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Shape Workshop:
Notes and Questions
Topics:
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Angles – in parallel lines and polygons
Area and perimeter –including circles and compound area
Basic Trigonometry
Bearings
Circle Theorems
Constructions
Plans and Elevations
Pythagoras’ Theorem
Similar Shapes
Transformations
Volume and Surface Area
Angles
Revision Notes
Angle Facts
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Acute angles – less than 90o.
Right angle – equal to 90o.
Obtuse angles – greater than 90o but less than 180o.
Reflex angles – greater than 180o but less than 360o.
The angles on a straight line equal 180o.
The angles around a point or in a full turn equal 360o.
Angles in Parallel Lines
Vertically opposite angles –
Vertically opposite angles are equal.
Corresponding angles –
Corresponding angles are equal.
Alternate angles –
Alternate angles are equal.
Interior angles –
Interior angles add up to 180o.
e.g. Find the missing angles giving reasons for your answers.
a = 45o because it is on a straight line with 135o
b = 45o because it is alternate to angle a and interior to 135 o
c = 135o because it is corresponding with 135o and on a straight line with angle b
Interior and exterior angles in Polygons
To find what the interior angles of a polygon add up to, use the following formula -
𝐀𝐧𝐠𝐥𝐞 𝐬𝐮𝐦 𝐨𝐟 𝐚 𝐏𝐨𝐥𝐲𝐠𝐨𝐧 = (𝑵𝒐 𝒐𝒇 𝒔𝒊𝒅𝒆𝒔 − 𝟐) × 𝟏𝟖𝟎
e.g.
The angle sum of a Hexagon(6 sided shape) would be –
𝐀𝐧𝐠𝐥𝐞 𝐬𝐮𝐦 𝐨𝐟 𝐚 𝐏𝐨𝐥𝐲𝐠𝐨𝐧 = (𝟔 − 𝟐) × 𝟏𝟖𝟎
= 𝟒 × 𝟏𝟖𝟎
= 𝟕𝟐𝟎𝒐
You can use the angle sum of a polygon to find unknown angles in that polygon.
The exterior angle of a polygon and its corresponding interior angle always add up to 180°
(because they make a straight line).
e.g. Find the size of the interior and exterior angles in this regular polygon.
Interior angle – Angle sum = (5 – 2) X 180 = 540o
Each interior angle therefore =
540
5
= 108𝑜
Exterior angle = 180o – 108o = 72o
For any polygon, regardless of the number of sides it has,
the sum of its exterior angles is always 360°.
Calculating the number of sides in a regular polygon, given the interior angle
To calculate the number of sides in a regular polygon you need to use the following formula-
𝟑𝟔𝟎
𝑵 𝒐𝒇 𝒔𝒊𝒅𝒆𝒔 =
𝒔𝒊𝒛𝒆 𝒐𝒇 𝒆𝒙𝒕𝒆𝒓𝒊𝒐𝒓 𝒂𝒏𝒈𝒍𝒆
𝒐
e.g.
The interior angles of a regular polygon are each 120°. Calculate the number of sides.
First you need to work out the exterior angles of this polygon.
Remember that the sum of an interior angle and the corresponding exterior angle is 180 o.
Therefore, for this example the exterior angle will be 180o – 120o = 60o.
We know that the exterior angles of any polygon add up to 360o.
𝑵𝒐 𝒐𝒇 𝒔𝒊𝒅𝒆𝒔 =
𝟑𝟔𝟎
= 𝟔 𝒔𝒊𝒅𝒆𝒔
𝟔𝟎
Exam Questions
ABC, PQR and AQD are straight lines.
ABC is parallel to PQR.
Angle BAQ = 35°
Angle BQA = 90°
Work out the size of the angle marked x.
Give reasons for each stage of your working.
x=......................°
(Total for Question is 4 marks)
The diagram shows a square and 4 regular pentagons.
Work out the size of the angle marked x.
.......................................................................................................................................
(Total for Question is 3 marks)
ABC is parallel to DEF.
EBP is a straight line.
AB = EB.
Angle PBC = 40°.
Angle AED = x°.
Work out the value of x.
Give a reason for each stage of your working.
(Total for Question is 5 marks)
The interior angle of a regular polygon is 160°.
Diagram NOT accurately drawn
(i) Write down the size of an exterior angle of the polygon.
......................°
(ii) Work out the number of sides of the polygon.
....................................................................................................................................
(Total for Question is 3 marks)
CDEF is a straight line.
AB is parallel to CF.
DE = AE.
Work out the size of the angle marked x.
You must give reasons for your answer.
(Total for Question is 4 marks)
Area and Perimeter
Perimeter is the length around the shape
Exam Questions
Circles
Exam Questions
Basic Trigonometry
hyp = hypotenuse(side opposite to the right angle)
opp = opposite(side opposite to the non-right angle given in the question or opposite to the
angle you are trying to find).
adj = adjacent(side next to the non-right angle given in the question or opposite to the angle
you are trying to find).
