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!