Download Geo reasoning

Angles on a straight line add to 180
a  b  c  180
b
a
c
Angles round a point add to equal 360
e
a  b  c  d  e  360
a
d
c
b
Vertically Opposite Angles are Equal
ab
cd
c
b
a
d
Vert opp 's 
Angles in a triangle add to equal 180
b
a  b  c  180
a
 sum  add to 180
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
b
a
c
d
a  b  c  180
c  d  180
a  b  c  c  d
c c
a  b  d
Ext of   sum opp 2int s
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
35
x
x
35
38
2a
a a  50
a  20
x
*
x
x
42
75
42
x
*
x
123
x
*
67
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
37
x
y
z
We can say x  143 cointerior to 37
y  37 corresponding to 37
or s on st ln with x
z  143 Vertically opposite x
or s on st ln with y
57  27  30
x
57
30° AltÐsof P lns are =
27
27° AltÐsof P lns are =
So we can say x = 30° (AltÐsof P lns are = ) Twice
x
a
65
a = 65°, CorrespÐs P lns are =
57
65  57  x  180
x  180  65  57
 x  58
sumadd to 180
x
x = 36°
oppÐs of a P gram are =
36
x
42
72
a
a = 72° , AltÐs P lns are =
42  72  x  180
x  180  42  72
x  66
sum  add to180
4x 10
3x  5
We make a statement
3x + 5 = 4 x - 10 (CorrespÐs P lns are = )
x = 15
No
Name
of
sides
No
Of
Degrees
in
polygon
3
Triangle
1
180
4
Quadrilateral
2
180×2
=360
5
Pentagon
3
180×3
=540
6
Hexagon
4
180×4
=720
Each interior angle Sum of exterior
for regular
angles
polygons
(sides are equal)
180÷3=60
360÷4=90
360
540÷5=108
360
720÷6=120
360
etc
12
360
360
Dodecagon
10
180×10
=1800
1800÷12=150
360
For ANY polygon
Sum of the angles in a Polygon  (No of sides -2) 180
For regular polygons only
Sum of interior angles
Each interior angle 
Number of sides
For regular polygons only
Sum of interior angles
Number of sides 
Each interior angle
155
135
x
95
95
135
Degrees in the polygon :
130
x
80
130
135
Degrees in the polygon:
(5  2)  3  no of sides-2
(6  2)  4  no of sides-2
3180  540
4 180  720
 95  95  135  135  x  540
 460  x  540
 x  540  460
 x  80
15  130  80  130  135  x  720
 630  x  720
 x  720  630
 x  90
A regular polygon has 9 sides what is the interior angle?
9  2 180  1260  sum of all angles
1260
 140  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?
2340
 15  no of sides
156
Eg Interior angle is 150. Find the number of sides.
Ext Angle  180  150
=30
 No of sides 

360
Ext Angle
360
30
 12
This is a regular Polygon
Similar Triangles
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
4m
x
20m
25m
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
little little

big
big
x
4

25 20
4
x   25
20
x5
If the angles of two triangles are the same,
they are similar triangles.
48 m
12 m
y
15 m
x
48 m
12 m
x
Start with unknown on the top
x 48

15 12
48  15
x
12
x  60m
15 m
 y  60  15
y  45m
#11
20 m
1.5m
4.5m
x
2m
20 m
lx
3m
2m
xl
2

20 3
2  20
xl 

3
xl  13 1 m
3
Lesson 6
Circle Language and Angle at
Centre
Equal Radii: Two radii in a circle always form an
isosceles triangle
Isos
, = radii
x
76
x
37
Base
‘s isos Δ, = radii
Base
‘s isos Δ, = radii
Sum of Δ = 180°
*
Angle at the centre is twice the angle at the
circumference
a
a
2a
2a
2a
2a
a
a
at centre  2  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
y
62
x
59
x
y
y
18
x
*
z
x
42
85
y
38
x
y
*
66
x
85
y
*
Find unknowns and give reasons
y
z
63
x
*
Find unknowns and give reasons
57
43
z
x
*
y
Cyclic Quadrilaterals
Cyclic Quadrilaterals
A quadrilateral which has all four vertices on the circumference of a circle is called a
Cyclic quadrilateral
Rule 1:
opps cyclic quad
c
b
a
d
a  b  180
c  d  180
opposite angles of a cyclic
quadrilateral add to equal 180
opposite angles of a cyclic
quadrilateral are supplementary
Rule 2:
The exterior angle of a
cyclic quadrilateral is
equal to the opposite
interior angle
a
ext , cyclic quad
b
x
95
67
x
x
y
140
*
y
34
37
x
x
*
y
Find unknowns and give reasons
43
y
x
*
Find unknowns and give reasons
x
110
*
Tangents
Tan  radius
When two tangents are
drawn from a point to a
circle, they are the same
length
 tangents
y
z
140
x
w
*
25
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
4m
x
20m
25m
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
little little

big
big
x
4

25 20
4
x   25
20
x5
If the angles of two triangles are the same,
they are similar triangles.
48 m
12 m
y
15 m
x
48 m
12 m
x
Start with unknown on the top
x 48

15 12
48  15
x
12
x  60m
15 m
 y  60  15
y  45m
#11
20 m
1.5m
4.5m
x
2m
20 m
lx
3m
2m
xl
2

20 3
2  20
xl 

3
xl  13 1 m
3
Revision
Geometric reasoning revision
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 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 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.
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.
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.