Transcript Slide 1

Unit 3 Revision Notes - Higher
Angles: Parallel Lines
a
Parallel lines are shown by
having arrows drawn on
them
b
c
Corresponding angles
‘a’ and ‘b’ are equal
Alternate angles
‘c’ and ‘d’ are equal
d
Notes: 1) Use the correct words: Alternate and Corresponding in the exam
2) Useful for Bearings questions. The Bearing of A from B is 120 degrees what is the
Bearing of B from A?
N
Bearing of B
N
from A = 300
N
degrees
B
120
B
A 120
A
180
3) Used when showing that two triangles are similar/congruent – remember to explain
why each angle is the same!
E
B
Show that ABC is similar to CED
Angle A = Angle E (alternate)
Angle B = Angle D (alternate)
Angles at C are the same
(vertically opposite)
C
A
D
Angles: Circle Thoerems
Chord my not always be visible!
The angle subtended by a
chord at the circumference
is half that at the centre.
The angles subtended (made)
by a common chord at the
circumference are equal.
The angle made by a
diameter at the
circumference is 90 degrees.
Opposite angles in a cyclic quadrilateral
add to 180.
The angle between a tangent and
chord is equal to the angle
subtended (made) by a chord in the
alternate segment.
Note when doing circle theorems questions watch out for isosceles triangles – in particular
triangles with two sides as a radius of the circle.
Also remember that a tangent to a circle will always make a 90 degree angle with the radius at
the point where they meet.
Eg.
What are the missing angles?
Give reasons.
A = 120 (angle at centre = twice that at
circumference)
B = 60 (angles at circumference are constant)
C = 120 (opposite angles in a cyclic quad add to
180)
Right Angled Trigonometry
opposite
Sin 
hypotenuse
adjacent
Cos 
hypotenuse
Tan 
opposite
adjacent
THESE FORMULA ONLY WORK FOR RIGHT ANGLED TRIGONOMETRY!!
IF YOU SEE A RIGHT ANGLED TRIANGLE (INDEED A RIGHT ANGLE!) ANYWHERE IN THE
EXAM THINK PYTHAGORAS AND (SOHCAHTOA)
Eg1 Find the angle X
8 cm
X
12 cm
We know 8 = adjacent
12 = hypotenuse
Cos X = 8/12
X = Cos-1 (8/12)
X = 48.2 degrees (1 d.p.)
Remember to use shift button on calculator to get the angle using cos inverse)
Eg2 Find the side x
We know 12 = opposite
70 = angle
12 cm
70o
x
Eg3 Find the side x
6 cm
x cm
70o
Tan 70 = 12/x
x = 12/tan(70)
x = 4.4 cm
This is what I calla ‘hard side’ question. You need to get the rearrangement
correct from the first to the second line.
We know 6 = opposite
70 = angle
Sin 70 = x/6
6 × Sin 70
x = 5.6 cm
This is what I call an ‘easy side’ question. You need to get the rearrangement
correct from the first to the second line – but it is easier to do than in eg 2
above.
Trigonometry for general Triangles
These rules work for any triangle. If you see a non right angled triangle you have to use them.
On the formula sheet in the exam:
a  b  c  2bcCosA
2
2
2
a
b
c


SinA SinB SinC
Useful rearrangements but you need
to learn them if you want to use
them
2
2
2
CosA 
b c a
2bc
SinA SinB SinC


a
b
c
Both of these
are Useful for
finding an
angle
Use the Cosine rule when:
You want to find a third side and you
have 2 sides and an angle in between
50O
12
10
a
You want to find an angle and you
have 3 sides
Use the Sine rule when:
The Cosine rule doesn’t work!
This will be when you have a side and
an opposite angle as a pair.
10
A
12
8
10
50O
B
12
Eg 3. A yacht sets off from A and sails 5km on a bearing of 067 degrees
to a point B so that it can clear the headland before it turns onto a
bearing of 146 degrees. It then stays on that course for 8km until it
reaches point C.
Find the distance AC.
The trick here is to find the angle at B.
Clearly this is a cosine rule question. Use a diagram and corresponding angles
to do this.
N
N
67
A
5
B
N
146
8
C
2 sides but need
angle in between so
redraw and use angle
facts with parallel
lines to find it!
