Download Geometry Notes - Mrs A`s Weebly

Do Now
Factorize x2 + x - 12
= (x + 4)(x – 3)
Expand (2x + 3)(x – 4)
= 2x2 – 5x – 12
Fun with Reasoning
Two tribes live on an island. Members of one tribe always tell the truth
and members of the other tribe always lie.
You arrive on the island and meet two islanders named Flora and
Fred.
Flora says, “Only one of us is from the tribe that always lies.”
Which tribe does Fred come from?
Flora
Fred
T
T
~ X not possible because Flora would have lied
T
L
~ Possibly correct
L
T
~ X not possible because Flora would have been
telling the truth when she was actually a liar
L
L
~ Possibly correct
Therefore, either correct option makes Fred a liar.
Geometry
IGCSE – Chapter 4
ANGLE PROPERTIES OF LINES
AND TRIANGLES
Angle Definitions
• Acute
• Obtuse
• Reflex
• Less than 90º
• More than 90º, less
than 180º
• More than 180º
Angle Notation
Use capital letters at vertex of angle.
Use lower case for opposite side.
Angles can also be described as:
BÂC or BAC.
Angle Rules
•1 Straight line
5• Parallel lines
2• Vertically Opposite
6• angles of polygons
3
• At a point
4• Triangle
Angles on a Straight Line
‘s on line
• Angles on a straight line add to 180o
• x + 117o = 180o ( ‘s on line)
• x = 63o
Vertically Opposite
Vert opp ‘s
• Vertically Opposite angles are equal
• xo = 40º (Vert opp ‘s)
• yo + 40o = 180º ( ‘s on line)
• yo = 140º
Angles at a Point
 ‘s at pt
• Angles at a point add to 360o
u + 100º + 90º + 75º = 360º
u + 265º = 360º
u = 360º - 265º
u = 95º ( ‘s at pt)
Angles of a triangle
 Sum of
• The sum of all angles in a triangle = 180º
50º + 70º +s = 180º
120º + s = 180º
s = 180º -120º
s = 60º ( Sum of
)
Exterior Angles of a Triangle
Ext

of
• The exterior angle of a triangle is the sum of the
two interior opposite angles
tº = 50º + 70º
tº = 120º (Ext

of
)
Special Triangles
Base  ’s isos
• Isosceles – 2 sides are equal
•
2 base angles are equal
22 + i + j = 180º but i = j (isosceles)
22 + 2 i = 180º
2i = 180º - 22º
2i = 158º
i = 79º , j = 79º
Equilateral Triangles
equilat
• 3 equal sides → 3 equal angles
180º / 3 = 60º
n + p + o = 180º
But as equilateral, n = p = o
So 3n = 180º
n = 60º = p =o
IGCSE
Ex 1 Pg 115-116
Starter
3x5 y 3
Simplify
9x2 y6
x3
 3
3y
Simplify
3x  2
6x  4
3x  2