Using Basic Trigonometry to find the length of a missing side
e.g.
opp
a
hyp
14.6cm
37o
adj
Label the sides.
Select the rule. To do this you need to look at what side you are trying to find. In this
example it is the opposite side. You then need to see which other side you know the length
of. In this example you know that the hypotenuse is 14.6cm. You need to select the rule
that contains these two sides. For this example therefore you need to use –
You’re trying to find the opposite so cover that up in the triangle and write down what you
can see.
Opposite = sinθ x hyp
Fill in the known values
Opposite = sin(37) x 14.6 =
Using Basic Trigonometry to find a missing angle
e.g.
12.2cm
hyp
opp
x
7.8cm
adj
Label the sides.
Select the rule. To do this you need to look at what sides you know the length of. In this
example you know that the adjacent is 7.8cm and hypotenuse is 12.2cm. You need to select
the rule that contains these two sides. For this example therefore you need to use –
You’re trying to find the angle so cover up the cosθ and write down what you can see
𝐜𝐨𝐬 𝜽 =
𝒂𝒅𝒋
𝒉𝒚𝒑
𝒄𝒐𝒔𝜽 = (
𝟕. 𝟖
)
𝟏𝟐. 𝟐
Fill in the known values
𝜽 = 𝒄𝒐𝒔−𝟏 (
𝟕. 𝟖
) = 𝟓𝟎. 𝟐𝟔𝒐 (𝟐𝒅𝒑)
𝟏𝟐. 𝟐
Key Points
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Basic Trigonometry can only be used on right angle triangles. If the question does
not contain a right angle triangle, then you cannot use Basic Trigonometry to answer
it!
Basic Trigonometry can be used to find a missing side if you are given another side
and at least one of the angles.
Basic Trigonometry can also be used to find a missing angle if you are given at least
two of the sides.
Remember if you are finding a missing side and are given the other two sides, you
should use Pythagoras’ Theorem to find the answer.
Don’t forget to include units on your answer.
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Be careful when using your calculator. Make sure you enter the values in the correct
order depending on your calculator.
Exam Questions
Calculate the value of x.
Give your answer correct to 3 significant figures.
.......................................................................................................................................
(Total for Question is 3 marks)
ABCD is a parallelogram.
DC = 5 cm
Angle ADB = 36°
Calculate the length of AD.
Give your answer correct to 3 significant figures.
.......................................................................................................................................
(Total for Question is 4 marks)
* The diagram shows a ladder leaning against a vertical wall.
The ladder stands on horizontal ground.
The length of the ladder is 6 m.
The bottom of the ladder is 2.25 m from the bottom of the wall.
A ladder is safe to use when the angle marked y is about 75°.
Is the ladder safe to use?
You must show all your working.
(Total for Question is 3 marks)
Bearings
Bearings
Exam Questions
Circle Theorem
x
Theorem 1: Angle at the Centre
Fact: The angle at the centre is twice as big as
the angle at the circumference made by the
same arc or chord
.
How to spot it: Start with two points (could be the
ends of a chord). If you go point-centre-point, the
2x
angle you make will be twice as big as if you go
point-circumference-point
Theorem 2: Angles in a Semi-Circle
Fact: The angle made at the circumference in a semi circle
0
is a right angle (90 )
How to spot it:
180˚
.
Look for a triangle whose base is the diameter of the circle
(a line going through the centre). The angle at the
circumference in this triangle will always be a right angle
Note: This theorem is just a special case of Theorem 1,
because the angle at the centre when you have a straight
0
line is 180 , so the angle at the circumference must be half
of this!
Theorem 3: Angles in the Same Segment
Fact: Angles in the same segment of a circle are equal to
each other
How to spot it:
Start with two points (could be the ends of a chord).
If you go point-circumference-point, the angle you make will
be exactly the same as if you go point-circumference-point
again…so long as you stay in the same segment of the circle!
a
a = b
b
Theorem 4: Cyclic Quadrilateral
Fact: The opposite angles in a cyclic quadrilateral add up to 180
0
How to spot it: Look for a four-sided shape with each of the
corners on the circumference.
The opposite angles in this shape will always add up to 180
0
Note: Just like any other quadrilateral, the sum of all the interior
0
angles is still 360
a + b = 180
c + d = 180
0
0
Theorem 5: Tangent
Fact: The angle made by a tangent and the radius is a right0
angle (90 )
.
How to spot it:
A tangent is a straight line that only touches a circle in one
place. If you draw a line from that one place to the centre
of a circle, then the angle you form is always a right-angle!
Theorem 6: Alternate Segment Theorem
x
Fact: The angle between a tangent and a chord at the point of
contact is equal to the angle made by that chord in the other
y
segment of the circle.
How to spot it:
Look for a tangent and a chord meeting at the same point. The
y
x
angle they make is exactly the same as the angle at the
circumference made by that chord – imagine the chord is the
base of a triangle, and the angle you want is at the top of the
triangle!