67
A
b2 = c2 + a2 - 2×c×a×Cos B
Be careful with the double
2
2
2
negative!!!!! (Cos of a value
b =5 + 8 - 2×5×8×Cos 101
above 90 is always negative)
b2 =25 + 64 - 80×Cos 101
b2 =25 + 64 - -15.2647
b2 = 104.267
So
b = 10.2 km
B
67
146 – 67 = 79
C
101
Internal and External Angles of Regular Polygons
Regular shapes are
shapes where the
internal angles are
all the same and
where the side
lengths are all the
same. This is a
Regular Pentagon.
Internal angles are those on the inside of the
polygon
For a triangle the internal angles sum to 180o
For a quadrilateral they sum to 360o
For a pentagon 540o.
(Each extra side adds on 180o)
External angles always add up to 360 degrees
and are the amount you turn at each corner.
The external angle of a regular polygon with
‘n’ sides is 360 ÷ n
The internal angle of a regular polygon
= 180 – external angle
Answer:
External angle in a regular octagon =
360 ÷ 8 = 45o
Internal angle = 1800 – 45 = 135o
At Angle here
= 360 – 135 – 135 = 90o
Graph Transformations
Move up: Add on the end
y = f(x) + 1
Vertical stretch: Times
whole expression
y = 2f(x)
Original Graph
y = f(x)
This is drawn as dotted
line in each example
Move sideways: Add to x
(but works ‘wrong’ way)
y = f(x + 2)
Stretch sideways:– Times the x
(but works ‘wrong’ way)
y = f(3x)
y= 2f(x)
Look at the graph of
y = f(x) opposite.
On the same axes plot
the graph of:
a) y = f(x)-3s
Translate down 3
b) y= 2f(x)
Stretch by factor 2
vertically – all the
points are twice as
Far away from the
X-axis as they
Were to begin with
a)
y = f(x)-3
Look at the graph of
y = f(x) opposite.
On the same axes plot
the graph of:
a) y = f(2x)
Stretch by factor
½ horizontally – all the
points are half as
far away from the
y-axis as they
were to begin with
a)
y = f(2x)
a)
y = f(x - 4)
b) y = f(x - 4)
Translate +4 horizontally
Note: The graphs y = f(x) and y = -f(x) are related by reflecting the shape in the x axis
Similarly y = -2f(x) is a reflection and a stretch of scale factor 2 vertically
Using Graphs to solve equations
An important principle here is that if you want to solve any equation:
Eg 2x + 1 = 3x – 5
Then you can do this by plotting the graphs of what is on each side and then seeing where
they meet. In this case plot the graphs y = 2x + 1 (y = what is on the left)
and y = 3x – 5 (y = what is on the right)
Where the graphs meet the x values will give the same y vale ie the equation will be sovled.
There may be several values for x.
Eg.
Use the graph of y = cos Ө shown below to find all the solutions to
cos Ө = - 0.7 in the range 0o ≤ Ө ≤ 720o
Find one solution using cos inverse:
cos Ө = - 0.7
Then Ө = cos-1 (- 0.7)
= 134.4
Draw the line y = -0.7
134
225.6
134
225.6
Then use the fact that the graph is
symmetrical about 180 to find 3 other
360-134
360+134.4 and 360+134.4
solutions:
(Remember the graph repeats every 360 degrees, so adding
225.6 or 494.4 or 585.6
or taking 360 degrees to any solution will get another)
This is the graph of
y = x2 – 5x
Use it to solve
a) x2 – 5x
=0
b) x2 – 5x
= -4
c) x2 – 6x + 5 = 0
Easy just see where y = x2 – 5x meets the graph y = 0 (ie the x axis)
Clearly 0 and 5
b) Again just see where y = x2 – 5x meets the graph y = -4
Clearly 1and 4 by drawing the line y = -4 on the graph
c) This is the hard one!
