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

1Press Ctrl-APress Ctrl-A

©G Dear2008 – Not to be sold/Free to use

Stage 6 - Year 12Stage 6 - Year 12

Mathematic(HSC)

2

Types of Angles

1. Acute angles 2. Right Angle

3. Obtuse Angle

4. Straight Angle 5. Reflex Angle

6. Angle of Revolution

(0o < θ < 90o θ = 90o

(90o < θ < 180o)

(θ = 360o)

(θ = 180o)

(180o < θ < 360o)

3

1. Vertically Opposite Angle are equal.

2. Complementary Angles ao

bo add to 90o.

2. Supplementary Angles ao

boadd to 180o.

Pairs of Angles

4

Transversal1. Alternate Angles

2. Corresponding Angles

3. Co-Interior angles

Makes a Z shape.

Makes a F shape.

Makes a C shape.

andare equal.

andare equal.

andAdd to 180o

Angles between Parallel Lines

5

Based on SidesBased on Angles1. Equilateral triangle.

•All sides equal•All angles equal (60O)

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 90o

•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.

ao

bo

co

ao + bo + co = 180oao

bo co

do

ao + bo + co + do = 360o

bo

co

ao

ao = bo + coao

bo

co

ao + bo + co = 360o

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

αβ

γγ

2. Corresponding sides are in the

same ratio.

a x

ax

b

y

by

= = cz

c z

3. Two pairs of sides are in proportion and their included

angles are equal.

pq

s

rθ

φ

pr

= qs

=

Similar Triangles

9

AB : BC = DE : EF

A

F

E

D

C

B

ABBC EF

DE=

Ratio of Intercepts

10

ca

b

c2 = a2 + b2

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. Rectangle 2. Square 3. Rhombus

4.Parallelogram 5. Trapezium

6. Kite

You must know their properties

Types of Quadrilaterals

12

1. Triangle 2. Square 3. 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 360o.

Regular Polygons

14

Area Formulae

1.SquareA = s2 ss

2. RectangleA = LB L

B

3.TriangleA=½bh

b

h 6.TrapeziumA=½(a+b)h

b

a

h

7.CircleA=πr2

r

4. ParallelogramA=bh

b h

5.Rhombus/KiteA=½xyx y

xy

15

1. Rectangular Prism

lb

h

SA = 2(bh + hl + lb)

2. Cube

s

SA = 6s2

3. Sphere

SA = 4 π r2

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

lb

h

V = lbh

2. Cube

s

V = s3

3. Sphere

V = 4 π r3

34. Cylinder

V = π r2 h

5. Cone

h

r

h

rV = 1 π r2 h 3

V = Ah

Volume Formulae

17

2006 HSC Question 6

18

BCA = CAD [Alternate angles between parallel lines.]

BAC = CAD [Given]

(i) Prove that BAC = BCA 1

BAC = BCA

2006 HSC Question 6

19

BP [Common]

PBA = PBC [Given]

(ii) Prove that ∆ABP ≡ ∆CBP 1

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 = 180o [straight angle]

2 x BPC = 180o

BPC = 90o = APB

Diagonals bisect at 90o [ Square or Rhombus ????]

2006 HSC Question 6

21

2005 HSC Question 5

22

2004 HSC Question 2

23

2004 HSC Question 6