Angles on a straight line add to 180

180a b c

a

bc

d

e

ab

c

Angles round a point add to equal 360

360a b c d e

a b

c

d

Vertically Opposite Angles are Equal

a b

c d

a

b

c

Vert opp 's

sum add to 180

Angles in a triangle add to equal 180

180a b c

Base angles of an isosceles triangle are equal

Base 's isosc are

Exterior angle of a triangle equals the sum

of the opposite two interior angles

180

180

a b c

c d

a b c c

b

c c

d

a d

2intExt of sum opp s

a c

b

d

Complementary Angles add to 90o

The complement of 55o is 35o because these add to 90o

Supplementary Angles add to 180o

The supplement of 55o is 125o because these add to 180o

C before S90 before 180

35

x 35x

2aa 50a

20a

38

x

*

x

42 x

x

x42

75

*

123

67x

*

Rules of Parallel Lines

Corresponding angles of parallel

lines are equal

Alternate angles of parallel

lines are equal

Co-interior angles of parallel

lines add to 180

We can say 143 cointerior to 37x

37 corresponding to 37

or s on st ln with

y

x

143 Vertically opposite

or s on st ln with

z x

y

37x

y

z

x

57

27

27 Alt of lns ares° Ð =P

57 27 30

30 Alt of lns ares° Ð =P

So we can say 30 (Alt of lns are )x s Twice= ° Ð =P

57

x

65

65 , Corresp s lns area= ° Ð =P

65 57 180

180 65 57

58

sum add to 180

x

x

x

a

36

x

36

opp of a gram are

x

s

= °

Ð =P

42

72x

72 ,Alt s lns area= ° Ð =P

42 72 180

180 42 72

66

x

x

x

sum add to180

a

3 5x

4 10x

We make a statement

3 5 4 10 (Corresp s lns are )

15

x x

x

+ = - Ð =

=

P

No ofsides

Name NoOf

Degrees in

polygon

Each interior angle for regular

polygons(sides are equal)

Sum of exterior angles

3 Triangle

4 Quadrilateral

5 Pentagon

6 Hexagon

etc

12 Dodecagon

1

2

10

180

180×2=360180×3=540180×4=720

180×10=1800

180÷3=60

360÷4=90

540÷5=108

720÷6=120

1800÷12=150

360

360

360

360

360

360

3

4

Sum of interior anglesEach interior angle

Number of sides

Sum of the angles in a Polygon (No of sides -2) 180

Sum of interior anglesNumber of sides

Each interior angle

For regular polygons only

For ANY polygon

For regular polygons only

(5 2) 3 no of sides-2

Degrees in the polygon : Degrees in the polygon:

(6 2) 4 no of sides-2

95

135

x

13595

3 180 540

95 95 135 135 540

460 540

540 460

80

x

x

x

x

x

155 130

80

135130

4 180 720

15 130 80 130 135 720

630 720

720 630

90

x

x

x

x

A regular polygon has 9 sides what is the interior angle?

9 2 180 1260 sum of all angles

1260140 size of each interior angle

9

The sum of all the angles in a polygon is 2340 .If each interior

angle is 156 , how many sides does the polygon have?

234015 no of sides

156

Eg Interior angle is 150 . Find the number of sides.

Ext Angle 180 150

=30

360No of sides

Ext Angle

360

30

12

This is a regular Polygon

Similar Triangles

Similar TrianglesIf triangles are similar:Corresponding side lengths are in proportion. (One triangle is an enlargement of the other)Corresponding angles in the triangle are the same

25m20mx4 m

It doesn’t matter which way round you make the fraction BUT you must do the same for both sides

little little

big big

4

25 204

25205

x

x

x

It is sensible to start with the x so it is on the top

If the angles of two triangles are the same, they are similar triangles.

