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Stand SW 100
© 2004 - 2007 All rights reserved
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3. This is a short demo
that auto-runs.
Drawing the graph of
a quadratic function?

4. 8
y
7 Equation Drawing quadratic graphs of
LoS
6
of Line of the form y = ax2 + bx + c
symmetry
5 is x = 1
Example 1.
4
3
y = x2 - 2x - 8
2
x -3 -2 -1 0 1 2 3 4 5
1
x
-3 -2 -1 0 1 2 3 4 5 x2 9 4 1 0 1 4 9 16 25
-1
-2x 6 4 2 0 -2 -4 -6 -8 -10
-2
-3 -8 -8 -8 -8 -8 -8 -8 -8 -8 -8
-4
y 7 0 -5 -8 -9 -8 -5 0 7
-5
-6
Minimum point
-7
-8
at (1, -9)
-9

5. Look at graphs of
some trig functions?

6. 90o
The -360 -270 -180 -90 0 90 180 270 360
Trigonometric
Ratios for any
180o 0o
angle -360 -270 -180 -90 0 90 180 270 360
-450o -360o -270o -180o -90o 0o 90o 180o 270o 360o 450o
270o
1
sinx + circle
0o 90o 180o 270o 360o
θ
-1

7. y = f(x)
f(x) = cosx f(x) = cos2x f(x) = cos3x f(x) = cos ½ x
2
1
x
-360 -270 -180 -90 0 90 180 270 360
-1
-2

8. Introducing addition
of fractions with
different
denominators?

9. 2 1
+
3 4
Equ
t
len
3 va
qui
ival
+
E
8
ent
Multiples of
12
3 and 4
12 3 4
6 8
11
9 12
=
12 16
12
15 20
12 is the LCM

10. Probability for
Dependent Events

11. Conditional Probability: Dependent Events
When events are not independent, the outcome of earlier
events affects the outcome of later events. This happens in
situations when the objects selected are not replaced.

12. Conditional Probability: Dependent Events
A box of chocolates contains twelve chocolates of three different types.
There are 3 strawberry, 4 caramel and 5 milk chocolates in the box. Sam
chooses a chocolate at random and eats it. Jenny then does the same.
Calculate the probability that they both choose a strawberry chocolate.
P(strawberry and strawberry) = 3/12 x

13. Conditional Probability: Dependent Events
A box of chocolates contains twelve chocolates of three different types.
There are 3 strawberry, 4 caramel and 5 milk chocolates in the box. Sam
chooses a chocolate at random and eats it. Jenny then does the same.
Calculate the probability that they both choose a strawberry chocolate.
P(strawberry and strawberry) = 3/12 x 2/11 = 6/132 (1/22)

14. Enlarge on object?

15. Enlargements from a Given Point
Centre of Enlargement
To enlarge the kite by
B
scale factor x3 from the
point shown.
A Object C B/
1. Draw the ray lines
through vertices.
D 2. Mark off x3 distances
C/ along lines from C of E.
A /
Image
3. Draw and label image.
No Grid 2 D/

16. Do some Loci?

17. Loci (Dogs and Goats)
Q2
Billy the goat is tethered by a 15m long chain to a tree at A. Nanny the goat is
tethered to the corner of a shed at B by a 12 m rope. Draw the boundary locus
for both goats and shade the region that they can both occupy.
Wall Scale:1cm = 3m
A
Shed
B
Wall
1. Draw arc of circle of radius 5 cm
2. Draw ¾ circle of radius 4 cm
3. Draw a ¼ circle of radius 1 cm 4. Shade in the required region.