Theorem 7: Two Tangents
Fact: From any point outside the circle, you can only
draw two tangents to the circle, and these tangents will
be equal in length.
How to spot it: Look for where the tangents to a circle
meet. The lengths between where they touch the circle
and the point at which they meet will always be the
same
Note: More often than not, this theorem leads to some
isosceles triangles, so be on the look out!
PA = PB
Exam Questions
Constructions
Exam Question
Plans and Elevations
Exam Questions
Pythagoras’ Theorem
Revision Notes
Using Pythagoras’ Theorem to find the hypotenuse
e.g.
c
a
3.2cm
4.4cm
b
First label the sides ensuring the hypotenuse is labelled ‘c’.
Then substitute the values from the question into Pythagoras’ Theorem.
𝑎2 + 𝑏 2 = 𝑐 2
3.22 + 4.42 = 𝑐 2
10.24 + 19.36 = 𝑐 2
𝑐 2 = 29.6
𝑐 = √29.6 = 5.44𝑐𝑚(2𝑑𝑝)
Using Pythagoras’ Theorem to find a short side
e.g.
c
a
13.5cm
7.8cm
b
Again, first label the sides ensuring the hypotenuse is labelled ‘c’.
Then substitute the values from the question into Pythagoras’ Theorem.
𝑎2 + 𝑏 2 = 𝑐 2
𝑎2 + 7.82 = 13.52
𝑎2 + 60.84 = 182.25
𝑎2 = 182.25 − 60.84 = 121.41
𝑎 = √121.41 = 11.02𝑐𝑚(2𝑑𝑝)
Key points
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Pythagoras’ Theorem can only be used on right angle triangles. So if the question
does not contain a right angle triangle, then you cannot use Pythagoras’ Theorem to
answer it!
Pythagoras’ Theorem can only be used to find the length of a missing side when you
are given the lengths of the other two sides. If you are asked to find an angle then
you cannot use Pythagoras’ Theorem to answer it!
The hypotenuse(side opposite to the right angle) should always be labelled ‘c’. It
does not matter which way round the others are labelled ‘a’ and ‘b’.
Remember to include the correct units on your answer.
Exam Questions
.
Calculate the length of AB.
Give your answer correct to 1 decimal place.
...............................................................................................................................
(Total for Question is 3 marks)
XYZ is a right-angled triangle.
Calculate the length of XZ.
Give your answer correct to 3 significant figures.
.......................................................................................................................................
(Total for Question is 3 marks).
ABC is a right-angled triangle.
AC = 6 cm
AB = 13 cm
(a) Work out the length of BC.
Give your answer correct to 3 significant figures.
.....................................................................................................................................
(Total for Question is 3 marks)
ABCD is a trapezium.
AD = 10 cm
AB = 9 cm
DC = 3 cm
Angle ABC = angle BCD = 90°
Calculate the length of AC.
Give your answer correct to 3 significant figures.
.......................................................................................................................................
(Total for Question is 5 marks)
Similar Shapes
Exam Question
Transformations
Revision Notes
There are four types of transformation :1. Reflection
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To describe a reflection you must state the following things :
Reflection
Mirror line (as an equation)
e.g ‘reflection, in the line x = 3’ would earn 2 marks if it was a correct description
2. Rotation
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To describe a rotation you must state the following things :
Rotation
Angle and direction (e.g. 90° clockwise)
Centre of rotation (as a coordinate)
e.g. ‘rotation, 90° anti-clockwise about (0, 0)’ would earn 3 marks if it was a correct
description
3. Translation
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To describe a translation you must state the following things :
Translation
Distance and direction (using words or vectors)
−𝟑
e.g. ‘translation, 3 left, 2 up’ or ‘translation ( )’ would earn 2 marks if it was a correct
𝟐
description
4. Enlargement
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To describe an enlargement you must state the following things :
Enlargement
Scale factor (e.g. ‘scale factor 2’ NOT ‘multiply by 2’ or ‘it doubles’)
Centre of enlargement (as a coordinate)
e.g ‘enlargement, scale factor 2, centre (1, 2)’ would earn 3 marks if it was a correct
description
Examples
1.
Reflection in the line x = 1
2.
From R to S, rotation 90° clockwise
about (-3, 2)
3.
From A to B, translation 7 right, 3 down OR
𝟕
translation( )
−𝟑
4.
From A to B, enlargement, scale factor 3,
centre (2, 1)
Key Points
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You must only describe ONE transformation, e.g. ‘enlargement scale factor 2 and
then translated 4 right, 1 up’ would earn ZERO marks!! ‘enlargement scale factor 2’
would earn 2/3 marks
Use tracing paper to help with describing rotations
Learn basic equations of lines e.g x = 2, y = -1, y = x for describing reflections
Draw in the ray lines (example 4) to help find the centre of enlargement
You do not have to use vectors to describe translations, use words if you are not
confident with vectors
You need to be able to understand vectors for drawing translations
Exam Questions
1
(3 marks)
(2 marks)
Volume and Surface Area
Exam Questions