Draw/Sketch/Interpret a Diagram using Bearings – SS6
To save paper this section has been moved to p79 after SS18.
Calculate the Surface Areas of Prisms, & Cylinders – SS7
Surface area is the total area of all the outer surfaces of the object. If you were painting it, it’s all the sides
you would paint.
Strategy
1. Consider the outer surface or draw the net if this helps
2. Look for faces of the same size and shape
3. Find the area of each different face
4. Total together the areas to find the surface area of the solid
surface area of the shape can often be the area of the net
A net is the surfaces of a solid shape folded out flat
The nets of the following 3D shapes have been drawn, and the individual shapes that make up the nets listed:
cylinder
cuboid
2
10
2
10
2
10
2
triangular prism
2
2
2
2
2
circumference of
the circle = 2πr
10
10
10
2
2 small rectangles
4 large rectangles
1 rectangle
2 circles
2 triangles
3 rectangles
Example
What is the surface area of the cylinder above?
Solution
Starting at stage 3 of the strategy:
3. Area of circle = πr 2 = π × 2 2 = 4π
Area of rectangle = 2πr × h = 2 × π × 2 × 10 = 40π
4. 2 × 4π + 40π = 48π units2 or 151 units2 to 3 s.f.
Your Turn!!
a)
Find the surface area of the cuboid above.
b)
Find the surface area of a cuboid with dimensions 2cm by 3cm by 4cm.
c)
Find the surface area of a closed cylinder 5cm long and with a radius of 4cm.
Give your answer exactly in terms of π.
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
1.
Find the Surface Area of the two solid shapes:
Give your answer to ii) to 3s.f. and exactly in terms of π.
i)
ii)
1
2
3
3
6
© ZigZag Education, 2004
Calculate the Volume of Prisms, & Cylinders – SS8
The volume of a solid object is the amount of ‘3D space’ it occupies.
A) The Volume of Prisms - ‘Blobs’, Cuboids and Cylinders
A prism is a solid (3D) object which has a constant area of
cross-section. This means it has the same shape and size from
one end to the other.
Volume of prism = Area of cross-section × length
Strategy 1 for cuboids and cylinders
1. Identify the cross-sectional shape
2. Recall the appropriate area formula.
3. Recall/work out the appropriate
volume formula.
Your formulae sheet
gives you this formula
cross
section
length
Use the V = Area of cross-section × length formula to remind you of the formula for cylinders and cuboids.
This is Strategy 1. Watch…
A cylinder is a
prism with a
uniform
circular crosssection
Remember…
r
Area of a circle = πr2
Therefore…
Volume of Cylinder
= Area of cross-section × length
= πr2 × l = πr2l
length, l
constant area of cross-section
A cuboid is a
prism with a
uniform
length, c
Remember…
b
Area of a Rectangle = ab
a
rectangular
Therefore…
Volume of Cuboid
cross-section
= ab×c = abc
constant area of cross-section
Exam style Question
Strategy 2 for blobs and other prisms
A blob of ink marks an area of 0.2cm2 on some paper. 1. Identify the shape of the cross-section
The blob penetrates the paper uniformly by 0.1cm.
2. Calculate the area of this cross-section.
Calculate the volume of the blob of ink.
3. Use Volume of prism = Area of cross-section × length
Solution (Strategy 2)
1. & 2. The cross section is a blob with an area of 0.2cm2.
3.
V = Area of cross-section × length = 0.2 × 0.1 = 0.02cm3.
Your Turn!!
a) Use the above formula to work out the volume of a cuboid with
dimensions 1cm by 2cm by 3cm.
b) Use the above formula to work out the volume of a cylinder with
radius 3cm and height 10cm. Note: the word ‘height’ in this
question represents the length in the formula above.
c) Calculate the volume of the following triangular prism
using Strategy 2.
