~ x not possible because flora would have lied ~ possibly correct
DESCRIPTION
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.” - PowerPoint PPT PresentationTRANSCRIPT
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?
~ X not possible because Flora would have lied
~ Possibly correct
~ X not possible because Flora would have been
telling the truth when she was actually a liar
~ Possibly correctTherefore, either correct option makes Fred a liar.
Fun with Reasoning
Flora Fred
T T
T L
L T
L L
StarterFactorize x2 + x - 12
= (x + 4)(x – 3)
Expand (2x + 3)(x – 4)
= 2x2 – 5x – 12
Geometric Reasoning
ANGLE PROPERTIES OF LINES AND TRIANGLES
Angle Definitions• Acute• Obtuse
• Reflex
• Less than 90º• More than 90º, less
than 180º• More than 180º
Angle NotationUse Capital Letters at vertex of angle Use lower case for case for opposite side
Angles can also be described as
BÂC or BAC
Angle Rules
• Straight line
• Vertically Opposite
• At a point
• Triangle
• Parallel lines
• angles of polygons
1
3
4
5
2 6
Angles on a Straight Line
• Angles on a straight line add to 180o
• x + 117o = 180o ( ‘s on line)• x = 63o
‘s on line
Vertically Opposite
• Vertically Opposite angles are equal
• xo = 40º (Vert opp ‘s)• yo + 40o = 180º ( ‘s on line)• yo = 140º
Vert opp ‘s
Angles at a Point
• Angles at a point add to 360o
u + 100º + 90º + 75º = 360ºu + 265º = 360ºu = 360º - 265ºu = 95º ( ‘s at pt)
‘s at pt
Angles of a triangle
• The sum of all angles in a triangle = 180º
50º + 70º +s = 180º120º + s = 180ºs = 180º -120ºs = 60º ( Sum of )
Sum of
Exterior Angles of a Triangle
• The exterior angle of a triangle is the sum of the two interior opposite angles
Ext of
tº = 50º + 70ºtº = 120º (Ext of )
Special Triangles
• Isosceles – 2 sides are equal • 2 base angles are equal
22 + i + j = 180º but i = j (isosceles)22 + 2 i = 180º2i = 180º - 22º2i = 158º i = 79º , j = 79º
Base ‘s isos
Equilateral Triangles
• 3 equal sides → 3 equal angles 180º / 3 = 60º
n + p + o = 180ºBut as equilateral, n = p = oSo 3n = 180º
n = 60º = p =o
equilat
Practice Problems
GAMMA Text - Exercise 31.01 – pg. 448-450• Q #1 ~ basic (you can skip this if you want)
• Q #2-17 ~ good achievement questions• Q #18-25 ~ gets increasingly more difficult
IWB Gamma Mathematics Ex 18.01 pg 447
Starter
Simplify
Simplify
62
35
93yxyx
3
3
3yx
4623
xx
21
Note 2: – Properties of Parallel Lines
58°
58°x y
103°
103°
sr
Parallel line angles• Corresponding angles on parallel
lines are equalw = 55o
• Alternate angles on parallel lines are equalg = 38o
• Co-interior angles on parallel lines add to 180o
y + 149º =180º y = 180º -149º y = 31º
Example – Parallel Lines
A walkway and its hand rail both slope upwards at an angle of 6º. Calculate the size of the co-interior angles of the bars, base and handrails
x = 90º – 6º = 84º
6º Risex
yy = 90º + 6º = 96º
Practice Problems - GAMMA
• Ex 31.02 pg. 451 # 6a, c, 7a, c, 8• Ex 31.03 pg 453 # 3, 5
• Ex 31.04 ALL
IWB GAMMA Mathematics pg 457 Ex 18.04
StarterSolve for x
5x + 4 = 3x -16 2x = -20 x = -10
x2 + x - 2 = 0 (x + 2)(x-1) = 0x = -2 or x = 1
75o
50o x
1
Find x
130o
y2
Find y
v3
Find v
40oa
bc
4
Find a, b & c
40o
p40o
5
Find p
135o
75o 45o
j6
Find j
x = 180 – 75 – 50 ( sum of ∆ = 180)x = 180 – 125x = 55
y = 40° v = 60°
a = c = 140b = 40° p = 30°
j = 165°
50°
140°
40° 70°70°110° 60°
sum ے in ∆ = 180s on \ add to 180‘ ے in equil ∆ are ے
equal
s on \ add to 180‘ ے Vert opp ے ‘s are equal
s at a pt add to 360°‘ ے
sum ے in ∆ = 180
Practice Problems - GAMMA
Textbook Ex. 31.05 pg 456 # 1 - 14
IWB Gamma Mathematics Ex 18.05 pg 461-462
Starter1.) Solve these simultaneous equations 2x + 3y = 11 (using substitution or elimination)
x + y = 4
2.) Solve for x11x
90 – 3x4x + 6
11x + 4x + 6 + 90 – 3x = 18012x + 96 = 18012x = 84 x = 7 (ےsum of ∆)
x = 1, y = 3
Note 3: Polygons
~ many sided figures that are closed and lie on a plane.