Start with the equation you want to solve x2 – 6x + 5 = 0
Add and subtract things to both sides to rearrange the equation to make one side
look like the equation of the graph you have been given (in this case x2 – 5x).
x2 – 6x + 5 = 0
Add x and take 5 from both sides give us:
x2 – 5x = x - 5
Now plot the graph of y = x – 5 and see where the lines meet. In this case x = 1 & 5.
a)
Graphs you need to know:
General quadratic y = ax2 +bx + c
(usually have two solutions which
are the x values where the graph
crosses the x axis)
General cubic y = ax3 +bx2 + cx + d
(usually have three solutions which
are the x values where the graph
crosses the x axis)
Inverse Proportion graph y = k/x
Straight line graph y = mx + c straight line graph
The direct proportion graph y = kx looks like this but always
Go through the origin
y = sin(x) y = cos(x) and y = tan(x) you should be familiar with but you don’t need to be able to
sketch them for GCSE maths
Exponential graphs like
y = 2x, y = 3x , y = 4x etc..
These all look like the shape on the right, which is the
Graph of y = 2x
when x = 1 y = 21 = 2
when x = 2 y = 22 = 4
when x = 3 y = 23 = 8 etc
Transformations
Transformations are ways to change a shape. There are 4 main transformations:
Enlargements
Translation
Rotation
Reflection
(shape’s size is changed so that is still in proportion)
(shape’s position is changed)
(shape’s orientation is changed)
(shape’s orientation is changed)
You can get other transformations by combining these together. You can also define new
transformations (like a stretch which enlarges a shape in 1 direction only – these are not on
the syllabus apart from as part of graph transformations)
Reflections
A reflection is obtained by ‘folding’ the space over a mirror line.
The shortest distance of any point on the original object to the mirror
line should be equal to the shortest distance of its image to the mirror
line.
OBJECT
IMAGE
To describe a reflection you must have:
• An Object
• A line of reflection described as an equation
(Remember vertical lines are x = lines, horizontal lines are y = lines, the line y = x goes through (0,0), (1,1), (2,2) etc
When reflecting in a diagonal line it is easier to turn the line of reflection so that it is horizontal or vertical)
Rotations
A rotation is obtained
by turning a space
around a fixed point
(we call this the
centre of rotation).
The distance of every
point from the centre
of rotation will
remain constant as
the space is turned.
To describe a rotation you must have:
• An Object
• An Centre of Rotation (an axis) described as a coordinate
• An Angle of Rotation
• A Direction to rotate in (usually clockwise/anticlockwise)
(Note: Tracing the object onto scrap paper and rotating the paper using a pencil nib as an axis is a good way to avoid
making mistakes. Ask for tracing paper if you need it!)
Translations
A translation is obtained by moving a shape in space. Each corner of
the object is moved by an amount described in a vector.
The image should be congruent to the object and of the same
orientation.
In this case the Object has
been translated by 4 to
the right and 2 up
As described in the vector
shown.
IMAGE
OBJECT
To describe a translation you must have:
• An Object
• A Vector
The top line of the vector describes the horizontal movement
The bottom line describes the vertical movement
(+ve right, -ve left)
(+ve up, -ve down)
Method:
1) Draw lines from the centre of enlargement through all the corners of the object
2)
Find the distance of the first corner from the centre of enlargement (counting
squares is best)
3)
Multiply this distance by the scale factor
4)
Measure this new distance from the centre of enlargement and mark it as the first
corner in the enlarged image
5) Repeat for every corner until the enlarged shape is ready to be drawn in.
NOTE:
•If the Scale Factor is –ve, measure
in the opposite direction when
plotting the image. The image will
be upside down and on the other
side of the centre (essentially it has
been rotated by 180 degrees at the
same time as being enlarged).
•If the Scale Factor is between -1
and 1 the shape will actually get
smaller!
Eg Enlargement
Scale Factor 3 with
(2,1) as the centre
of enlargement
OBJECT
B
IMAGE
Enlarge the
shape shown
by scale
factor -2
using the
point B as the
centre of
enlargement
OBJECT
IMAGE
C
Enlarge the
shape shown
by scale
factor ½
using the
point C as the
centre of
enlargement
Area and perimeter of 2d shapes
a
Area of a trapezium = ½ (a + b) × h
Remember: This is on the formula page!