2cm
6cm
length,
10cm
constant area of cross-section
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
1. Find the volume of a cuboid with dimensions 2cm by 3cm by 4cm.
2. Find the volume of a cylinder with radius 5cm and height 6cm.
3. Find the volume of an ink blob that marks an area of 0.4cm2 on some paper and penetrates uniformly 0.2cm.
© ZigZag Education, 2004
Know basic Circle Properties, Know & Use the Circle Theorems – SS9
An arc
is a small part of
the circumference of the circle
A chord is the line joining
two points on the
circumference of a circle
A segment is the area
bounded by an arc and a
chord
Basic Circle Facts…
Tangents drawn from the same point are equal in length.
The perpendicular from the centre to a chord bisects (divides equally) the chord
Tangents and radii at the same point on the circumference are perpendicular.
Segment
The circumference is the
distance around the circle
Chord O
The diameter is the distance across the circle through the
centre of the circle, O, and is twice the length of the radius.
Your Turn!!
a) Read each of the theorems below three times and carefully study each diagram.
b) Write down each theorem.
c) Now draw a diagram to represent each theorem.
i) The angle at the centre is twice the angle at the circumference
C
x°
This is sometimes written as the angle subtended (made) by an arc (or
chord) at the centre of the circle is twice the angle subtended at any
point on the circumference (in the same segment).
2x°
To help you spot these look for two points on the circumference (in this case A and B)
which are both connected to the centre (O) and to a third point on the circumference (C).
ii)
The angle from a diameter is a right angle (90°)
B
O
A
iii) Angles in the same segment are equal
C
C
x°
90°
A
O
D
To help you spot these look for the
common chord, in this case AB.
B
major
segment
x°
B
A
iv) Opposite angles of a cyclic quadrilateral sum to 180°
In the diagram shown this means ∠BAD + ∠BCD = x° + y° = 180o
A cyclic quadrilateral is a 4-sided shape
with every corner touching the circle.
A
Your Turn!!
d) State two other angles from the diagram that
must sum to 180o.
x°
O
B
minor
segment
D
y°
C
Your Turn!!
e) Giving a reason,
state whether
EFGH is a cyclic
quadrilateral.
E
F
H
G
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
In the diagram, BD is the diameter of the circle and O is the centre of the circle.
C
D
If ∠ABC is 110o, giving a reason, what is angle ADC?
Explain why ∠BAD is 90o
Giving a reason, which angle is twice ∠ADB?
i)
Giving a reason, which angle is equal to ∠ACB?
ii)
Is this angle acute (less than 90°) or obtuse (>90, <180°)?
Hint: Giving a reason in this question means stating which theorem you are using.
1.
2.
3.
4.
O
E
B
A
© ZigZag Education, 2004
Draw Basic Constructions – SS12
Perpendicular Bisector, Angle Bisector and Equilateral Triangle
Learn the following three key constructions using a pair of compasses, a straight edge and a pencil.
Your Turn!! a) Copy the 3 key constructions as shown.
Perpendicular bisector of a line AB
C
With compasses
centred at A, draw
arcs above and
below the line.
A
B
C
With the same radius,
and with compasses
centred at B, draw
crossing arcs above
and below the line.
A
Join C to D. This is
the perpendicular
bisector of AB.
B
A
B
D
D
Angle bisector of ∠ABC
With the same radius,
draw arcs centred on D
and E to cross at F.
With the compasses centred on
B, draw arcs to cut the lines at
A
D and E.
D
Join the points B and F.
This is the bisector of
the angle ABC
A
D
E
D
F
E
F
E
C
B
C
B
A
B
C
Equilateral Triangle
F
F
Set the compasses to the
same length as AB.
Draw arcs, centred on
A and B to cross at F.
A
Join the points A to F
and B to F.
B
A
B
Constructing Angles of 90°, 60°, 45° or 30°
Notice how the perpendicular construction can be used to construct an angle of 90°. The equilateral triangle
construction can be used to make an angle of 60°. By bisection of 90°/60° you can construct angles of 45°/30°.
