Ppt for geometry

Pythagoras’ theorem is used when you know two sides in a right-angled
triangle and you want to find the length of the third side.
a
b
c
The longest side in a right-angled triangle
is called the hypotenuse.
Pythagoras discovered that….

The sum of the areas of the squares on the two shorter sides
is equal to the area of the square on the hypotenuse.
a
b
c
a2
 b2
 c2

Example
1 Calculate the value of x.
6 cm
2 cm
x
22
 62
 x2
4  36  x2
40  x2
x  40
x  6.32 cm (to 3 s.f.)

Example
10 cm
5 cm
x
52
102
 x2
25 100  x2
125  x2
x  125
x  11.2 cm (to 3 s.f.)

Example
9 cm
4 cm
x
x2
 42
 92
x2
16  81
x2
 65
x  65
x  8.06 cm (to 3 s.f.)

Example
7.6 cm
3.2 cm
x
x2
 3.22
 7.62
x2
10.24  57.76
x2
 47.52
x  47.52
x  6.89 cm (to 3 s.f.)

Example
8 cm
x
x
x2
 x2
 82
2x2
 64
x2
 32
x  32
x  5.66 cm (to 3 s.f.)

Example
6 The equilateral triangle has sides of length 6 cm.
Calculate a the height of the triangle
b the area of the triangle. 6 cm h
h2
 32
 62
h2
 9  36
h2
 27
h  5.196...
h  5.20 cm (to 3 s.f.)
3 cm
Area 
1
2
 base  height

1
2
 6  5.196...
 15.6 cm2
(to 3 s.f.)
a
b

x
y
2
-2
2
-2
-4
4
4
-4 0
Example
7 Find the length of the line joining (−4, 4) and (3, 1).
3
7
x
32
 72
 x2
9  49  x2
58  x2
x  58
x  7.62 (to 3 s.f.)

Pythagoras in 3-D
To find the length of
AG you need to look
at triangle AGC.
A B
C
E
G
H
D
F
7 cm
6 cm
5 cm
A B
C
E
G
H
D
F
7 cm
6 cm
5 cm
A C
G
5 cm
First you need to
calculate AC using
Pythagoras on
triangle ABC.
AC2
 72
 62
AC2
 85
AC  85
85
Now use
Pythagoras on
triangle ACG.
AG2
 AC2
CG2
AG2
 110
AG2
 85  25
AG  10.5 cm (to 3 s.f.)

It is useful to remember that AG can be calculated directly using:
A B
C
E
G
H
D
F
7 cm
6 cm
5 cm
AG2
 72
 62
 52
AG2
 49  36  25
AG2
 110
AG  10.5 cm (to 3 s.f.)

Names of angles
ACUTE angles
RIGHT angles
OBTUSE angles
REFLEX angles
angles between 0o
and 90o
angles of 90o
angles between 90o
and 180o
angles greater than180o

Names of triangles
equilateral
triangle
isosceles
triangle
right-angled
isosceles
triangle
right-angled
triangle
scalene
triangle

Vertically opposite angles
a
b
Angles at a point
a
b
c
Angles on a straight line
a b
Angle properties
a  b  180o
a  b  c  360o
a  b

Angle properties
Exterior angles of a
triangle
a
c
b
Angles in a quadrilateral
c
a
b
d
Angles in a triangle
a
b c
a  b  c  180o
a  b  c a  b  c  d  360o

Parallel and perpendicular lines
Parallel lines Perpendicular lines

Angle properties of parallel lines
Alternate angles
a
b
Corresponding angles
a
b
Interior angles
a
b
a  b a  b a  b  180o

Examples
1 Calculate the size of each lettered angle.
a
b
c
127o
a  127o
a  b  180
b 127  180
b  53o
c  127o
opposite angles
angles on a straight line
opposite angles

Examples
2 Calculate the size of each lettered angle.
a  72o
a  b  72  180
b 144  180
b  36o
isosceles triangle
angles in a triangle
a
b
72o

Examples
3 Calculate the size of angle a.
a  61143 102  360
a  306  360
a  54o
angles in a quadrilateral
a
102o
61o
143o

Examples
a
128o
a  128o
corresponding angles

Examples
a
131o
a 131 180 interior angles
a  49o

Examples
a
132o
a  132o
alternate angles

Number of sides Name of polygon
3 triangle
4 quadrilateral
5 pentagon
6 hexagon
7 heptagon
8 octagon
9 nonagon
10 decagon
A polygon is a shape enclosed by straight lines.

