…add up to
interior angle
@lhmaths
3600
Lesley Hall @ Soar Valley College
height
height
rectangle
height
a triangle is half the area of a rectangle
base
parallelogram
height
AREA
height
Area = base x height
2
Area = base x height
Always use the
perpendicular
base
Area = base x height
Area = (a + b) x h
2
height
a
height
trapezium
@lhmaths
base
base
base
circle
radius
Area = πr2
b
Lesley Hall @ Soar Valley College
Y=3x+2
Y = 4x + 3
Quadratic Graphs
Y=3x-4
y = x2
intercept
3
gradient
3
U shaped –
parabola.
y = mx + c
Y=3x+2
y = -x + 5
3
Parallel lines have
the same gradient.
GRAPHS
Perpendicular lines
y = tanx
have gradients with
a product of -1.
y = sinx
Sine Curve
@lhmaths
Trigonometric Graphs
y = cosx
Cosine Curve
Tangent
Reciprocal
y = ax2 + bx + c
Cubic Graphs
Linear Graphs
Square
numbers.
4
3
y = x2 – 3x - 2
y= 1
x
Cube
numbers.
y = x3
Lesley Hall @ Soar Valley College
Angle Sum
3
(n – 2) x 1800
triangle
5
pentagon
number of
Polygons
8
octagon
9 - nonagon
10 - decagon
exterior angle
interior angle
angle sum
number of sides
3600
number of sides
OR
OR
exterior
180 – angle
0
@lhmaths
6 hexagon
7 - heptagon
triangles
4 x 1800 = 7200
4 quadrilateral
interior
1800 – angle
Lesley Hall @ Soar Valley College
Completing the square:
Solving:




Factorising
Formula
Completing the square
Drawing a graph
x
2
x
x2
2x
2
2x
4
half of 4x
x2 + 4x - 3 = 0
(x + 2)2 – 4 – 3 = 0
2
x + 4x
subtract 22
(x + 2)2 – 7 = 0
= (x + 2)2 - 4
x + 2 = ±7
x = ±7 - 2
complete
the square
Factorising:
The formula:
Quadratic Equations
x2 + 9x + 20 = 0
x
x
5
x2
5x
4 4x
20
ax2 + bx + c
fill in the blue
squares first
work out the factors
(red numbers)
(x + 5)(x + 4) = 0
x = -5 x = -4
3x2 - 10x + 8 = 0
x
-2
the white
squares make
the x term
Difference of Two Squares:
x2 - 16
(x – 4)(x + 4)
x squared subtract
4 squared
3x
3x
2
-4 -4x
-6x
Graphs:
draw lines to
find solutions
8
(3x - 4)(x - 2) = 0
x =
x=2
@lhmaths
x = -b ±b2 – 4ac
2a
Parabola – u shaped graph
Lesley Hall @ Soar Valley College
y = -fx
y = fx + a
y = kfx
plus a - up
reflection
minus a – down
in x-axis
k
0
a
a
y=sinx
y=x
2
y = f(x + a)
plus a - left
minus a – right
a
@lhmaths
scale
factor k
stretch in
y = f(-x)
x-axis
reflection
0
in y-axis
might think!
y-axis
y = f(Kx)
-a
opposite to what u
stretch in
y=x3
scale factor 1/k
Lesley Hall @ Soar Valley College
Rotation
Translation
Rotation of 900,
Describe with
clockwise, about
To describe a rotation
a vector
centre (2,-1)
you need:
3
4
3
4
squares right
squares up

the angle of rotation

the direction

the coordinates of
anticlockwise
the centre
Transformations
Reflection
clockwise
Centre
Centre of rotation
Negative enlargements
3
– HIGHER only!
1
Enlargement
Enlargement,
scale factor 3,
centre (0,7)
x=2
2
Describe by naming
the line of symmetry
Reflection in the
line x = 2.
@lhmaths
Always use TRACING
PAPER for translation,
reflection & rotation.
To describe an enlargement
you need:
Enlargement of

