~ X not possible because Flora would have lied ~ Possibly correct

1. 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 T L L T L L ~ 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.

2. Starter Factorize x2 + x - 12 = (x + 4)(x – 3) Expand (2x + 3)(x – 4) = 2x2 – 5x – 12

3. Geometric Reasoning ANGLE PROPERTIES OF LINES AND TRIANGLES

4. Acute Obtuse Reflex Less than 90º More than 90º, less than 180º More than 180º Angle Definitions Angle Notation Use Capital Letters at vertex of angle Use lower case for case for opposite side Angles can also be described as BÂC orBAC

5. Straight line Vertically Opposite At a point Triangle Parallel lines angles of polygons Angle Rules 5 1 2 6 3 4

6. Angles on a Straight Line ‘s on line • Angles on a straight line add to 180o • x + 117o = 180o ( ‘s on line) • x = 63o

7. Vertically Opposite Vert opp ‘s • Vertically Opposite angles are equal • xo = 40º(Vert opp ‘s) • yo + 40o = 180º( ‘s on line) • yo = 140º

8. 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)

9. 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 )

10. 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 )

11. 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

12. 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

13. 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

14. Starter Simplify Simplify

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

16. Parallel line angles • 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 y + 149º =180º y = 180º -149º y = 31º

17. 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 y = 90º + 6º = 96º y 6º Rise x

18. 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

19. 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

20. 1 3 2 75o y a b 40o 50o x 130o c Find v Find x Find y x = 180 – 75 – 50 ( sum of ∆ = 180) x = 180 – 125 x = 55 5 6 4 v j 135o 40o 75o 45o Find j Find p Find a, b & c 40o p 50° y = 40° v = 60° ے ‘s on \ add to 180 ے in equil∆ are equal sum ے in ∆ = 180 140° 70° 40° 60° 110° 70° j = 165° a = c = 140 b = 40° sum ے in ∆ = 180 p = 30° ے ‘s on \ add to 180 ے ‘s at a pt add to 360° Vertoppے ‘s are equal

21. Practice Problems - GAMMA Textbook Ex. 31.05 pg 456 # 1 - 14 IWB Gamma Mathematics Ex 18.05 pg 461-462

22. Starter 1.) Solve these simultaneous equations 2x + 3y = 11 (using substitution or elimination) x + y = 4 2.) Solve for x x = 1, y = 3 11x 11x + 4x + 6 + 90 – 3x = 180 12x + 96 = 180 12x = 84 x = 7 (ﮮsum of ∆) 4x + 6 90 – 3x

23. Note 3: Polygons ~ many sided figures that are closed and lie on a plane. A polygon is a regular polygon when it has equal sides and equal interior angles. Eg.

24. 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

25. 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

26. 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

27. 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, // linesco-int. angles on parallel lines add to 180 isos Δ, base ’s =angles at the base of a isosceles triangle are equal sum Δ =180ºsum of the angles of a triangle add to 180 vert opp ’s = vertically opposite angles are equal ext sum of polygon = 360ºsum 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

28. Practice Problems Textbook GAMMA 31.07 page 461 # 1 – 15 odd IWB Gamma Mathematics Ex 18.07 pg 470

29. Starter Solve for x 3x = 5 • 6 (x+1) = 3 2 4 x = 1/2 x = 20/9

30. SOLUTION The sum of all angles in a pentagon is ______ Each interior angle of a regular pentagon is ______ Angles at a point add to____ Starter 540° 108° 360° 2( ) + x = 360° 108° x = 144° (n-2) 180 = 144 n (n-2) 180 = 144n 180n -360 = 144n 36n = 360 n = 10

31. 1 2 3 35° 85° z 45 y x 115 w Find y Find x Find w & z 5 4 d 35 50 60 95 110 x Find x Find d 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 b y w z 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°

32. 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

33. The ratios of the sides are equal PQ = QT = PT PR RS PS

34. Example GAMMA Math IWB Ex 19.01 pg 486-489 Ex 19.02 pg 491-494 GAMMA Text Ex 32.03 pg 473-5 AB = CB = AC DE DF FE x = 8 = y 9 12 6 * Not to scale Solving for x 12x = 72 x = 6 6 12 x Solving for y 12y = 48 y = 4 9 y 8

35. Starter 95 120 z 110 Find 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 Solve for a Factorize & Simplify

36. Angles in a Circle

37. ANGLES IN A SEMI-CIRCLE The angle in a semi-circle is always 90o A = 90o ( in semi-circle) Applet

38. 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)

39. Examples Angle at the centre is twice the angle at the circumference. Applet GAMMA Mathematics IWB Ex 22.02 pg 537-539

40. 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 B = 108 – 47= 61o The exterior angle of a triangle equals the sum of the two opposite interior angles Applet

41. Practice ProblemsAngles on the Same Arc Angles at the centre of a circle are twice the angle at the circumferenceon the same arc Angles from the same arc to different points on the circumference are always equal GAMMA Math IWB Ex 22.02 pg 537-539 Ex 22.03 pg 541-543 GAMMA Text Ex 33.02 pg 479 Ex 33.03 pg 481

42. 1 2 35 a y x b 55 150 Find A Find x Find y 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) 4 40 s = 80 (<‘s at centre, = 2x circ) 2p = 180 – 80 (base <‘s isos are =) p = 50 s p 3 a = 75 (<‘s at centre, = 2x circ) x = 55 (corresp <‘s, // lines are =)

43. Starter Find the missing angles 26° 26° x = 26° (isosceles Δ) y = 26° (alt. angles || lines) z + 26° + 105° = 180° z = 49 (co-int add to 180 || lines) x + 90 + 67 = 180 x = 23° (angles in a Δ) x = y = 23° (angles on same arc) z = 23° (isosceles Δ)

44. Tangents to a Circle A tangent to a circle makes a right-angle with the radius at the point of contact.

45. Tangents to a Circle • When two tangents are drawn from a point to a circle, they are the same length.

46. Example 1– Give Geometric Reasons x = 90 – 68 = 22o y = 90o z = 180 – 90 – 22 = 68o Angle between tangent and radius is 90º Angle in a semi circle is a right angle Angle sum in a triangle add to 180º

47. Example 2 – Give Geometric Reasons x = ½(180 – 62) = 59o y = 90 – 59 = 31o Base angles in an isosceles tri are equal Angle between tangent and radius is 90º GAMMA Math IWB Ex 22.06 pg 555-557 Ex 22.07 pg 561-562 GAMMA Text Ex 33.06 pg 487

48. 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

49. Starter Solve for angle x and side length y x = 90º y2 + 122 = 182 y2 + 144 = 324 y2 = 180 y = 13.41 cm tgt | rad A 18 cm O y cm 12 cm xº B

50. Cyclic Quadrilaterals