# of sides name3 Triangle4 Quadrilateral5 Pentagon6 Hexagon7 Heptagon8 Octagon
A polygon is a regular polygon when it 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 = 360ºa + 290º = 360ºa = 70º
• The interior angles of any polygon add to (n-2) x 180º, where n is the number of sidesHere, n = 5 So, angle sum = (5-2) x 180º
= 3 x 180º = 540º90º + 114º + 89º + 152º + r = 540º445º + r = 540ºr = 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 = 144ºK = 72º
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
Practice Problems
Textbook GAMMA 31.07 page 461 # 1 – 15 odd
IWB Gamma Mathematics Ex 18.07 pg 470
StarterSolve for x
3x = 58 6
(x+1) = 3 2 4
x = 20/9 x = 1/2
Starter SOLUTION
The sum of all angles in a pentagon is ______
Each interior angle of a regular pentagon is ______
Angles at a point add to____
540°
108°
360°
(n-2) 180 = 144 n
2( ) + x = 360°108°
x = 144°
(n-2) 180 = 144n180n -360 = 144n 36n = 360 n = 10
45 x
1
Find x
35°y
2
Find y
z
w115
85°3
Find w & z
95
4 d 35
Find d x
6050110
5
Find x
x = 45° (alt <‘s are =)
y = 180 – 90 – 35y = 55° (alt <‘s are =) (<‘s of ∆ = 180)
w = 180 – 115 = 65°z = 180 – 85 = 95° (co-int <‘s = 180°)
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°
w = 120 & y = 130 (<‘s on a line = 180)z = 60° (alt <‘s are =)x = (5 – 2)180 – 110 – 130 – 120 – 60x = 540 – 420 (<‘s of poly = (n-2)180) = 120°
ab y w
z
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 = PTPR RS PS
Example
AB = CB = ACDE DF FE
x
9
12
8y
* Not to scale
6
x = 8 = y 9 12 6
Solving for x
12x = 72 x = 6
Solving for y
12y = 48 y = 4
GAMMA Text Ex 32.03 pg 473-5
GAMMA Math IWB Ex 19.01 pg 486-489Ex 19.02 pg 491-494
StarterSolve for a
Factorize & Simplify
7311
a
10a
41628
yy
21
Find z
z12095
110
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
Angles in a Circle
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)
Examples
Angle at the centre is twice the angle at the circumference.
AppletGAMMA Mathematics IWB Ex 22.02 pg 537-539
ANGLES ON THE SAME ARCAngles 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
B = 108 – 47 = 61o
The exterior angle of a triangle equals the sum of the two opposite interior angles
Applet
Practice Problems Angles on the Same Arc
Angles at the centre of a circle are twice the angle at the circumference on the same arc
Angles from the same arc to different points on the circumference are always equal
GAMMA Text Ex 33.02 pg 479Ex 33.03 pg 481
GAMMA Math IWB Ex 22.02 pg 537-539Ex 22.03 pg 541-543
55
x
1
Find x
x = 55 (corresp <‘s, // lines are =)
Find y
35y
2
a = 35 (alt <‘s, // lines are =)b = 35 (base <‘s isos ∆ are =)y = 180 – 35 – 35 (<‘s of ∆ = 180)y = 110ORy = 180 – 35 – 35 (co-int <‘s, // lines = 180)
a
b
3
150Find A
a = 75 (<‘s at centre, = 2x circ)
40
p
4
s = 80 (<‘s at centre, = 2x circ)2p = 180 – 80 (base <‘s isos are =)p = 50
s
StarterFind the missing angles
x + 90 + 67 = 180
x = y = 23° (angles on same arc)x = 23° (angles in a Δ)
z = 23° (isosceles Δ)
x = 26° (isosceles Δ)
26°
y = 26° (alt. angles || lines)
26°
z + 26° + 105° = 180° z = 49 (co-int add to 180 || lines)
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.
