1 Press Ctrl-A ©G Dear2008 – Not to be sold/Free to use Stage 6 - Year 12 Mathematic (HSC)

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)

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

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

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

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

7 1. Side, Side, Side.2. Side, Angle, Side. 4. Right angle, Hypotenuse, Side. 3. Angle, Angle, Side. SSS SAS AAS RHS Congruence

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

9 AB : BC = DE : EF A F E D C B AB BCEF DE = Ratio of Intercepts

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

11 1. Rectangle2. Square3. Rhombus 4.Parallelogram 5. Trapezium 6. Kite You must know their properties Types of Quadrilaterals

12 1. Triangle2. Square3. Pentagon 4. Hexagon 5. Octagon Types of Regular Polygons

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

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

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

16 1. Rectangular Prism l b h V = lbh 2. Cube s V = s 3 3. Sphere V = 4 π r Cylinder V = π r 2 h 5. Cone h r h r V = 1 π r 2 h 3 V = Ah Volume Formulae

HSC Question 6

18  BCA =  CAD[Alternate angles between parallel lines.]  BAC =  CAD[Given] (i)Prove that  BAC =  BCA1   BAC =  BCA 2006 HSC Question 6

19 BP[Common]  PBA =  PBC[Given] (ii)Prove that ∆ABP ≡ ∆CBP1  BAC =  BCA[See part (i)]  ∆ABP ≡ ∆CBP[AAS] 2006 HSC Question 6

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

HSC Question 5

HSC Question 2

HSC Question 6