TRIANGLE QUADRILATERAL
PENTAGON HEXAGON
sum of interior angles =
180°
sum of interior angles
= 2 x 180° =
360°
= 4 x 180° =
720°
= 3 x 180° =
540°

An n-sided polygon can be split into (n – 2) triangles.
The sum of the interior angles in an n-sided polygon
= (n – 2) x 180°
Example
Find the sum of the interior angles in a nonagon.
Sum of interior angles = (n – 2) x 180°
= 7 x 180°
= 1260°
= (9 – 2) x 180°

Example Find the value of x.
105°
107°
140°
75°
x
Sum of interior angles = (5 – 2) x 180° = 540°
x + 75 + 105 + 107 + 140 = 540
x + 427 = 540
x = 113°

 All the sides are the same length
AND
 All the angles are the same size
In a regular polygon

Regular or not?
 







60°
120°
Sum of exterior angles = 360°
Exterior angle = 360° ÷ 6
= 60°
Interior angle = 180° – 60°
= 120°
Regular hexagon

72°
108°
= 7°
= 108°
Regular pentagon

45°
135°
= 45°
= 135°
Regular octagon

In a regular n-sided polygon:
Exterior angle =
360o
n
Interior angle = 180o

360o
n
Example
The interior angle of a regular polygon is 156°.
How many sides does the polygon have?
Interior angle = 180o

360o
n
 156o
360
n
 24
n 
360
24
 15 The polygon has 15 sides.

72°
72°
Example ABCDE is a regular pentagon.
Find the size of angle x .
x
A
B
C
D E F
Angle AEF = exterior angle = 360 ÷ 5 = 72°
Angle EAF = 72°
x = 180 – (72 + 72) = 36°

Congruency
Two shapes are congruent if one of the shapes fits exactly on top of the other shape.
In congruent shapes:
• corresponding angles are equal
• corresponding lengths are equal
These three triangles
are all congruent

To prove that two triangles are congruent you must show that they satisfy one
of the following four sets of conditions:
ASA: two angles and the included
side are the same
SSS: three sides are equal SAS: two sides and the included
angle are the same
RHS: right-angled triangle with
hypotenuse and one other side the
same

A
Which triangles are congruent to triangle A?

Similar shapes
In similar shapes:
• corresponding angles are equal
• corresponding sides are in the same ratio
These two quadrilaterals are similar.
PQ
AB

QR
BC

RS
CD

SP
DA
A B
C
D
P Q
R
S
To show that two triangles are similar it is sufficient
to show that just one of the above conditions is satisfied.

Which triangles are similar to triangle A?
A

The triangles are similar. Find the values of x and y.
(All lengths are in cm.)
Examples
1
12
9
x
8
5.5
y
y
9

8
12
Using ratio of corresponding sides:
12y  72
y  6
x
5.5

12
8
8x  66
x  8.25

The triangles are similar. Find the values of x and y.
Examples
2
y
7

12
8
8y  84
y 10.5
x
15

8
12
12x  120
x  10
7
8
x
15
12
y
Turn one of the triangles so that
you can see which are the
corresponding sides.
7

3 Find the values of x and y.
x
2.5
Examples
9
y
6
2
8
x +2.5
y
x 6
9
x  2.5
x

8
6
6(x  2.5)  8x
6x 15  8x
2x  15
x  7.5
y
9

8
6
6y  72
y  12
Separate the two triangles.

4 4
Examples
x
9
x
4
6
y
6
9
6
6
y
y
6

6
9
9y  36
y  4
x
4

9
6
6x  36
x  6
Find the values of x and y.
Turn the top triangle so that you can see
which are the corresponding sides.