Example 1 – Give Geometric Reasons
x = 90 – 68 = 22o
y = 90o
z = 180 – 90 – 22 = 68o
Angle between tangentand radius is 90º
Angle sum in a triangle add to 180º
Angle in a semi circle is a right angle
x = ½(180 – 62) = 59o
y = 90 – 59 = 31o
Example 2 – Give Geometric Reasons
Base angles in anisosceles tri are equal
Angle between tangentand radius is 90º
GAMMA Text Ex 33.06 pg 487
GAMMA Math IWB Ex 22.06 pg 555-557Ex 22.07 pg 561-562
Angle between a Chord and a Tangent
The angle between a chord and a tangent equals the angle in the alternate (opposite) segment.
x = 550 y = 1160
Starter
A
B
O 18 cm
12 cmy cm
xº
x = 90º
y2 + 122 = 182
y2 + 144 = 324
y2 = 180
y = 13.41 cm
Solve for angle x and side length y
tgt | rad
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
AB
C
DAngle ADB = Angle ACB
Example 1
• Find angle A with a geometrical reason.
A
47o
A = 180 – 47 = 133o
(Opp. ے,cyc. Quad add to 180)
Example 2
• Find angle B with a geometrical reason
B
47o
B = 470
The exterior angle of a cyclic quadrilateral equals the opposite interior angle.
(ext ے cyc quad = opp int)
Example 3
Find, with geometrical reasons the unknown angles.
105o
C
D
41o
C = 41o
D = 180 – 105 = 75o
ext ے,cyc quad = opp int
Opp. ے cyc quad add to 180.
Which of these is cyclic?
A is not cyclic.
Opposite angles do not add to 180o B is cyclic
because 131 + 49 = 180o
GAMMA Text Ex 33.04 pg 483Ex 33.05 pg 486
GAMMA Math IWB Ex 22.04 pg 546-548Ex 22.05 pg 552-553
54º
q
E
B
C
Proof
90°
tgt | rad
cyclic
180°
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 DCA = 36º BAC = ACD DCA & ABD are equal alternate
Angles, therefore AB ll CD
on same arc =
base of isos Δ =
GAMMA Text Ex 33.09 pg 491
GAMMA Math IWB Ex 22.10 pg 567-571
Proof
The diagram shows two parallel lines, DE and AC. CBE = 54ºے ABD = 63º and ے
Prove that ∆ABC is isosceles.
54º63º pD
q
E
B
A
C
q = 63ºp = 63º
Therefore, p = q
Therefore, ے ABC = ے BAC BC = AC
Therefore ∆ABC is isosceles
Proof
alt. ے’s, || lines =
Angles on line add to 180o
Starter – Solve for the missing angles
66°
66°
x + 66° = 180 x = 114°
114°
72°32°
76°
104°
50°50°
80°
100°
43°
129°
28°
67°
Do Now1.) The angle in a semi circle is ___ degrees.2.) A _______is perpendicular to the radius at the point of
contact.3.) The angle at the centre of a circle is ______ the angle at
the circumference.4.) An ___ is part of the circumference of a circle.5.) In a cyclic quadrilateral, an interior angle is equal to the
_____________ angle.6.) Angles on the same arc of a _______ are equal.7.) A set of points all on the circumference of a circle are
said to be _______.8.) An ________ triangle has 2 equal sides.
circle, arc, 90, isoceles, concyclic, tangent, exterior opposite, twice
90
tangent
twice
arc
exterior oppositecircle
concyclicisoceles
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 !
Starter
B C
DA
O 53
yz
v
u
xw
w + x = 90º (ے in a semi circle = 90)v = 180 – 90 – 53 = 37º (sum ے in ∆ )t = 90 – 37 = 53ºx = t = 53º (from the same arc =)u = 53º (ے from the same arc =)z = 180 – 53 – 53 = 74º (sum ے in ∆ )y = 180 – 74 = 106º (ےon a line)t
Notice that these are all isoceles triangles – however we did not need to know this to solve for these angles – EXAMPLE of a PROOF!