Circles:
h – perpendicular
b
distance between the
two parallel sides
Radius = distance from the middle to a circle to the circumference
Diameter = width through the middle of a circle
Diameter = 2 × radius
Sector = a pizza slice from a circle
Arc = part of the circumference of a circle
Segment = pieces sliced across a circle
Area of a circle = ½ π × radius2
radius
Area of a sector = ½ π ×
radius2 ×

360
θ
Circumference of circle
= π × diameter
Arc length
= π × diameter ×
Eg 1 Find the Perimeter of the
sector below which is made
from a circle of perimeter 5 cm
diameter

360
θ
Eg 2 Find the Area of the segment shown if
the radius of the circle is 8 cm
100o
300o
Perimeter consists of 2 radii
(each 5 cm) and an arc
Arc length = π × diameter × 300
360
= 26.1666
Total perimeter = 10 + 26.17 = 36.17 cm
Area of segment = Area of sector– Area of triangle
Arc of sector = π × 82 × 100
360
= 55.85053
Area of Triangle = ½ × 8 × 8 × Sin 100
= 31.51384
Area of Segment = 55.85053 – 31.51384 = 24.3 cm2
Triangles:
Area of a triangle = ½ base × height
OR
Remember to use the vertical height
(ie from the base to the tip of the
triangle when using this formula)
Area of a triangle= ½ a × b × Sin C
Use this formula when you have two
sides of the triangle and the angle in
between – it is on the formula page!
Eg Find the area
Eg 1
3cm
10 cm
50O
12 cm
10 cm
Area = ½ 10 × 3 = 15 cm2
Area = ½ 10 × 12 × Sin 50
= 91.9 cm2
Eg 2
4cm
8 cm
Area = ½ 8 × 3 = 16 cm2
Eg
Answer:
The key to this question is that the areas are equal so start by writing down an expression for
the area of each shape:
(3x +2)(x +1) = Area of rectangle
(3x + 5)x + 5x = Area of L shape
Areas are equal: (3x +2)(x +1) = (3x + 5)x + 5x
3x2 + 5x + 2 = 3x2 + 5x + 5x
5x + 2 = 10x
x = 2/5
Volume and Surface Area of 3d shapes
A prism is a 3d shape where the cross section is uniform (unchanged)
Examples of prisms – cuboids, cylinders
Examples of non-prisms – pyramids, cones, spheres
To find the volume of any prism calculate the area of the cross-section and multiply it by the
length of the prism:
8 cm
Eg 1 Find the Volume of the prism below
Area of cross-section (blue)
= 6 x 8 + 4 x 4 = 64
4 cm
6 cm
Volume = 64 x 10 = 640 cm3.
12 cm
Eg 2 The prism below has a volume of 200 cm3.
If it is 12 cm long what is the area of the cross – section?
Volume = Area of cross-section x length
200 = Area x 12
so Area =200 ÷ 12 = 16.67 cm2
10cm
Volume of a sphere
= 4/3 π × radius3
Surface area of a sphere
= 4 π × radius2
Volume of a pyramid
= 1/3 Area of Base x height
Volume of a cone
=1/3 π × radius2 x height (because it is a pyramid with a circular
base, so the area of the base is π × r2 )
Surface area of a cone
= Area of base + Area of curved surface
π × radius2 + π × radius x slope length
Note:
These are in the
formula booklet
The formula for the curved surface
of a cone is the only formula that
uses the actual slope length rather
than the height.
Slope
length
radius
The curved surface part of the formula (π × radius x slope length)
is on the formula page.
Surface area of a cylinder
= Area of top and bottom + Area of curved surface
2 x π × radius2
+ π × diameter x height
(Two circles)
(Rectangle that has equal to the
circumference of the top/bottom)
Answer
Radius of sphere = 3
so height of cylinder = 9cm
Cylinder = π x 32 x 9
= 81π
Hemisphere = 4/3 π x 33 ÷ 2
= 18 π
Volume = 99 π
---------------------------------------Volume of liquid is 49.5 π
18π in the hemisphere
So 31.5π = Volume of cylinder
31.5π = π x 32 x d
d= 3.5 cm
Eg A cone is made by cutting a 60 degree sector
from a circle of radius 8 cm as show in in the
picture.
A
300
Join these
edges
together
O
This is a classic A* grade question.
Method:
Find the length of the arc from A to B.
This length is the same as the circumference of
the base of the cone. Use this fact to find the
radius of the base of the cone.
