Download GCSE Mathematics – Revision

GCSE
Mathematics – Revision
Notes
Topic ‘Geometry and
Algebra’
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AQA – Geometry and Algebra
Contents
Proportionality ........................................................................................................................................ 3
Map Scales .............................................................................................................................................. 8
Trial and improvement ......................................................................................................................... 10
Quadratic Formula ................................................................................................................................ 13
Sketching Graphs ..................................................................................... Error! Bookmark not defined.
6. Transformations ................................................................................... Error! Bookmark not defined.
7. Area and Perimeter .............................................................................. Error! Bookmark not defined.
8. Points, Lines and Angles and Triangles ................................................ Error! Bookmark not defined.
9. Circles ................................................................................................... Error! Bookmark not defined.
10. Enlargements ..................................................................................... Error! Bookmark not defined.
11. Trigonometry ..................................................................................... Error! Bookmark not defined.
Bisector of a given Angle...................................................................... Error! Bookmark not defined.
Division of a line segment into equal parts.......................................... Error! Bookmark not defined.
Line perpendicular to a given line , passing through a point on ...... Error! Bookmark not defined.
Line parallel to a given line, through a point not on ......................... Error! Bookmark not defined.
14. Volume ............................................................................................... Error! Bookmark not defined.
15. Cylinders............................................................................................. Error! Bookmark not defined.
16. Spheres and Cones ............................................................................. Error! Bookmark not defined.
1. Proportionality
Proportionality describes the relationship between two variables where the relationship
between the two variables can be expressed in terms of a proportionality constant.
Direct proportionality
The variable y is said to be directly proportional to the variable x if their relationship can be
described as:
y=cx
Where c is a constant.
The above can be re-written as
, i.e. the ratio y/x is constant.
Direct proportionality can be observed when plotting y against x.
y = 3x
13
12
11
10
9
y
8
7
6
5
4
3
2
1
0
0
1
2
3
4
x
As x increases y increases.
The rate of the increase is the proportionality constant, 3.
Inverse proportionality
The variable y is said to be inversely proportional to the variable x if their relationship can
be described as
This can be written as yx = c, i.e. the product yx is constant.
If y is inversely proportional to x then y decreases as x increases.
Proportion to powers
Direct and inverse proportionality can also be to the powers of x.
For example we could write:
y is directly proportional to x
y is directly proportional to
y is directly proportional to
√
y is directly proportional to
and
y is inversely proportional to x
y is inversely proportional to
y is inversely proportional to
√
y is inversely proportional to
Examples

Question 1.1
y is directly proportional to x.
We know that if y = 28 then x = 7.
Write the equation connecting y and x.
Find y when x is 12.
Answer
y is directly proportional to x:
Also
Therefore
So that we can write
And if x = 12 then

Question 1.2
y is inversely proportional to x.
We know that if x = 3 then y = 8.
Find the value of y when x = 5.
Answer
y is inversely proportional to x:
Determine c by using known values:
therefore
So
Finally, when x = 5:
therefore

y = 4.8
Question 1.3
W is inversely proportional to x.
If W = 6 then x = 20.
Find W for x = 24.
Answer
W is inversely proportional to x:
Again, use known values given:
therefore
So that
and for x = 24


Question 1.4
The variable M is directly proportional to
It is known that if r = 5 then M = 200.
Determine the value of M when r = 8.
Find r when M = 3125.
.
=5
Answer
M is directly proportional to
:
Again, if:
we have that
So we can write the equation
Now, if r = 8 we have:
Similarly, if M = 3125:
and

Question 1.5
The cost of a bucket of ice cream, C, depends on its weight w.
The price of ice cream, p, is 16 euro/kg.
Describe the relationship between C, w and p.
Write the equation that describes this relation.
What would you spend if you wanted to buy a bucket of 8kg of ice cream?
Answer
The cost of a bucket of ice cream C is directly proportional to the weight w.
The constant of proportionality is the price of 1 kg of ice cream, namely 16 euro/kg.
We then have:
The cost of a bucket og 8kg would then be:

Question 1.6
The circumference C of a circle is given by the formula
Where d is the diameter of the circle.
Describe in words the proportionality relation, specify the constant of proportionality.
Find the diameter of a circle with circumference C=35.
Answer
The circumference of a circle is directly proportional to the diameter d.
The constant of proportionality is , the constant ratio between the circumference and the
diameter of a circle.
If the circumference C is 35 then:
2. Map Scales
Map scales are ratios used to represent distances. They can be used as an alternative representation
of distances when the original set of distances cannot be conveniently displayed.
When drawing a map of a large area the original distances cannot be represented as they are.
We use map scales to relate original distances to new, more convenient distances so that it is
possible to draw a map.
A certain unit in our map represents a certain unit of land or sea. For example we can decide that a
distance of 1 meter = 100 cm corresponds to 1cm in our map.
We would write that as 1:100, meaning that 1 unit in the map represents 100 units of territory.
Examples

Question 2.1
The scale on a map is 1:250000.
What is the actual distance corresponding to 1cm in the map?
Express the answer in kilometres.
Answer
1 unit in the map represents 250000 units.
1cm in the map represents 250000cm.
250000cm = 2.5km.

Question 2.2
The scale on a map is 1:500000.
Two towns are 8cm apart on the map. What is the actual distance between the towns in
kilometres?
Answer
1cm on the map corresponds to 500000cm.
8cm on the map corresponds to 8*500000cm = 4000000cm.
4000000cm = 40km

Question 2.3
The scale on a map is 1:300000. What is the actual distance in kilometers represented by
11cm in the map.
Answer
1cm on the map corresponds to 300000cm.
11cm on the map corresponds to 11*300000cm = 3300000cm.
3300000cm = 33km.

Question 2.4
The scale on a map is 1:200000. What distance in the map represents 5km?
Answer
5km = 500000cm.
Also:
Therefore:
And

Question 2.5
In a map 1 inch represents 4 miles.
A road on the map measures 6 inches to the nearest inch.
What is the shortest possible distance of the road?
Answer
The road measures 6 inches to the nearest inch. This implies that the shortest measure could
be 5.5 inches before rounding up.
1 inch represents 4 miles; therefore 5.5 inches represent 5.5*4 miles = 22 miles.
3. Trial and improvement
Trial and improvement involves testing an unknown variable with different values; the value that
produces the result closest to the solution is selected.
When solving by trial and improvement the first value you test should be one that you think is close
to the answer. Using this as a starting point you then test different values until you find the one that
is closest.
Examples

Question 3.1
Use trial and improvement to find a solution to the equation
Give your solution to 1 decimal place.
Answer
Guess a value for x and evaluate
We want the result to be close to 25.
x
2
3
2.5
2.7
2.8
2.9
+x
10
30
18.125
22.38
24.75
27.29
.
Comment
Result is too small, try with a bigger number
Too big, try a number between 2 and 3
Too small, try a larger value
Still too small, try a larger value
Quite close to 25. Try next value, then choose the best
Too big. Previous result was closer to 25
To 1 decimal place the best solution to the equation is given by

Question 3.2
Use trial and improvement to find a solution to the equation
Give your answer to 1 decimal place.
Answer
Re-arrange the expression as
Again guess values for x, then improve.
We want to find the value of x that yield results closest to 26.
– 5x
3
4
3.5
3.6
12
44
25.37
28.65
Comment
Too small. Try larger value
Too big. Try a value between 3 and 4
Still small, but quite close to 26. Try next value and compare
Result too big. Previous value was closer to 26
To 1 decimal place the best solution to the equation is given by

Question 3.3
The perimeter of a square is given by the formula
Where l is the length of the side of the square.
If the perimeter of a square is P = 49, find by trial and improvement the value of l when l is a
whole number (i.e. no decimals).
Answer
It is known that 4l = P = 49.
We want the whole number that is closest to the solution.
Comment
15
14
13
12
60
56
52
48
Result too large. Try a smaller value
Still too large. Try smaller again
Still large but close to 49. Try next value and compare
Closer to 49. Best solution
Our answer is then l = 12.

Question 3.4
The area A of a circle is given by the formula
Where r is the radius of the circle.
Find r to 1 decimal place when A = 786.
Answer
Write
Then again look for the value for which the result is closest to 786.
Comment
14
15
16
15.9
15.8
615.75
706.85
804.25
794.23
784.27
Result is small. Try larger
Still small. Try larger again
Result bigger than 786. Try slightly smaller
Closer to 786. Check next smaller value
Best solution
To 1 decimal place the solution is r = 15.8