Your Turn!!
b) Use your right-angle you constructed from a), and bisect this to make an angle of 45°.
If you want you could bisect the 45° to make an angle of 22.5°.
RAPID ‘ACID’ TEST
1.
– Blank out the page above before answering these!
Construct an angle of 30°.
© ZigZag Education, 2004
Construct Basic Loci – SS13
Circle, Race Track, Angles Bisector & Perpendicular Bisector
A locus is a point, line or shading that shows all the points that satisfy a given rule. You need to know the
general shape of the solution and how to draw the constructions.
The most common loci rules are:
The fixed distance from a single point
The equidistance from two points
The equidistance from two lines
The fixed distance from a line
1.
2.
3.
4.
The solution shape is a:
Æ
Æ
Æ
Æ
Circle
Perpendicular Bisector
Angle Bisector
Race Track Shape
Construction Method:
Compasses centred on the point
See ‘Constructions’
See ‘Constructions’
This construction is shown below
Examples
1. A goat is attached to a
rope of length 5m, which
is attached to a post A.
Draw the maximum area
that the goat can move.
4. Draw the area of the
shape 2cm away from
the line AB
5m
A
2 cm
A
2. A ship must sail a
route equidistant
from two buoys A
and B. Draw its
route.
Draw a line touching the top of the two ¾ circles
and the top arc. Draw a line touching the bottom
of the two ¾ circles and the bottom arc.
B
B
B
A
ships route
3. A shot-putter throws a shot to a
point equidistant between the two
lines AB and BC. Draw the
line showing all the points
where the shot may land.
Draw a rough ¾ circle
centre A and B. With the
compasses near the centre
of the line, draw an
arc above
and below
the line.
A
B
A
C
Your Turn!!
a) Draw a line AB 5cm long. Draw all the points that are 5cm from AB.
b) Draw a line AB 8cm long. Shade the locus of points P, where P is less than 5cm from A and less than 4cm
from B.
B
RAPID ‘ACID’ TEST
– Blank out the page above before answering these!
Draw all the points that are 5cm from the triangle ABC, where ABC is
5cm
an equilateral triangle of side 5cm.
5cm
Start by constructing (see constructions) or tracing,
or roughly drawing the equilateral triangle.
A
5cm
C
© ZigZag Education, 2004
Describe Transformations – SS14
The 3 key transformations that you are asked to describe are Reflections, Rotations and Translations. You
will also need to describe Enlargements but these are easier to spot because the shape gets bigger, what a
surprise, or smaller – see Enlargements on the next page.
A Reflection is described by a line of reflection (or mirror line).
Example “The shape is reflected in the line y = x.”
The top number indicates
the change in the x direction
A Rotation is described by 3 properties: 1. Angle 2. Direction 3. Centre.
Example “The shape is rotated 90°, clockwise
, about (–1, 0).”
Meaning the
⎛3 ⎞
Translation means move and is described by a vector. Example ‘The shape is translated ⎜⎜ ⎟⎟ ’ shape has
⎝ 4 ⎠ moved 3
the
bottom
indicates
the
written
like
this
−
3
⎛ ⎞
units right,
⎜ ⎟ means 3 units left, 4 units down
change in the y direction
4
units up.
4
−
⎝ ⎠
Finding a Mirror Line - If you can find a mirror line so that each corner of the
‘object’ triangle is the same distance
Hint…
as the corresponding corner from the
Mirror lines are usually y = x, y = –x,
‘image’ triangle, then you have a line
the y-axis, the x-axis, vertical lines
of reflection.
y
4
A’’
y= x
A
3
2
(eg. x = 3) or horizontal lines (eg. y = 3.)
A’
1
Your Turn!!
a) Describe the reflection of A to A’’.