300
Arc length = 16 × π × 360 =
Then
40
3
π × diameter =
40
3
diameter =
40
3
B
radius of cone base =
O
Use the fact that the slant height of the cone OA
will be equal to the radius of the original circle.
Then use Pythagoras with this length and the
radius of the base of the cone to find the cone’s
height.
2
A&B
What is the volume of the cone?
20
3
 20 
h  8     4.422166
 3 
8
2
Volume = 2
1/ × π × 20  × 4.422..
3
 3 
= 205.8 cm3
h
40
6
Eg The diagram shows a solid made from a cone and a hemi-sphere.
The radius of both shapes is r
The slant height of the cone is l.
The perpendicular height of the cone is h
The curved surface area of the cone and the curved surface area of
the hemisphere are equal
(a) Show that l = 2r
(b) Find the perpendicular height, h, of the cone in terms of r.
(c) Find the ratio of the volumes of the cone and the hemisphere.
(2)
(2)
(2)
Be prepared to use algebra to solve problems involving volume and surface area
(a) If the curved surfaces of the two shapes are equal then we can write an equation relating
them. Curved surface of hemisphere = 4 π × radius2 ÷ 2
Curved surface of cone = π × radius× l
Hence 2 π × radius2 = π × radius× l
divide both side s by π and by the radius
2 × radius = l - result we need
2
2
(b) Using pythagoras r2 + h2 = l2 so h  l  r
(c) Volume of cone = π × radius2 × h
Volume of hemisphere = 4/3 π × radius3 ÷ 2
= π × radius2 × l 2  r 2
= 2/3 π × radius3
= π × radius2 × 4r 2  r 2
Ratio of Cone to hemisphere
= π × radius2 × 3r 2
= (π × radius3 × 3 ) ÷ (2/3 π × radius3)
3
= π × radius ×
= 3 3
3
2
Enlargement Scale Factors
If two objects are similar it means that their side lengths are related by a multiplying
factor called the enlargement scale factor.
To find the enlargement scale factor divide a length in the image by a corresponding
length in the object.
Note: If the enlargement scale factor for length is x
then the enlargement scale factor for corresponding areas is x2
and the enlargement scale factor for corresponding volumes is x3
E
If triangles ABC and CED are similar, find:
a) The length BC
B
b) The area of triangle CED given that the
area of triangle ABC is 8 cm2
4 cm
C
A
6 cm
12 cm
D
Answer:
a) Enlargement scale factor = 6÷4 = 1.5
So BC = 12 ÷ 1.5 = 8 cm
b) Area scale factor from ABC to CED = 1.52
So area of CED = 8 x 1.52 = 18cm2
Answer
This is a volume scale factor question.
Enlargement scale factor for
length = 30÷20 = 1.5
So volume scale factor = (1.5)3
Small bottle holds 480 ml so bigger bottle
holds 480 x (1.5)3 = 1620 ml
Eg Two solids are similar.
The area of one of the faces is 20 cm2 on the larger shape and 5 cm2 on the smaller shape. If
the smaller shape has a volume of 8 cm3 what is the volume of the larger shape?
Area scale factor = 20 ÷ 5 = 4
So length scale factor will be the square root of the area scale factor, so = 2
Volume scale factor is the length scale factor cubed, so is 23 = 8 in this case.
Hence volume of larger solid is 8 x 8 = 64 cm3
Answer:
You can use enlargement scale factors to
make this question a lot quicker.
The volume of the cone missing from the
top is = 1/3 × π × 22 × 3 = 4π
The larger cone is similar to the smaller
one. The length scale factor is 3 by
comparing the radii. This makes h = 9
cm.
5 marks
So volume of the whole cone including
the missing top which shouldn’t really be
there is 1/3 × π × 62 × 3 = 36π
The volume of the frustum = 36π - 4π
= 32 π
Vector Geometry - Important principles:
1) If you have a situation like this
The vector between the
end points of the vectors
f and a:
FA = a - f
F
A
f
a
O
2) The vector between BA is the negative of the vector from AB
BA =- AB
3) They often ask you to find easier paths at the start of the
question and you need to use these paths to help you with the
harder paths at the end of the question.
4) If you asked to prove that two vectors are parallel you need to
show that one of them is a multiple of the other
Answer
A) OS = OP + ½ of PQ
We have OP = p
We need PQ in terms of p and q.
