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 correct Therefore, either correct option makes Fred a liar. Fun with Reasoning FloraFred T T L L T L

Starter Factorize x 2 + x - 12 = (x + 4)(x – 3) Expand (2x + 3)(x – 4) = 2x 2 – 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 Notation Use Capital Letters at vertex of angle Use lower case for case for opposite side Angles can also be described as BÂCor BAC

Angle Rules Straight line Vertically Opposite At a point Triangle Parallel lines angles of polygons

Angles on a Straight Line Angles on a straight line add to 180 o x o = 180 o ( ‘s on line) x = 63 o ‘s on line

Vertically Opposite Vertically Opposite angles are equal x o = 40º ( Vert opp ‘s ) y o + 40 o = 180º ( ‘s on line ) y o = 140º Vert opp ‘s

Angles at a Point Angles at a point add to 360 o u + 100º + 90º + 75º = 360º u + 265º = 360º u = 360º - 265º u = 95º ( ‘s at pt) ‘s at pt

Angles of a triangle The sum of all angles in a triangle = 180º 50º + 70º +s = 180º 120º + s = 180º s = 180º -120º s = 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 + 70º tº = 120º (Ext of )

Special Triangles Isosceles – 2 sides are equal 2 base angles are equal 22 + i + j = 180º but i = j (isosceles) i = 180º 2i = 180º - 22º 2i = 158º i = 79º, j = 79º Base ‘s isos

Equilateral Triangles 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 equilat

Practice Problems GAMMA Text - Exercise – pg 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 pg 447

Starter Simplify

Note 2: – Properties of Parallel Lines x y s r

Parallel line angles Corresponding angles on parallel lines are equal w = 55 o Alternate angles on parallel lines are equal g = 38 o Co-interior angles on parallel lines add to 180 o 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 Rise x y y = 90º + 6º = 96º

Practice Problems - GAMMA Ex pg. 451 # 6a, c, 7a, c, 8 Ex pg 453 # 3, 5 Ex ALL IWB GAMMA Mathematics pg 457 Ex 18.04

Starter Solve for x 5x + 4 = 3x -16 2x = -20 x = -10 x 2 + x - 2 = 0 (x + 2)(x-1) = 0 x = -2 or x = 1

75 o 50 o x 1 Find x 130 o y 2 Find y v 3 Find v 40 o a b c 4 Find a, b & c 40 o p 5 Find p 135 o 75 o 45 o j 6 Find j x = 180 – 75 – 50 ( sum of ∆ = 180) x = 180 – 125 x = 55 y = 40° v = 60° a = c = 140 b = 40° p = 30° j = 165° 50° 140° 40° 70° 110° 60° sum ے in ∆ = 180 ے ‘s on \ add to 180 ے in equil ∆ are equal ے ‘s on \ add to 180 Vert opp ے ‘s are equal ے ‘s at a pt add to 360° sum ے in ∆ = 180

Practice Problems - GAMMA Textbook Ex pg 456 # IWB Gamma Mathematics Ex pg

Starter 1.) Solve these simultaneous equations 2x + 3y = 11 (using substitution or elimination) x + y = 4 2.) Solve for x 11x 90 – 3x4x x + 4x – 3x = x + 96 = x = 84 x = 7 (ﮮsum of ∆) x = 1, y = 3

Note 3: Polygons ~ many sided figures that are closed and lie on a plane. # of sidesname 3Triangle 4Quadrilateral 5Pentagon 6Hexagon 7Heptagon 8Octagon 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 360 o 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 = 95 o

The exterior angles of any polygon add to 360 o 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

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 page 461 # 1 – 15 odd IWB Gamma Mathematics Ex pg 470

Starter Solve for x 3x = (x+1) = 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 = 144n 180n -360 = 144n 36n = 360 n = 10