2 (3 x  2 )
1

2
Note 2: Polygons
~ many sided figures that are closed and lie on a plane.
# of sides
3
4
5
6
7
8
name
Triangle
Quadrilateral
Pentagon
Hexagon
Heptagon
Octagon
A regular polygon
has equal sides
and equal interior
angles.
Eg.
Angles on a Polygon
• Exterior angle – one side is extended
outwards, to make an the angle – H
• Interior angle – inside the shape - G
G
H
Quadrilaterals and other Polygons
• The interior angles of a quadrilateral add
to 360o
a + 130º +75º + 85º = 360º
a + 290º = 360º
a = 70º
• The interior angles of any polygon add to
(n-2) x 180º, where n is the number of sides
Here, n = 5
So, angle sum = (5-2) x 180º
= 3 x 180º = 540º
90º + 114º + 89º + 152º + r = 540º
445º + r = 540º
r = 95o
The exterior angles of any polygon add to 360o
G = 360º/10 (reg. poly)
G = 36º
H = 180º – 36º = 144º (adj.  )
10J = 360º (  at a pt)
J = 36º
2K +36º = 180º ( of isos ∆)
2K = 144º
K = 72º
IGCSE
Ex 2 Pg 117-118
Starter
Solve for x
5x + 4 = 3x -16
2x = -20
x = -10
x2 + x - 2 = 0
(x + 2)(x-1) = 0
x = -2 or x = 1
(quadratics always have 2 solutions)
Note 3: – Properties of Parallel Lines
58°
y
103°
x
58°
r
s
103°
Angles on Parallel Lines
• Corresponding angles on parallel
lines are equal
w = 55o
• Alternate angles on parallel lines
are equal
g = 38o
• Co-interior angles on parallel lines
add to 180o (allied angles)
y + 149º =180º
y = 180º -149º
y = 31º
e.g. Parallel Lines
A walkway and its hand rail both slope upwards
at an angle of 6º. Calculate the size of the cointerior angles of the bars, base and handrails
y
x
x = 90º – 6º = 84º
y = 90º + 6º = 96º
6º Rise
IGCSE
Ex 3 Pg 119
1
2
75o
50o
Find x
x = 180 – 75 – 50 ( sum of ∆ = 180)
Find y
Find v
y = 40°
v = 60°
angles on \ add to 180
sum ‫ ے‬in ∆ = 180
x = 55
4
5
40o
140°
a
40o
b 40°
c
Find a, b & c
70°
110°
40o
p
Find p
‫‘ ے‬s on \ add to 180
Vert opp ‫‘ ے‬s are equal
70°
‫ ے‬in equil ∆ are
equal
6
j
135o
60°
75o
45o
Find j
j = 165°
a = c = 140
b = 40°
v
3
50°
130o
x
x = 180 – 125
y
p = 30°
sum ‫ ے‬in ∆ = 180
‫‘ ے‬s at a pt add to 360°
Starter
1.) Solve these simultaneous equations
2x + 3y = 11 (using substitution or elimination)
x+y=4
x = 1, y = 3
2.) Solve for x
4x + 6
11x
11x + 4x + 6 + 90 – 3x = 180
12x + 96 = 180
12x = 84
x=7
(‫ﮮ‬sum of ∆)
90 – 3x
Starter
Solve for x