5 6
Examples
y
20
y
6
x
12
16
20
16
x
12
y
6

16
12
12y  96
y  8
x
20

12
16
16x  240
x  15
Turn the top triangle so that you can see
which are the corresponding sides.

enlarge with a length
scale factor of 2
scale factor of 3
2 cm
1 cm
2 cm
1 cm
4 cm
2 cm
6 cm
3 cm
The diagram shows that:
If the length scale factor is 2, then the area scale factor is 4.
If the length scale factor is 3, then the area scale factor is 9.
General rule:
If the length scale factor is k, then the area scale factor is k2.

1 The two shapes are similar.
The area of the smaller shape is 5 cm2.
Find the area of the larger shape.
3 cm
6 cm
Examples
Length scale factor =
6
3
 2
Area scale factor = 22
Area of larger shape = 5  22
 20 cm2

The area of the larger shape is 13.5 cm2.
Find the area of the smaller shape.
4 cm
12 cm
Examples
12
4
 3
Area scale factor = 32
Area of smaller shape = 13.5  32
 1.5 cm2

The area of the smaller shape is 12 cm2.
The area of the larger shape is 27 cm2.
Find the value of x.
4 cm
x cm
Examples
Length scale factor = 
9
4
Area scale factor =
27
12

So x = 4 
3
2
 6 cm
9
4
3
2

4 Area of triangle CDE = 10 cm2.
a Calculate the area of triangle ABC.
b Calculate the area of ABDE.
4 cm
2 cm
Examples
6
4

Area scale factor =
 
 
 
2
3
2
Area of triangle ABC =
 
 
 
2
3
10
2
 22.5 cm2
3
2
A
E D
B
C
a Triangles ABC and EDC are similar.
b Area of ABDE = area of Δ ABC − area of Δ CDE
 22.5 10  12.5 cm2
6 cm
A B
C
4 cm
E D
C

5 Find the area of triangle CDE. 15 cm
Examples
15
21

Area scale factor =
 
 
 
2
5
7
Area of triangle CDE =
 
 
 
2
5
147
7
 75 cm2
5
7
A
E
D
B
C
Triangles ABC and EDC are similar.
21 cm
147 cm2
A B
C
21 cm
147 cm2
D
E
C
15 cm

scale factor of 2
scale factor of 3
If the length scale factor is 2, then the volume scale factor is 8.
If the length scale factor is 3, then the volume scale factor is 27.
General rule:
If the length scale factor is k, then the volume scale factor is k3.
1 cm
1 cm
1 cm
1 cm
1 cm
1 cm
2 cm
2 cm
2 cm
3 cm
3 cm
3 cm

1 The two pyramids are similar.
The volume of the small pyramid is 24 cm2.
Find the volume of the larger pyramid.
4 cm
8 cm
Examples
8
4
 2
Volume scale factor = 23
Volume of larger pyramid = 24  23
 192 cm3

x cm
2 The two prisms are similar.
The volume of the small prism is 27 cm3.
The volume of the large prism is 64 cm3.
Find the value of x. 20 cm
Examples
Volume scale factor =
27
64
Length scale factor = 
3
27
64
x  20 
3
4
 15 cm
3
4

This shape is symmetrical.
The shape has one line of symmetry.

Some shapes have more than one line of symmetry.
The shape has two lines of symmetry

How many lines of symmetry?
The shape has three lines of symmetry.

The shape has five lines of symmetry.

The shape has infinite lines of symmetry.

There are no lines of symmetry but it does have another type of symmetry.

X
It has rotational symmetry of order 4.

There are 2 lines of symmetry.

What is the order of rotational symmetry?
X

There are 8 lines of symmetry.

X
What is its order of rotational symmetry?

SYMMETRY PROPERTIES OF
QUADRILATERALS

X
Square
Number of lines of symmetry = 4
Order of rotational symmetry = 4

X
Rectangle

Parallelogram
X

X
Kite

X
Rhombus

Plane of symmetry or not?


 

How many planes of
symmetry are there in
a cuboid?
There are three planes
of symmetry.

