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1 Press Ctrl-A ©G Dear2008 – Not to be sold/Free to use Stage 6 - Year 12 Mathematic (HSC)
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2 Types of Angles 1. Acute angles 2. Right Angle 3. Obtuse Angle 4. Straight Angle5. Reflex Angle 6. Angle of Revolution (0 o < θ < 90 o θ = 90 o (90 o < θ < 180 o) (θ = 360 o) (θ = 180 o) (180 o < θ < 360 o)
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3 1. Vertically Opposite Angleare equal. 2. Complementary Angles aoao bobo add to 90 o. 2. Supplementary Angles aoao bobo add to 180 o. Pairs of Angles
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4 Transversal 1. Alternate Angles 2. Corresponding Angles 3. Co-Interior angles Makes a Z shape. Makes a F shape. Makes a C shape. and are equal. and are equal. and Add to 180 o Angles between Parallel Lines
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5 Based on SidesBased on Angles 1. Equilateral triangle. All sides equal All angles equal (60 O ) 2. Isosceles triangle. Two sides equal Two base-angles equal 3. Scalene triangle. No sides equal No angles equal 1. Acute angled triangle. 2. Right angled triangle. All angles acute One angle 90 o One Obtuse angle. 3. Obtuse angled triangle. Types of Triangles
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6 3. Exterior Angle of a Triangle. 1. Angle Sum of a Triangle 2. Angle Sum of a Quadrilateral 4. Angles at a point. aoao bobo coco a o + b o + c o = 180 o aoao bobo coco dodo a o + b o + c o + d o = 360 o bobo coco aoao a o = b o + c o aoao bobo coco a o + b o + c o = 360 o Angle Sums
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7 1. Side, Side, Side.2. Side, Angle, Side. 4. Right angle, Hypotenuse, Side. 3. Angle, Angle, Side. SSS SAS AAS RHS Congruence
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8 1. Corresponding angles are all equal. α β α β γ γ 2. Corresponding sides are in the same ratio. a x axax b y byby == czcz c z 3. Two pairs of sides are in proportion and their included angles are equal. p q s r θ φ prpr = qsqs = = Similar Triangles
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9 AB : BC = DE : EF A F E D C B AB BCEF DE = Ratio of Intercepts
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10 c a b c 2 = a 2 + b 2 You need to be able to: 1. Find the length of the hypotenuse. 2. Find the length of the shorter side. 3. Prove you have a right angle. Pythagoras Theorem
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11 1. Rectangle2. Square3. Rhombus 4.Parallelogram 5. Trapezium 6. Kite You must know their properties Types of Quadrilaterals
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12 1. Triangle2. Square3. Pentagon 4. Hexagon 5. Octagon Types of Regular Polygons
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13 1. Angle Sum of a Polygon. 2. Interior angle. 3. Exterior angle a f c b e d = (n – 2) x 180 (n is the number of angles) Divide the angle sum by the number of angles. The exterior angles of add to 360 o. Regular Polygons
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14 Area Formulae 1.Square A = s 2 s 2. Rectangle A = LB L B 3.Triangle A=½bh b h 6.Trapezium A=½(a+b)h b a h 7.Circle A=πr 2 r 4. Parallelogram A=bh b h 5.Rhombus/Kite A=½xy x y x y
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15 1. Rectangular Prism l b h SA = 2(bh + hl + lb) 2. Cube s SA = 6s 2 3. Sphere SA = 4 π r 2 4. Cylinder SA = 2 π r (r + h) 5. Cone h r h r l SA = π r (r + l) Surface Area Formulae
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16 1. Rectangular Prism l b h V = lbh 2. Cube s V = s 3 3. Sphere V = 4 π r 3 3 4. Cylinder V = π r 2 h 5. Cone h r h r V = 1 π r 2 h 3 V = Ah Volume Formulae
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17 2006 HSC Question 6
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18 BCA = CAD[Alternate angles between parallel lines.] BAC = CAD[Given] (i)Prove that BAC = BCA1 BAC = BCA 2006 HSC Question 6
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19 BP[Common] PBA = PBC[Given] (ii)Prove that ∆ABP ≡ ∆CBP1 BAC = BCA[See part (i)] ∆ABP ≡ ∆CBP[AAS] 2006 HSC Question 6
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20 (iii)Prove that ABCD is a rhombus.3 APB = BPC [corresponding angles in congruent triangles – part ii] APB + BPC = 180 o [straight angle] 2 x BPC = 180 o BPC = 90 o = APB Diagonals bisect at 90 o [ Square or Rhombus ????] 2006 HSC Question 6
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21 2005 HSC Question 5
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22 2004 HSC Question 2
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23 2004 HSC Question 6
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