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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 ~ X not possible because Flora would have lied T L ~ Possibly correct L T ~ X not possible because Flora would have been telling the truth when she was actually a liar L L ~ Possibly correct Therefore, either correct option makes Fred a liar. Geometry IGCSE – Chapter 4 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 opposite side. Angles can also be described as: BÂC or BAC. Angle Rules •1 Straight line 5• Parallel lines 2• Vertically Opposite 6• angles of polygons 3 • At a point 4• Triangle 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 3x5 y 3 Simplify 9x2 y6 x3 3 3y Simplify 3x 2 6x 4 3x 2 2 (3 x 2 ) 1 2 Note 2: Polygons ~ many sided figures that are closed and lie on a plane. # of sides 3 4 5 6 7 8 name Triangle Quadrilateral Pentagon Hexagon Heptagon Octagon A regular polygon 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 = 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) Note 3: – Properties of Parallel Lines 58° y 103° x 58° r s 103° Angles on Parallel Lines • 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 cointerior angles of the bars, base and handrails y x x = 90º – 6º = 84º y = 90º + 6º = 96º 6º Rise IGCSE Ex 3 Pg 119 1 2 75o 50o Find x x = 180 – 75 – 50 ( sum of ∆ = 180) Find y Find v y = 40° v = 60° angles on \ add to 180 sum ےin ∆ = 180 x = 55 4 5 40o 140° a 40o b 40° c Find a, b & c 70° 110° 40o p Find p ‘ ےs on \ add to 180 Vert opp ‘ ےs are equal 70° ےin equil ∆ are equal 6 j 135o 60° 75o 45o Find j j = 165° a = c = 140 b = 40° v 3 50° 130o x x = 180 – 125 y p = 30° sum ےin ∆ = 180 ‘ ےs at a pt add to 360° Starter 1.) Solve these simultaneous equations 2x + 3y = 11 (using substitution or elimination) x+y=4 x = 1, y = 3 2.) Solve for x 4x + 6 11x 11x + 4x + 6 + 90 – 3x = 180 12x + 96 = 180 12x = 84 x=7 (ﮮsum of ∆) 90 – 3x Starter Solve for x 3x = 5 8 6 x = 20/9 (x+1) = 3 2 4 x = 1/2 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.g. E 4 AB = CB = AC EF DF DE 5 x A x 5 = y 6 Solving for x 4x = 10 x = 2.5 Solving for y 4y = 12 y=3 = 2 4 D 2 B y C 6 F * Not to scale Example IGCSE Ex 6 pg 125 AB = CB = AC DE DF FE x 9 = 8 = 12 y 6 * Not to scale Solving for x 12x = 72 x=6 6 12 x Solving for y 9 12y = 48 y=4 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 X S A bisector cuts an angle in half ^ ^ SPQ + PQR = 180° R ^ ÷2 ^ ½ SPQ + ½ PQR = 90° Therefore, because the sum of all angles in a triangle add to 180°, X must be right-angled IGCSE Ex 7 pg 127 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 1 2 45 85° z 3 35° y x 115 w Find y Find x x = 45° (alt <‘s are =) 4 d 95 35 y = 180 – 90 – 35 Find w & z w = 180 – 115 = 65° y = 55° (alt <‘s are =) z = 180 – 85 = 95° a (<‘s of ∆ = 180) 5 b Find d 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° (co-int <‘s = 180°) 50 y w 60 110 x z Find x 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° Starter Solve for a a 11 7 3 95 120 z 110 a 10 Find z Factorize & Simplify 8y 2 16 y 4 1 2 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 Circle Theorems 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 Applet o B = 108 – 47 = 61 The exterior angle of a triangle equals the sum of the two opposite interior angles Examples Angle at the centre is twice the angle at the circumference. Applet IGCSE Ex 10 pg 135-136 2 1 35 y x 3 a b 55 150 Find y Find x x = 55 (corresp <‘s, // lines are =) a = 35 (alt <‘s, // lines are =) b = 35 (base <‘s isos ∆ are =) Find A a = 75 (<‘s at centre, = 2x circ) y = 180 – 35 – 35 (<‘s of ∆ = 180) y = 110 OR 4 y = 180 – 35 – 35 (co-int <‘s, // lines = 180) 40 s = 80 (<‘s at centre, = 2x circ) s p 2p = 180 – 80 (base <‘s isos are =) p = 50 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 C B v t w y x w + x = 90º ( ﮮin a semi circle = 90) v = 180 – 90 – 53 = 37º (sum ﮮin ∆ ) t = 90 – 37 = 53º x = t = 53º (from the same arc =) u = 53º ( ﮮfrom the same arc =) z = 180 – 53 – 53 = 74º (sum ﮮin ∆ ) y = 180 – 74 = 106º (ﮮon a line) z u A O 53 D 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 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 C ﮮACB = ﮮADB 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. (extﮮ, cyc quad) 47o B Example 3 • Find, with geometrical reasons the unknown angles. C = 41o (ext ﮮ,cyc quad) D = 180 – 105 = 75o (Opp. ﮮcyc quad.) 105o 41o C D Which of these is cyclic? A is not cyclic. Opposite angles do not add to 180o B is cyclic because 131 + 49 = 180o IGCSE Ex 11 pg 137-138 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. Do Now 90 degrees. 1.) The angle in a semi circle is ___ tangent perpendicular to the radius at the point of 2.) A _______is contact. twice the angle at 3.) The angle at the centre of a circle is ______ the circumference. arc is part of the circumference of a circle. 4.) An ___ 5.) In a cyclic quadrilateral, an interior angle is equal to the exterior opposite angle. _____________ circle are equal. 6.) Angles on the same arc of a _______ 7.) A set of points all on the circumference of a circle are said to be _______. concyclic isoceles triangle has 2 equal sides. 8.) An ________ circle, arc, 90, isoceles, concyclic, tangent, exterior opposite, twice 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 ! e.g. Find x, y and z x = 90 – 68 = 22o (tgt | rad) y = 90o ( ﮮin semi-circle) z = 180 – 90 – 22 = 68o (sum ﮮin ∆) e.g. Find x and y x = ½(180 – 62) = 59o (base ﮮ, isoc ∆) y = 90 – 59 = 31o (tgt | rad) IGCSE Ex 12 pg 139-141 Starter Solve for angle x and side length y x = 90º (tgt | rad) y2 + 122 = 182 y2 + 144 = 324 y2 = 180 y = 13.4 cm 18 cm O y cm xº B 12 cm A Proof 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 (base of isos Δ) DCA = 36º ( on same arc) BAC =ACD are equal alternate Angles, therefore AB ll CD Proof • The diagram shows two parallel lines, DE and AC. ﮮABD = 63º and ﮮCBE = 54º • Prove that ∆ABC is isoceles. q = 63º (alt. ’ﮮs, || lines) p = 63º ( ’ﮮs on a line) E B 54º Therefore, p = q Therefore, ﮮABC = ﮮBAC BC = AC D 63º p C q A Therefore ∆ABC is isoceles