Download Presentation
## Do Now

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**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 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.**GeometryIGCSE – Chapter 4**ANGLE PROPERTIES OF LINES AND TRIANGLES**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**Straight line**Vertically Opposite At a point Triangle Parallel lines angles of polygons Angle Rules 5 1 2 6 3 4**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**Simplify Simplify**Note 2: Polygons**~ many sided figures that are closed and lie on a plane. A regular polygonhas 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)**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 (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 co-interior angles of the bars, base and handrails x = 90º – 6º = 84º y = 90º + 6º = 96º y 6º Rise x IGCSE Ex 3 Pg 119**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**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**Starter**Solve for x 3x = 5 • 6 (x+1) = 3 2 4 x = 1/2 x = 20/9**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**e.g. 4 AB = CB = AC EFDF DE 5 D x A B x = y = 2 564 6 2 y C F Solving for x 4x = 10 x = 2.5 * Not to scale Solving for y 4y = 12 y = 3**Example**IGCSE Ex6pg125 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**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 A bisector cuts an angle in half X ^ ^ ^ ^ ½ SPQ + ½ PQR = 90° SPQ + PQR = 180° ÷ 2 R S Therefore, because the sum of all angles in a triangle add to 180°, X must be right-angled IGCSE Ex7 pg127**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**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°**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 Starter Solve for a Factorize & Simplify**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 B = 108 – 47= 61o The exterior angle of a triangle equals the sum of the two opposite interior angles Applet**Examples**Angle at the centre is twice the angle at the circumference. IGCSE Ex10 pg135-136 Applet**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 =)**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**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) C B v w x t y z u O 53 D A 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**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 ﮮACB = ﮮADB C**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. 47o B (extﮮ, cyc quad)**Example 3**Find, with geometrical reasons the unknown angles. 105o 41o C D C = 41o (ext ﮮ,cyc quad) D = 180 – 105= 75o (Opp. ﮮ cyc quad.)**Which of these is cyclic?**A is not cyclic. Opposite angles do not add to 180o Bis cyclic because 131 + 49 = 180o IGCSE Ex11 pg137-138**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.