18. Investigate some
Properties of
Pascal’s Triangle

19. Pascal’s 1 1. Complete the rest of the triangle.
Triangle 1 1
1 2 1
1 3 3 1 Counting/Natural Numbers
1 4 6 4 1
Blaisé Pascal
(1623-1662) 1 5 10 10 5 1
Triangular Numbers
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1 Tetrahedral Numbers
1 10 45 120 210 252 210 120 45 10 1
1 Pascal’s
11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495Triangle
792 924 792 495 220 66 12 1 Pyramid
Numbers
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 (square
base)

20. 1 The Fibonacci Sequence
Add the numbers shown
along each of the 1 1 1 1 2 3 5 8 13
shallow diagonals to find
Leonardo of Pisa
another well known 1 2 1 21 34 55 89 144 233 377
1180 - 1250
sequence of numbers. 1 3 3 1
The sequence first appears as a
Fibonacci travelled 1 4 6 4 1 recreational maths problem
extensively throughout about the growth in population
the Middle East and 1 5 10 10 5 1 of rabbits in book 3 of his
elsewhere. He strongly famous work, Liber – abaci (the
recommended that
1 6 15 20 15 6 1 book of the calculator).
Europeans adopt the 1 7 21 35 35 21 7 1
Indo-Arabic system of
numerals including the 1 8 28 56 70 56 28 8 1
use of a symbol for
zero “zephirum” 1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 Fibonacci
11 55 165 330 462 462 330 165 55 11 1
1 12 Sequence
66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1

21. National Lottery Jackpot? Row 0 49 balls choose 6
12 49 11 33 15
21 38
7 30 36
31 25
17
24
20 1 35
45 3 14
13
49
C6
37
19
43 39 22
16 40 44
9 4
46 There are 13 983 816 ways of 47 32
34 choosing 6 balls from a set of 8
2
49. So buying a single ticket
29 means that the probability of a 23 26
5
42 win is 1/13 983 816 6
10 41
18
28 Choose 6
27
Row 49
13 983 816 48

22. The Theorem of
Pythagoras?

23. A 3 rd Pythagorean Triple
In a right-angled triangle,
the square on the
625
hypotenuse is equal to
the sum of the squares 7, 24, 25
on the other two sides.
25
49 7
24
576 7 2
+ 24 2
= 25 2
49 + 576 = 625

24. The Theorem of Pythagoras: A Visual Demonstration
In a right-angled triangle,
the square on the
hypotenuse is equal to the
sum of the squares on the
other two sides.
Henry Perigal
(1801 – 1898)
Perigal’s
Dissection
Gravestone
Inscription
Draw 2 lines through the centre of the middle square, parallel to the sides of the large square
This divides the middle square into 4 congruent quadrilaterals
These quadrilaterals + small square fit exactly into the large square

25. Look at one of the 6 proofs
of the Theorem from the
Pythagorean Treasury.

26. President James Garfield’s Proof(1876) To prove that a2 + b2 = c2
We first need to show that the angle between
angle x and angle y is a right angle.
•This angle is 90o since x + y = 90o (angle sum of a
triangle) and angles on a straight line add to 180o 
Draw line:The boundary shape is a trapezium Area of trapezium
= ½ (a + b)(a + b) = ½ (a2 +2ab + b2)
yo Area of trapezium is also equal to the
areas of the 3 right-angled triangles.
= ½ ab + ½ ab + ½ c2
c b So
xo ⇒ ½ (a2 +2ab + b2) = ½ ab + ½ ab + ½ c2
c
a ⇒ a2 +2ab + b2 = 2ab + c2
yo xo ⇒ a 2 + b2 = c2 QED
b a
Take 1 identical copy of this right-angled triangle and arrange like so.

27. Sample some material
from the Golden section
presentation.

28. THE GOLDEN SECTION
Constructing a Golden Rectangle.
1. Construct a square and the
2. Extend the sides as shown.
perpendicular bisector of a side
to find its midpoint p.
L M Q
3. Set compass to
1 length PM and draw
an arc as shown.
O P N R
4. Construct a
LQRO is a Golden Rectangle. perpendicular QR.

29. THE GOLDEN SECTION
"Geometry has two great treasures: one is the Theorem of
Pythagoras, and the other the division of a line into extreme
and mean ratio; the first we may compare to a measure of
gold, the second we may name a precious jewel."
Johannes Kepler
1571- 1630

30. Or just simply ride
your bike!

31. Wheels in Motion
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The Cycloid Order Forms/New for V5
to view latest material
and other catalogues.
It’s true! The point at the bottom
of a moving wheel is not moving!
Wheel

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Stand SW 100
© 2004 - 2007 All rights reserved
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