Looking carefully PQ = -p + q or q – p
So OS = p + ½(q – p) = p + ½q - ½p = ½q + ½p
B) RS = RP + PS
RP = ½p and PS = ½PQ = ½(q – p) from part a
So RS = ½p + ½(q – p) = ½p + ½q – ½p = ½q
Since OQ = q clearly RS is a multiple of OQ ( ½OQ to be exact), hence parallel
Solving quadratic equations
You need to be able to solve quadratic equations by either factorising, using a graph or by
using the quadratic formula. The formula is on the formula page.
Expect to see quadratics popping up in strange places and don’t be concerned if you see
one as part of a question on area or volume – just rearrange to make the equation equal
to zero and solve.
If you can’t get the quadratic equation to work then check you r arithmetic – especially the
negative numbers
Solving simultaneous equations
If, as part of a question you write down an equation with two variables in it you will not be able
to solve the equation uniquely unless you find another equation. You need two equations to
solve uniquely for a system containing two variables.
IF BOTH EQUATIONS ARE LINEAR
Rearrange the equations and then add and subtract multiples of them together to eliminate one
of the variables. You can then solve.
Eg Solve 3x = 10 + 2y
2x + 3y = 11
Rearrange to get letters on the same side in both equations
3x – 2y = 10
1
2x + 3y = 11
2
Note:
Be careful when
subtracting/adding
the equations.
Multiply equation 1 by 2 and equation 2 by 3 to get the number of xs in both equations to
match:
6x – 4y = 20
1
6x + 9y = 33
2
Since the signs are the same on the x’s subtract equation 2 from equation 1
-13y = -13
now solve to get y = 1 and from that we can see that x = 4
IF ONE QUATIONS IS LINEAR BUT THE OTHER IS QUADRATIC
Rearrange one equation – the linear one! To get an explicit equation for one variable. Then
replace this variable in the second equation using the equation.
Eg Solve y2 + 16= 8x
3 + y = 2x
Answer
Rearrange the second equation to get y = 2x – 3 now replace y in the ‘harder’ equation with
2x – 3 - be careful to use brackets if they are necessary.
(2x – 3)2 + 16 = 8x
(2x – 3)(2x – 3) + 16 = 8x
4x2 – 12x + 9 + 16 = 8x
Its a quadratic equation so rearrange to make it equal to zero and then solve
4x2 – 20x + 25 = 0
(2x – 5)(2x - 5) = 0 x = 2.5 or 2.5
when x = 2.5 , y = 2
Note:
This question could be phrased by asking you where the curves y2 + 16= 8x and 3 + y = 2x
meet. Note also that you expect two values for x and 2 corresponding y values, but the
quadratic in this case only gives one solution (because it is repeated)
Answer
a) Parallel lines so
2x + y = 180 because:
This will be y as these
are alternate angles
b) 6x + 2y – 100 = 360
Since angles in a quad
Sum to 360 degrees
Adding 100
6x + 2y = 460
Dividing by 2
3x + y = 230
c) Solve simultaneously
(c) Find the values of ‘x’ and ‘y’
2x + y = 180
3x + y = 230
Subtracting
X = 50 so y = 80
Solving Equations by Trial and Improvement
Exam Question:
Eg Use Trial and Improvement to Solve x3 + x2 = 200
Find your answer to 2 decimal places.
Working:
Guess
7
6
5
5.7
5.6
5.4
5.5
5.55
5.54
5.53
You must find the number
to 2 decimal place that is
nearest to 200.
Answer
392
252
150
217.683
206.976
186.624
196.625
201.7564
200.7231
199.6933
Show your results clearly
and show how you are
using intelligence to home
in on the solution rather
than just random guesssing
Method:
1)
Guess a number that might work
2)
Try it out – if the answer is too low make a second guess that is higher
- if the answer is too high make a second guess that is lower.
3)
Keep guessing, each time going half way between your best two results
Note: Your final answer is the guess with the correct number of
decimal places that is the closest fit. You must demonstrate
for sure that your answer is the best.
Method:
1) Guess a number that might work
2) Try it out – if the answer is too low make a second guess that is higher
- if the answer is too high make a second guess that is lower.