Question 3.5
Find the solution to the equation
Correct to 1 decimal place by trial and improvement.
Answer
First we re-arrange the equation:
Again we look for the value of w for which
decimal place.
is closest to 145 when w has 1
Comment
4
3
3.3
3.2
352
111
164.07
144.79
Result is too large. Try smaller
Result too small. Try a value between 3 and 4
Result bigger than 145. Try slightly smaller
Closest to 145. Best solution
The solution to 1 decimal place is w = 3.2
4. Quadratic Formula
A quadratic equation has standard form.
The numbers a, b and c are the coefficients of the equation, while x is said to be a variable.
Note that in the standard form all non-zero terms appear in one side, and the other side is
always zero.
The solutions of an equation are the values of the variable x for which the equation is true.
Therefore, to find the solutions of a quadratic equation we must find the values for which
the expression
is equal to 0.
If solutions of the equations exist we usually find them by factorization. However, when
factorization is not an option we can solve the equation by using the quadratic formula.
√
The quantity
is called the discriminant of the quadratic equation, and is usually
represented by the symbol Δ (greek uppercase delta).
We can rewrite the above as:
√
Where
The discriminant Δ can be positive, negative or zero.
 If Δ > 0
There are two distinct roots (solutions)
 If Δ = 0
There is 1 root
 If Δ < 0
There are no real roots
Examples

Question 4.1
In the equation
Indicate a, b and c.
Determine Δ and indicate whether the equation has 1, 2 or no solutions.
Answer
We have:
a=1
b=5
c=3
And the determinant is
The determinant is positive; therefore there are two distinct roots.

Question 4.2
In the equation
Indicate a, b and c.
Determine Δ and indicate whether the equation has 1, 2 or no solutions.
Answer
We have:
a = 12
b=-4
c=2
And the determinant is
(
)
The determinant is negative; therefore there are no real roots.

Question 4.3
In the equation
Indicate a, b and c.
Determine Δ and indicate whether the equation has 1, 2 or no solutions.
Answer
We have:
a=2
b=-4
c=2
And the determinant is
(
)
The determinant is zero, therefore there is 1 real root.

Question 4.4
In the equation
Indicate a, b and c.
Determine Δ and indicate whether the equation has 1, 2 or no solutions.
Answer
We have:
a=-3
b=3
c=
And the determinant is
(
) (
)
The determinant is zero; therefore there is 1 real root.

Question 4.5
In the equation
Indicate a, b and c.
Determine Δ and indicate whether the equation has 1, 2 or no solutions.
Answer
Note that the term containing x is not present.
To say that there is no x is the same of saying that its coefficient is zero (0x).
We have:
a = 26
b=0
c=
And the determinant is
(
(
) (
)
)
The determinant is positive; therefore there are two distinct roots.

Question 4.6
In the equation
Indicate a, b and c.
Determine Δ and indicate whether the equation has 1, 2 or no solutions.
Answer
Here the constant term c does not appear. We write that as c = 0 .
We have:
a=-1
b=-5
c=
And the determinant is
(
)
(
) ( )
The determinant is positive; therefore there are two distinct roots.
We now use the quadratic formula to find roots (if any) of quadratic equations

Question 4.7
Find the roots (if any) of the equation
Answer
Here we have:
a=
b=-5
c=
Calculate the determinant
(
)
The determinant is zero so we know that there is one root.
√
To check if the solution is correct, just verify that it satisfies the equation:
The solution x = 2 does indeed satisfy the equation.

Question 4.8
Find the roots of the equation given in Question 4.6
Answer
The equation given in question 4.6 is
The determinant of the equation, calculated above, is Δ = 25 and there are two
solutions.
We call the solutions
and
.
√
( )
√
So we get
and
Check that
and
.
satisfy the equation:
For
(
)
(
)
For
( )

( )
Question 4.9
Find the roots (if any) of the equation
Answer
First of all we put the equation in standard form, so that all non-zero terms appear
in one side of the equation:
Then we continue as before and calculate the discriminant:
a=1
b=
c=
(
)
( ) (
(
)
)
The determinant is positive, therefore there are two distinct roots
√
So we get
and
Check that
and
(
)
and
√
.
satisfy the equation:
For
0=0
For
(
)
(
0.0676 + 3.9 4 = 0
0=0

)
Question 4.10
Solve the equation
Give your answer in the form
Answer
Again, first re-arrange into standard form, then calculate the discriminant:
a=1
b=
c=
(
)
.
The determinant is positive; therefore there are two distinct roots.
We are asked to present the answer in the form
(
√
)
√
√
We can simplify the above expression by noting that
√
√
√
√
And dividing across by 2
√
So our final answer is
√
√