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b. a. c. e. a. d. c. b. Angle Rules. b. a. c. c. b. a. d. b. a. c. d. Complementary Angles add to 90 o The complement of 55 o is 35 o because these add to 90 o Supplementary Angles add to 180 o The supplement of 55 o is 125 o because these add to 180 o. C before S

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slide1

b

a

c

e

a

d

c

b

Angle Rules

slide2

b

a

c

c

b

a

d

slide3

b

a

c

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 S

90 before 180

slide14

1

180÷3=60

360

180

180×2

=360

360

2

360÷4=90

180×3

=540

3

360

540÷5=108

180×4

=720

720÷6=120

4

360

360

180×10

=1800

1800÷12=150

10

360

slide15

For ANY polygon

For regular polygons only

For regular polygons only

slide16

155

135

130

x

95

x

80

135

130

95

135

Degrees in the polygon:

Degrees in the polygon :

similar triangles1
Similar Triangles

If 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

25m

20m

x

4 m

It doesn’t matter which way round you make the fraction

BUT you must do the same for both sides

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

slide24

#11

x

x

x

x

x

lesson 6

Lesson 6

Circle Language and Angle at Centre

slide27

x

Base ‘s isos Δ, = radii

Base ‘s isos Δ, = radii

Sum of Δ = 180°

*

lesson 7

Lesson 7

Tangent is perpendicular to the radius and Angles on Same Arc are equal

angles on the same arc are equal
Angles on the same arc are equal

‘s On the same arch equal

cyclic quadrilaterals
Cyclic Quadrilaterals

A quadrilateral which has all four vertices on the circumference of a circle is called a

Cyclic quadrilateral

Rule 1:

slide40

Rule 2:

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

tangents
Tangents

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

similar triangles2
Similar Triangles

If 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

25m

20m

x

4 m

It doesn’t matter which way round you make the fraction

BUT you must do the same for both sides

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

slide51

#11

x

x

x

x

x

2006 exam
2006 exam

QUESTION ONE

The 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 TWO

The 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 exam1
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.

slide56

QUESTION THREE

The diagram shows the design for a gate.

AE = 85 cm

BE = 64 cm

CD = 90 cm

Triangles ABE and ACD are similar.

Calculate the height of the gate, AD.

slide57

QUESTION FOUR

The 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.

slide58

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.

slide59
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.