−4 −3 –2 −1 0
B
Finding a Translation Vector - Translations are easy to spot as their object
⎛ 5⎞
and image have the same orientation and size! B to B’ is a translation of ⎜ ⎟ .
⎝ −2 ⎠
Your Turn!!
b) Describe the translation of i) B to B’’ ii) B’’ to B.
Finding the Centre of Rotation - Join two corresponding
corners y and construct, or roughly draw, the
A’
4
perpendicular bisectors – see constructions.
A’
The centre of rotation is where these two
perpendicular bisectors cross. Use
2
A
the 3rd corner and corresponding corner
to check – its perpendicular bisector −4 −3
should
go through the SAME point.
−4 −3 −2
1 2 3
4 x
−1
In this case through (–1, 0).
3
+5
−2
1
y
4
3
2
−1
A
1
2
3
2
4
−2
3
4 x
B’’
−4 −3 −2 −1 0
−2 −1
1
y
4
1
2
B’
3
4 x
The angle of
rotation can be
found by joining
a corresponding
corner on the
object and
image to the
x
centre of
rotation. In this
case the angle
of rotation is 90°
anticlockwise.
−3
Your Turn!!
−3
−4
c) Describe the rotation that takes i) A to A’ ii) A’ to A.
−4 the grid and draw the two triangles A and A’.
d) Copy out
Now without looking at the diagram repeat the construction to find the centre of rotation.
Draw Rotations/Reflections by using the facts above or follow the following tracing paper method
Reflections - Copy the shape and mirror line onto tracing paper. Turn the paper over so the image appears on the
opposite side of the mirror line and align the mirror lines. Mark the corners of the image and join these together.
Rotations - Copy the shape and mark the centre of rotation. Place a compass or pencil point on the centre of
rotation. Rotate the tracing paper the required amount. Mark the corners of the image and join these together.
RAPID ‘ACID’ TEST
Blank out the page above before answering these!
1.
y
4
A’
2
b) A to A’’
c) A to A’’’
A
1
Describe the single transformation that takes:
a) A to A’
3
-4 -3 -2 -1
-1
A’’
-2
1 2 3 4
x
A’’’
-4
© ZigZag Education, 2004
Describe Enlargements – SS15
Enlargement is described by a centre and a scale factor of enlargement. For example, the image A’ in fig1 is
an enlargement of object A from (–4, –3) by a scale factor of 3.
y
4
3
Finding the Centre of Enlargement
Join corresponding corners together and extend them until they cross;
this will be the centre of enlargement. In fig1 the corresponding corners
meet at (–4, –3).
2
1
-4 -3 -2
A
Deciding the Scale Factor of Enlargement
Measure the length from the centre of enlargement to a corner and a
corresponding corner. Divide image length by object length,
image length to centre of enlargement
. Alternatively simply divide two corresponding
object length to centre of enlargement
lengths of image and object,
image length
object length
-1
A’
-1
1
4
x
-2
-3
-4
centre (–4, –3) scale factor ×3
. This is the scale factor of
Hey Look…
enlargement. In fig1 the height of the object is 1 and the corresponding
length
image height is 3, image
= 13 = 3.
object length
The enlargement
backwards, so from A’
to A is an enlargement
about (-4,-3) by a scale
factor of ⅓.
Drawing Enlargements
Draw a line from the centre of enlargement to a corner of the object. Multiply this length by the scale factor
and extend your original line to this length. Repeat for the other corners and join these together to make the
enlarged shape. In fig1 one of the corners of the object is 3 left 1 down from the origin and so the
corresponding corner of the image will be 1 left and 3 up from the origin.
Your turn!! a) In fig1 another one of the corners of the object is 1 left and 2 down.
Describe the position of the corresponding image corner.
y
4
3
Your Turn!!
b) Describe the enlargement from A to A’.
c) Describe the enlargement from A’ to A.