45 x 1 Find x 35° y 2 Find y z w ° 3 Find w & z 95 4 d 35 Find d x Find x x = 45° (alt <‘s are =) y = 180 – 90 – 35 y = 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 – 60 x = 540 – 420 (<‘s of poly = (n-2)180) = 120° a byw 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 = PT PR RS PS

Example AB = CB = AC DE DF FE x y * Not to scale 6 x = 8 = y Solving for x 12x = 72 x = 6 Solving for y 12y = 48 y = 4 GAMMA Text Ex pg GAMMA Text Ex pg GAMMA Math IWB Ex pg Ex pg GAMMA Math IWB Ex pg Ex pg

Starter Solve for a Factorize & Simplify Find z z 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 90 o A = 90 o ( 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. Applet GAMMA Mathematics IWB Ex pg GAMMA Mathematics IWB Ex pg

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 = 47 o Angles on the same arc are equal B = 108 – 47 = 61 o 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 pg 479 Ex pg 481 GAMMA Text Ex pg 479 Ex pg 481 GAMMA Math IWB Ex pg Ex pg GAMMA Math IWB Ex pg Ex pg

55 x 1 Find x x = 55 (corresp <‘s, // lines are =) Find y 35 y 2 a = 35 (alt <‘s, // lines are =) b = 35 (base <‘s isos ∆ are =) y = 180 – 35 – 35 (<‘s of ∆ = 180) y = 110 OR y = 180 – 35 – 35 (co-int <‘s, // lines = 180) a b Find 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

Starter Find the missing angles x = 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 = 22 o y = 90 o z = 180 – 90 – 22 = 68 o Angle between tangent and radius is 90º Angle sum in a triangle add to 180º Angle in a semi circle is a right angle

x = ½(180 – 62) = 59 o y = 90 – 59 = 31 o Example 2 – Give Geometric Reasons Base angles in an isosceles tri are equal Angle between tangent and radius is 90º GAMMA Text Ex pg 487 GAMMA Text Ex pg 487 GAMMA Math IWB Ex pg Ex pg GAMMA Math IWB Ex pg Ex pg

Angle between a Chord and a Tangent The angle between a chord and a tangent equals the angle in the alternate (opposite) segment. x = 55 0 y = 116 0

Starter A B O 18 cm 12 cm y cm xºxº x = 90º y = 18 2 y = 324 y 2 = 180 y = 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 180 o 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 A B C D Angle ADB = Angle ACB

Example 1 Find angle A with a geometrical reason. A 47 o A = 180 – 47 = 133 o (Opp. ﮮ,cyc. Quad add to 180)

Example 2 Find angle B with a geometrical reason B 47 o B = 47 0 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. 105 o C D 41 o C = 41 o D = 180 – 105 = 75 o 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 180 o B is cyclic because = 180 o GAMMA Text Ex pg 483 Ex pg 486 GAMMA Text Ex pg 483 Ex pg 486 GAMMA Math IWB Ex pg Ex pg GAMMA Math IWB Ex pg Ex pg

54º q E B C 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 pg 491 GAMMA Text Ex pg 491 GAMMA Math IWB Ex pg GAMMA Math IWB Ex pg

The diagram shows two parallel lines, DE and AC. ﮮ ABD = 63º and ﮮCBE = 54º Prove that ∆ABC is isosceles. 54º 63º p D q E B A C q = 63º p = 63º Therefore, p = q Therefore, ﮮ ABC = ﮮ BAC BC = AC Therefore ∆ABC is isosceles alt. ﮮ’s, || lines = Angles on line add to 180 o

Starter – Solve for the missing angles 66° x + 66° = 180x = 114° 114° 72° 32° 76° 104° 50° 80° 100° 43° 129° 28° 67°

Do Now 1.) 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 opposite circle concyclic isoceles

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 D A O53 y z v u x w w + x = 90º (ﮮ in a semi circle = 90) v = 180 – 90 – 53 = 37º (sum ﮮ in ∆ ) t = 90 – 37 = 53º x = 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!