3) Keep guessing, each time going half way between your best two results
Note: Your final answer is the guess with the correct number of
decimal places that is the closest fit. You must demonstrate
for sure that your answer is the best.
Solve (to 2 decimal places)
1) x2 + 2x = 90
2) x2 + x = 80
4) x3 – x = 75
5) x3 + x2 = 100
Solve (to 2 decimal places)
1) x2 – x = 60
2) x2 – 3x = 100
3) x2 – 3x = 100
Proportion
Two things are in direct proportion if the value of the second can be found from
the value of the first by a multiplication.
In other words:
y  x  y  kx
(This means that the graph of y plotted against x would be a
straight line going through the origin)
(A quick check is to see if doubling one quantity would lead to an expected doubling in
the other quantity.
Eg If you travel for twice as long will the distance you travel double?)
The other types of direct proportion to watch out for are:
A is directly proportional to B
A  kB
A is directly proportional to B 2
A  kB
A is directly proportional to
the square root of B
2
Ak B
Eg The speed at which ball falls to the ground when dropped from a tall tower is directly
proportional to the square of the time it has been falling for. The ball reaches a speed of 12
m per second when it has been falling for 2 seconds.
a) How long will it take for the ball to reach a speed of 100 m per second?
b) When the ball has been falling for 20seconds how fast will it be travelling?
METHOD
Write down the
proportionality equation
with the constant k
Answer
v = speed t = time
v is directly proportional to the time squared so
When t = 2 v = 12
so
v = k × t2
12 = k × 22
12 = k × 4
k=3
Use the info in the
question to find a value
for k
Write down the formula
with k now known
v = 3 × t2
a) Now use the formula to answer the questions by solving the relevant equations
v = 100 what is t?
100 = 3 × t2 so t = 100
3
b) When t = 20s what is v
v = 3 × 202 v = 1200 m per second
If y and x are inversely proportional then we say:
1
k
y  y
x
x
But think of it like this:
y×x=k
A good way to think about inverse proportion is that as one thing doubles the other halves.
Eg If I run twice as fast will I get there in half the time? If so then it is likely to be inverse
proportional
Note that:
y is inversely proportional
to the square of x
1
k
y 2  y 2
x
x
But think of it as y × x2 = k
y is inversely proportional
to the square root of x
1
k
y
 y x
x
x
But think of it as y ×
=k
Eg If it take six people 8 hours to complete a job how many hours will it take 4 people?
Answer
Often proportionality questions mention the words inverse or direct proportion in the question,
but with this example you just need to recognise that this is likely to be an inverse proportion
situation. Twice as many people mean the job completing in half the time would seem
reasonable!
METHOD
So
p = number of people
Write down the
proportionality equation
with the constant k
h = number of hours
p is inversely proportional to h so
p×h =k
Use info from the question to find the value of the constant ‘k’
Use the info in the
question to find a value
for k
Write down the formula
with k now known
6 × 8 = k so k =48
Formula is
p × h = k = 48
So when p = 4 we need to solve for h
4 × h = 48 implies that h = 12 hours
Pythagoras
If you see a right angled triangle think Trigonometry and Pythagoras!
If you need to find side lengths and can’t see a way forwards sometime drawing in a
perpendicular to make a right angled triangle is helpful as you can then use pythagoras.
Eg Find the area of this isosceles triangle
13 cm
13 cm
13 cm
13 cm
h
10 cm
10 cm
Dropping a
perpendicular from the
top of the triangle makes
finding the height
obvious using pythagoras
132 = h2 + 52
169 = h2 + 52
169 = h2 + 25
144 = h2
12 = h can ignore -12!
Hence area =
12 x 5 = 60 cm2
Eg
Find the value of x.
Answer
If a right angled triangle then (x + 5)2 + (x – 2)2 = 102 (pythagoras theorem)
This is a quadratic so
make it equal to zero
and solve
x2 + 10x + 25 + x2 - 4x +4 = 100
2x2 + 6x + 29 = 100
2x2 + 6x – 71 = 0
Expand out each bracket
properly:
Remember (x + 5)2
(x + 5)(x + 5) = x2 + 10x + 25
The positive solution is 4.64 from the formula
Note: You can also use pythagoras to prove that a triangle is right angled.
It must be if the three sides fit the rule a2 + b2 = c2
Constructions