3x = 5
8 6
x = 20/9
(x+1) = 3
2
4
x = 1/2
Note 4: Similar Triangles
• One shape is similar to
another if they have
exactly the same shape
and same angles
• The ratios of the
corresponding sides are
equal
The ratios of the sides are equal
PQ = QT = PT
PR
RS
PS
e.g.
E
4
AB = CB = AC
EF
DF
DE
5
x
A
x
5
= y
6
Solving for x
4x = 10
x = 2.5
Solving for y
4y = 12
y=3
=
2
4
D
2
B
y
C
6
F
* Not to scale
Example
IGCSE
Ex 6 pg 125
AB = CB = AC
DE
DF
FE
x
9
= 8 =
12
y
6
* Not to scale
Solving for x
12x = 72
x=6
6
12
x
Solving for y
9
12y = 48
y=4
y
8
Note 5: Congruence
• Two plane figures are congruent if one fits
exactly in another
• Same size, same shape
e.g.
PQRS is a parallelogram in which the
bisectors of the angles P and Q meet at X.
Prove that the angle PXQ is right angled.
Q
P
X
S
A bisector cuts
an angle in half
^
^
SPQ + PQR = 180°
R
^
÷2
^
½ SPQ + ½ PQR = 90°
Therefore, because the sum of all angles in a
triangle add to 180°, X must be right-angled
IGCSE
Ex 7 pg 127
Shorthand Reasons - Examples
corr ’s =, // lines corresponding angles on parallel lines are equal
alt ’s =, // lines alternate angles on parallel lines are equal
coint ’s add to 180º, // lines co-int. angles on parallel lines add to 180
isos Δ, base ’s = angles at the base of a isosceles triangle are equal
sum Δ =180º sum of the angles of a triangle add to 180
vert opp ’s = vertically opposite angles are equal
ext sum of polygon = 360º sum of the ext. angles of a polygon = 360
int sum of polygon = (n – 2) × 180º the sum of interior angles of a
polygon = (n-2) x 180
ext  of Δ = sum of int opp  s exterior angles of a
triangle = the sum of the interior opposite angles
1
2
45
85° z
3
35°
y
x
115
w
Find y
Find x
x = 45° (alt <‘s are
=)
4
d
95
35
y = 180 – 90 – 35
Find w & z
w = 180 – 115 = 65°
y = 55° (alt <‘s are =)
z = 180 – 85 = 95°
a
(<‘s of ∆ = 180)
5
b
Find d
a = 180 – 95 (<‘s on a line = 180)
= 85°
b = 180 – 85 – 35 (<‘s of ∆ = 180)
= 60°
d = 180 – 60 (co-int <‘s = 180)
= 120°
(co-int <‘s = 180°)
50 y w 60
110
x
z
Find x
w = 120 & y = 130 (<‘s on a line = 180)
z = 60° (alt <‘s are =)
x = (5 – 2)180 – 110 – 130 – 120 – 60
x = 540 – 420 (<‘s of poly = (n-2)180)
= 120°
Starter
Solve for a
a  11
7
3
95
120
z
110
a  10
Find z
Factorize & Simplify
8y  2
16 y  4
1