the scale factor

coordinates of the centre
scale factor -2.
Lesley Hall @ Soar Valley College
Non-Prisms
Prisms
Cones
Prisms have a uniform
cross-section
Pyramids
2
Crosssection
height
V = πr h
3
length
Volume =
area of
cross-section
radius
Volume
× length
a cone is one third
of a cylinder
Frustrums
cylinders
cuboids
Volume = area of base x height
3
πr2
Spheres
a frustrum is a pyramid
with the top cut off.
length
Volume = length x width x height
height
height
width
You need to find
Volume = πr2h
radius
the volume of both
pyramids.
Often you need to use similar
V = 4πr3
3
shapes in frustrum problems.
@lhmaths
Lesley Hall @ Soar Valley College
@lhmaths
Lesley Hall @ Soar Valley College
Square
4 equal sides
opposite sides
are parallel
rotational
symmetry
of order 4
diagonals
meet at 900
diagonals of
equal length
4 lines of symmetry
Rhombus
4 equal sides
opposite sides
are parallel
rotational
symmetry
of order 2
kite
diagonals
meet at 900
rotational
symmetry
of order 1
2 pairs of
equal sides
Parallelogram
1 line of
symmetry
rotational
symmetry
of order 2
no line symmetry
Quadrilaterals
Trapezium
Rectangle
@lhmaths
opposite
sides are
equal &
parallel
one pair of
parallel sides
rotational
symmetry
of order 1
rotational
symmetry
of order 2
diagonals
meet at 900
2 lines of symmetry
opposite sides
are equal &
parallel
an isosceles trapezium has a line of symmetry
2 lines of symmetry
diagonals of
equal length
angles in a quadrilateral
add up to 3600
Lesley Hall @ Soar Valley College
10 millimetres = 1 centimetre
units
100 centimetre = 1 METRE
inches
yards
feet
MILES
Imperial
units
ounces
pounds (lbs)
1000 METRES = 1 kilometre
Metric units
1000 grams = 1 kilogram
Imperial
Metric units
STONES
Units
1 inch = 2.5 cm
1 kg = 2.2 pounds
units
1000 millilitres = 1 litre
(6 foot)
1 gallon = 4.5 litres
4 litres = 7 pints
1 litre = 1¾ pints
Metric units
metres tall.
1 mile = 1.6 km
Imperial
gallons
man is about
1.7 or 1.8
5 miles = 8 km
pints
An average
‘A litre of water is a pint
and three quarters’
@lhmaths
1 foot = 12 inches
That’s 30cm – the length of a ruler!
3 feet = 1 yard
A yard is almost 1 metre
(its 90cm).
Lesley Hall @ Soar Valley College
Brackets
The Power Zero
division, so
subtract the
powers
... but any number
divided by itself
Multiplying
equals 1 ...
Indices
add the
powers
Negative Indices
Dividing
power zero equals 1.
Fractional Indices
25
5
5–3=2
x 25 = 25
x 5
1
add
powers
= 25
must be equal
=
subtract the
powers
Any number to the
equal
25
2 – 5 = -3
8
@lhmaths
27
Lesley Hall @ Soar Valley College
increasing
on a calculator
39% of 82
0.39 x 82
fraction to %
15 = 75
= 75%
20
100
x
5
5
Change to a
decimal and
multiply
increase £60 by 12%
12% of 60 = 0.12 x 60 = £7.20
New amount = £60 + £7.20
= £67.20
Percentages
OR
15 ÷ 20 x 100 = 75%
without a calculator
50% - half
25% - half and half
75% - 50% + 25%
%
ADD
decreasing
decrease £60 by 12%
12% of 60 = 0.12 x 60 = £7.20
New amount = £60 - £7.20
= £52.80
10% - divide by 10
5% - half 10%
20% - double 10%
SUBTRACT
Lesley Hall @ Soar Valley College
@lhmaths