2
A
1
-4 -3 -2 -1
1
-1
2
A’
3
x
4
-2
-3
-4
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
1.
y
4
Describe the enlargement from
a) A to A’
A
3
2
b) A’ to A
A’
-4
1
2
3
4
x
-2
-3
-4
© ZigZag Education, 2004
Calculate Lengths in Similar Triangles – SS16
If one triangle is an enlargement of another, then the two are mathematically similar. Similar triangles
have all corresponding angles equal to each other – the size of the triangle is different! The sides of the
two triangles are in the same ratio.
‘Similar’ means same shape, different size
Example
BC is parallel to DE. Calculate the length AE
E
Strategy
1. Establish that triangles are similar if necessary.
2. Draw the two similar triangles with the same
orientation, and label the separate diagrams.
3. Find the ratio of two corresponding sides and so
calculate the scale factor of enlargement.
4. Apply this factor to the side required.
5. Check this is sensible in relation to the original diagram.
C
5 cm
A
4 cm
6 cm
B
D
Solution
1. ∠BAC = ∠DAE common to both
∠ACB = ∠AED F-angle in parallel lines
∠ABC = ∠ADE F-angle in parallel line. So ∆ABC is similar to ∆ADE.
C
2.
B
A
E
A
D
3. Side AB corresponds to side AD.
Ratio of corresponding sides AB to AD = 4 : 6
Scale factor of enlargement is 6 ÷ 4 = 1.5.
4. The side which needs to be found is AE, with a corresponding side on the smaller triangle AC.
AD = AB × 1.5 = 5 × 1.5 = 7.5 cm.
5. Checking: 7.5cm > 5cm, therefore answer is sensible.
Your Turn!!
I
J a)
H
∆FGH is similar to ∆HIJ because; ∠FHG = ∠IHJ (opposite angles),
∠GFH = ∠IJH (Z-angle), ∠FGH = ∠HIJ (Z-angle). Side FG
corresponds to IJ. Which side corresponds to side GH?
A
F
G
5 cm
4 cm
b)
Given that BC is parallel to DE, calculate
the length of sides AD and DE.
9 cm
C
B
6 cm
Hint: Use the ratio of AC : AE
D
E
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
1. These triangles are similar. Calculate the length XY.
∠BAC and ∠XZY are equal.
Do not calculate x.
4 cm
A
Z
B
xo
15 cm
5.6 cm
10 cm
xo
6 cm
C X
Y
© ZigZag Education, 2004
Decide if a Formula is a Perimeter, Area or Volume
by considering Dimensions - SS18
Constants are dimensionless and can be ignored – e.g. 2 and π are dimensionless constants.
In the following section a, b, c, l, h, w are all measures of length.
The formula for lengths involve lengths or added lengths.
E.g. Æ 2πl or 2π ( l + h ) or
l+h+w
The formula for areas involve the product of two lengths.
E.g. Æ 2πl or 2πl( l + h ) or
l 2 + h2 + w2
The formula for volumes involve the product of three lengths.
E.g. Æ 2πl 3 or 2πl 2 ( l + h ) or
l 3 + h 3 + w3
Strategy
1. Write without constants like 2, and π.
2. Replace lengths by cm, (areas by cm2).
3. cm Æ length. cm2 Æ area. cm3Æ volume.
2
Example 2(3a + πb)c2
1. (a + b)c2
2. (cm + cm) cm2 = (cm) cm2
cm + cm are more cm’s!
a 2 + b2
1.
a 2 + b2
2.
cm2 + cm2 = cm2 = cm
cm2 + cm2 are more cm2’s!
= cm3
3. Volume formula
Your Turn!! What types of formula are: a) 2πbc
Example
3. Length formula
2
2
b) π (bc +ac)
c) (πc + ½ab)
d)
πc2 − 12 ab
h
Extra note θ is a measure of length (angular length), but sinθ is a constant and can be ignored.
f) 3a sinθ
Extra Your Turn!! What types of formulae are e) 2aθ
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
Identify these formulae as perimeter (length), area or volume:
1.