2
w = 180 – 95 = 85º
(co-int <‘s, // lines = 180)
z = 180(5-2) – 95 – 120 –
110 – 85
(int <‘s poly = (n-2)180)
= 540 – 410
= 130º
Circle Theorems
ANGLES IN A SEMI-CIRCLE
• The angle in a semi-circle is always 90o
A = 90o

( in semi-circle)
Applet
ANGLES AT THE CENTRE OF
A CIRCLE
From the same arc, the
angle formed at the centre
is twice the angle formed at
the circumference.
C = 2A
(<‘s at centre, = 2x circ)
ANGLES ON THE SAME ARC
Angles extending to the
circumference from the
same arc are equal.
e.g. Find A and B giving
geometrical reasons for
your answers.
A = 47o Angles on the same arc are equal
Applet
o
B = 108 – 47 = 61
The exterior angle of a triangle equals the sum of the two
opposite interior angles
Examples
Angle at the centre is twice the angle at
the circumference.
Applet
IGCSE
Ex 10 pg 135-136
2
1
35
y
x
3
a
b
55
150
Find y
Find x
x = 55 (corresp <‘s,
// lines are =)
a = 35 (alt <‘s, // lines are =)
b = 35 (base <‘s isos ∆ are =)
Find A
a = 75 (<‘s at
centre, = 2x circ)
y = 180 – 35 – 35 (<‘s of ∆ = 180)
y = 110
OR
4
y = 180 – 35 – 35 (co-int <‘s, // lines = 180)
40
s = 80 (<‘s at centre, = 2x circ)
s
p
2p = 180 – 80 (base <‘s isos are =)
p = 50
Angles on the Same Arc
Angles at the centre of a circle are twice the
angle at the circumference
Angles from the same arc to different points
on the circumference are always equal!
Starter
C
B
v
t
w
y
x
w + x = 90º (‫ ﮮ‬in a semi circle = 90)
v = 180 – 90 – 53 = 37º (sum ‫ ﮮ‬in ∆ )
t = 90 – 37 = 53º
x = t = 53º (from the same arc =)
u = 53º (‫ ﮮ‬from the same arc =)
z = 180 – 53 – 53 = 74º (sum ‫ ﮮ‬in ∆ )
y = 180 – 74 = 106º (‫ﮮ‬on a line)
z
u
A
O
53
D
Notice that these are all isoceles triangles – however we
did not need to know this to solve for these angles –
EXAMPLE of a PROOF!
Cyclic Quadrilaterals
CYCLIC QUADRILATERALS
A cyclic quadrilateral has all four vertices on a
circle.
(concyclic points)
Opposite angles of a cyclic
quadrilateral add to 180o
The exterior angle of a cyclic
quadrilateral equals the opposite
interior angle.
Cyclic Quadrilaterals
If 2 angles extending
from the same arc
are equal, then the
quadrilateral is
cyclic
B
A
D
C
‫ﮮ‬ACB = ‫ﮮ‬ADB
Example 1
• Find angle A with a geometrical reason.
A = 180 – 47
= 133o
(Opp. ‫ﮮ‬,cyc. quad)
A
47o
Example 2
• Find angle B with a geometrical reason
B = 470
The exterior angle of a
cyclic quadrilateral equals
the opposite interior angle.
(ext‫ﮮ‬, cyc quad)
47o
B
Example 3
• Find, with geometrical
reasons the unknown
angles.
C = 41o
(ext ‫ﮮ‬,cyc quad)
D = 180 – 105 = 75o
(Opp. ‫ ﮮ‬cyc quad.)
105o
41o
C
D
Which of these is cyclic?
A is not cyclic.
Opposite angles do not
add to 180o
B is cyclic
because 131 + 49 = 180o
IGCSE
Ex 11 pg 137-138
Tangents to a Circle
• A tangent to a circle makes a right-angle
with the radius at the point of contact.
Tangents to a Circle
• When two tangents are drawn from a point
to a circle, they are the same length.
Do Now
90 degrees.
1.) The angle in a semi circle is ___
tangent perpendicular to the radius at the point of
2.) A _______is
contact.
twice the angle at
3.) The angle at the centre of a circle is ______
the circumference.
arc is part of the circumference of a circle.
4.) An ___
5.) In a cyclic quadrilateral, an interior angle is equal to the
exterior
opposite angle.
_____________
circle are equal.
6.) Angles on the same arc of a _______
7.) A set of points all on the circumference of a circle are
said to be _______.
concyclic
isoceles triangle has 2 equal sides.
8.) An ________
circle, arc, 90, isoceles, concyclic, tangent, exterior opposite, twice
Another interesting feature of
tangents and circles……..
When you form a
quadrilateral from 2
tangents and 2 radii,
the result is always a
cyclic quadrilateral !
e.g. Find x, y and z
x = 90 – 68
= 22o
(tgt | rad)
y = 90o
(‫ ﮮ‬in semi-circle)
z = 180 – 90 – 22
= 68o
(sum ‫ ﮮ‬in ∆)
e.g. Find x and y
x = ½(180 – 62)
= 59o
(base ‫ﮮ‬, isoc ∆)
y = 90 – 59
= 31o
(tgt | rad)
IGCSE
Ex 12 pg 139-141
Starter
Solve for angle x and side length y
x = 90º (tgt | rad)
y2 + 122 = 182
y2 + 144 = 324
y2 = 180
y = 13.4 cm
18 cm
O
y cm
xº
B
12 cm
A
Proof
This is where you use your knowledge of
geometry to justify a statement.
Prove: AB is parallel to CD
(like proving concylic points)
ABD = 36º (base of isos Δ)
DCA = 36º ( on same arc)
 BAC =ACD are equal alternate
Angles, therefore AB ll CD
Proof
• The diagram shows two parallel lines, DE and
AC. ‫ﮮ‬ABD = 63º and ‫ﮮ‬CBE = 54º
• Prove that ∆ABC is isoceles.
q = 63º (alt. ‫’ﮮ‬s, || lines)
p = 63º ( ‫’ﮮ‬s on a line)
E
B
54º
Therefore, p = q
Therefore, ‫ ﮮ‬ABC = ‫ ﮮ‬BAC
BC = AC
D
63º p
C
q
A
Therefore ∆ABC is isoceles