How many planes of symmetry are there in a cube?

note: an understanding of intersecting diagonal properties
is essential for later topics in shape and space

SQUARE RECTANGLE
Diagonals bisect each other
Diagonals cross at 90o

RHOMBUS PARALLELOGRAM
Diagonals cross at 90o

TANGENTS
A straight line can intersect a circle in three possible ways.
It can be:
A DIAMETER A CHORD A TANGENT
2 points of
intersection
2 points of
intersection
1 point of
intersection
A
B
O O O
A
B
A

TANGENT PROPERTY 1
O
The angle between a
tangent and a radius is a
right angle.
A

TANGENT PROPERTY 2
O
The two tangents drawn
from a point P outside a
circle are equal in length.
AP = BP
A
P
B

O
A
B
P
6 cm
8 cm
AP is a tangent to the circle.
a Calculate the length of OP.
b Calculate the size of angle AOP.
c Calculate the shaded area.
OP2
 62
 82
OP2
 100
OP  10 cm
tanx 
8
6
  
  
 
1 8
tan
6
x
 53.13o
AOP
c Shaded area = area of ΔOAP – area of sector OAB
a b
x

   
     
   
   
2
1 53.13
8 6 6
2 360
 24 16.69
 7.31 cm2
(3 s.f.)
Example

CHORDS AND SEGMENTS
major segment
minor segment
A straight line joining two points on the circumference of a
circle is called a chord.
A chord divides a circle into two segments.

SYMMETRY PROPERTIES OF CHORDS 1
O
A B
The perpendicular line from the
centre of a circle to a chord bisects
the chord.
ΙΙ
ΙΙ Note: Triangle AOB is isosceles.

SYMMETRY PROPERTIES OF CHORDS 2
O
A B
If two chords AB and CD are the
same length then they will be the
same perpendicular distance from
the centre of the circle.
ΙΙ
ΙΙ If AB = CD then OP = OQ.
C
D
P
Q
Ι
AB = CD

O
96o
x
2x  96  180
2x  84
x  42o
Triangle OAB is isosceles
because OA = OB (radii of circle)
Example
A
B
So angle OBA = x.

THEOREM 1
O
2x
x
The angle at the centre is
twice the angle at the
circumference.

O
96o
x
96  2x
x  96  2
x  48o
Angle at centre
= 2 × angle at circumference
Example

O
62o
x
x  2  62
x  124o
Angle at centre
Example

O
84o
x
84  2x
x  84  2
x  42o
Angle at centre
Example

O
104o
x
y  2 104
y  208
Angle at centre
y
x  360  208
x  152o
Example

THEOREM 2
O
An angle in a semi-circle
is always a right angle.

O
x  58  90  180
x  32o
Angles in a semi-circle = 90o
and
angles in a triangle add up to 180o
.
58o
x
Example

THEOREM 3
y
x
Opposite angles of a cyclic
quadrilateral add up to
180o
.
x  y  180o

x 132  180
x  48o
Opposite angles in a cyclic
quadrilateral add up to 180o
.
x
y
75o 132o
y  75  180
y  105o
Example

THEOREM 4
x
Angles from the same arc
in the same segment are
equal.
x
x

39o
x
x  39o
Angles from the same arc in the same
segment are equal.
Example

1 Construct triangle ABC in which AB = 10 cm,
AC = 12 cm and BC = 9 cm.
A B
With your compass point on A,
draw an arc radius 12 cm.
With your compass point on B, draw
an arc radius 9 cm.
C
Draw the lines AC and BC.
Label the point of intersection of
the two arcs as C.
10 cm
Draw AB =10 cm
12 cm 9 cm

2 Construct an angle of 600 at A.
A B
draw an arc to cut the line AB at X.
X
With your compass point on X, draw
an arc with the same radius to
intersect the first arc at Y.
Y
Draw the line AY.
Angle BAY = 60o
.
60o

B
3 Construct the bisector of angle ABC.
A
C
With your compass point on B,
draw an arc to intersect the
lines at P and Q.
With the same radius, draw
arcs from P and Q to intersect
each other at R.
Draw the line BR.
Angle ABR = Angle CBR.
P
Q
R

A
B
4 Construct the perpendicular bisector of the line AB.
draw an arc.
Using the same radius and
your compass point on B,
draw an arc to intersect the
first arc at P and Q.
Draw the line PQ.
PQ is the perpendicular
bisector of AB.
P
Q

Ppt for geometry

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