3(ac2 − abc)
4
2.
2πc 2 − ab
h
3. 2abc
From p71…
Draw/Sketch/Interpret a Diagram using Bearings – SS6
Pythagoras and Trigonometry questions are sometimes asked through a bearings style question. In such
questions you might need to draw or interpret a bearings diagram.
In a bearing of C from B; a
person would be standing at B!
Reminders
1. Bearings are always measured clockwise from North ↑.
2.
3.
4.
N2
Bearings are always given as three figures, so 005° not 5°.
Look carefully at the words ‘to’ and ‘from’ in questions.
Angles can be calculated using the fact that the Norths are parallel.
90°
For example ∠ABN2 is 180 – 45 = 135° see iSS1 for a reminder of Z-angles etc. B
Example
Draw a diagram where a ship travels from A to
B on a bearing of 045° a distance of 5km, then
to C from B on a bearing of 090° a distance of
5km. It then travels back to A from C.
Solution See DiagramÆ
5
N3
C
5
N1
45°
A
Your Turn!!
In the ∆ABC, ∠ABC = 360 – 90 – 135 = 135° (Angles at a point sum to 360°).
a) Calculate the other angles in the ∆ABC.
b) State the bearings of i) A from B
ii) B from C
iii) A from C.
1.
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
A ship travels to B from A on a bearing of 270° a distance of 5km and then from B to C a distance of 5km on
a bearing of 060°.
a) Sketch a diagram to represent this journey.
b) Calculate the bearing of i) C from A ii) A from C.
© ZigZag Education, 2004
Calculate Compound Measures like Speed & Density – SS19
Memorise the formulae: Average Speed =
Distance
Time
Acceleration =
Change in Speed
Time
Example
Calculate the average speed of a car that travels 21 km in 3 hours.
Solution
Sometimes you need to rearrange the
Distance 21
Speed =
=
km / hour
formula, making one of the other
Time
3
parts the subject of the formula.
Density =
Mass
Volume
Example
A car travels at 40km/hour for 3 hours.
How far does the car travel?
Solution
Distance
Speed =
.
Time
Your Turn!!
a) A giant potato has a mass of 0.414 kg and has a volume
of 90 cm3. Calculate the density of the potato.
So,
Note: Make sure you can read the calculator display, see NA7 parts e) & f).
Speed × Time
40 × 3
= Distance
= 120 km
b) A small orange has a mass of 0.05 kg and a density of 0.001 kg/cm3.
Calculate the volume of the orange.
Hint: Start by making Volume the subject of the equation Density =
Mass
.
Volume
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
1. A solid stone has a mass of 16 kg and has a volume of 400 cm3. Calculate the density of the stone.
2. Make ‘Time’ the subject of the equation Speed =
Distance
and hence calculate how long it takes an object
Time
travelling at 20m/s to cover 110 metres.
Convert between Volume Measures including cm3 and m3 – SS20
Converting Basic Units
Conversion Factor Reminder
cm Æ mm ×10
m Æ cm
×100
km Æ m
×1000
Example
Convert 3cm to m.
Solution
3 ÷ 100 = 0.03m
Converting Volume Measures
Example
Convert 3cm3 to m3.
Volume Measures so cube i.e. 3
the cm Æ m conversion factor
The cm Æ m conversion factor
Solution
3 ÷ (100)3 = 0.000003
× or ÷ as appropriate
This can be written 3 × 10–6 in Standard Form
Check that your answer is reasonable at the end; this will ensure you have used the correct sign.
It is not uncommon for students to use the incorrect symbol.
Your Turn!!
Convert
a) 30,000cm3 to m3
Convert
b) 12cm3 to mm3.
RAPID ‘ACID’ TEST – Blank out the page above before answering these!
1. 0.0025m3 to cm3
2. 120mm3 to cm3.
© ZigZag Education, 2004