48 m

12 m

15 m

x

y

48 m

12 m

15 mx

Start with unknown on the top

48

15 1248 15

1260

x

x

x m

60 15

45

y

y m

20 m

2m

1.5m

4.5m

20 m

2m3m

l

2

20 32 20

31

133

l

l

l m

#11

x

x

x

x

x

Lesson 6

Circle Language and Angle at Centre

Equal Radii: Two radii in a circle always form an isosceles triangle

Isos , = radii

76

x

*

37

Base ‘s isos Δ, = radii

x

Base ‘s isos Δ, = radii

Sum of Δ = 180°

Angle at the centre is twice the angle at the circumference

a

a

aa

2a

2a

2a

2a

2at centre at circum

Angle on the circumference of a semicircle is a right angle

in semi-circle

Lesson 7

Tangent is perpendicular to the radius and Angles on

Same Arc are equal

Tangent is perpendicular to the radius

Tan radius

Angles on the same arc are equal

‘s On the same arch equal

35

62

x

y

18

x

y

*

59x

y

42

x

y85 38 x

y

z

*

66x

85

y

*

63x

y

z

Find unknowns and give reasons

*

43

57

y

x

z

Find unknowns and give reasons

*

Cyclic Quadrilaterals

Cyclic Quadrilaterals

ab

c

d

opp s cyclic quad

180

180

opposite angles of a cyclic

quadrilateral add to equal 180

opposite angles of a cyclic

quadrilateral are supplementary

a b

c d

A quadrilateral which has all four vertices on the circumference of a circle is called a Cyclic quadrilateralRule 1:

Rule 2:

The exterior angle of a cyclic quadrilateral is equal to the opposite interior angle

,ext cyclic quad

a

b

67x 95

x

140

x

y

*

37

x

y 34

x y

*

43

x

y

*

Find unknowns and give reasons

110

Find unknowns and give reasons

*

x

Tangents

Tan radius

When two tangents are drawn from a point to a circle, they are the same length

tangents

25w

x

yz

140

*

Similar TrianglesIf triangles are similar:Corresponding side lengths are in proportion. (One triangle is an enlargement of the other)Corresponding angles in the triangle are the same

25m20mx4 m

It doesn’t matter which way round you make the fraction BUT you must do the same for both sides

little little

big big

4

25 204

25205

x

x

x

It is sensible to start with the x so it is on the top

If the angles of two triangles are the same, they are similar triangles.

48 m

12 m

15 m

x

y

48 m

12 m

15 mx

Start with unknown on the top

48

15 1248 15

1260

x

x

x m

60 15

45

y

y m

20 m

2m

1.5m

4.5m

20 m

2m3m

l

2

20 32 20

31

133

l

l

l m

#11

x

x

x

x

x

Revision

Geometric reasoning revision

2006 examQUESTION ONEThe diagram shows part of a fence.AD and BC intersect at E.Angle AEB = 48°.Angle BCD = 73°.Calculate the size of angle CDE.

QUESTION TWOThe diagram shows part of another fence.LM = LN.KL is parallel to NM.LM is parallel to KN.Angle LNK = 54°.

Calculate the size of angle LMN.

2006 exam

The points A, B, C and D lie on a circle with centre O.Angle OAD = 55°.Angle DOC = 68°.Calculate the size of angle ABC.You must give a geometric reason for each step leading to your answer.

QUESTION THREEThe diagram shows the design for a gate.

AE = 85 cmBE = 64 cmCD = 90 cm Triangles ABE and ACD are similar. Calculate the height of the gate, AD.

QUESTION FOURThe diagram shows a design for part of a fence.GHIJK is a regular pentagon and EHGF is a trapezium.AB is parallel to CD.Calculate the size of angle EHG.You must give a geometric reason for each step leading to your answer.

QUESTION FIVE

The diagram shows another fence design.ACDG is a rectangle.Angle CBA = 110°.CG is parallel to DE.DA is parallel to EF.Calculate the size of angle DEF. You must give a geometric reason for each step leading to your answer.

• In the above diagram, the points A, B, D and E lie on a circle.• AE = BE = BC.• The lines BE and AD intersect at F.• Angle DCB = x°.• Find the size of angle AEB in terms of x.• You must give a geometric reason for each step leading to your

answer.