Announcing the release of VERSION 6 •750 MB of Presentations •500+ files/10 000+ Slides •1200 Interactive SAT/GCSE Boosters •Huge Enrichment Area •New Functional Maths Element •100’s of Worksheets •100’s of teacher Q + A Sheets •1000’s of Teaching Hours •Database of Presentations This Demo shows just 20 of the 10,000 available slides and takes 7 minutes to run through. Please note that in the proper presentations the teacher controls every movement/animation by use of the mouse/pen. Click when ready Whiteboardmaths.com Stand SW 100 © 2004 - 2008 All rights reserved Click when ready In addition to the demos/free presentations in this area there are at least 8 complete (and FREE) presentations waiting for download under the My Account button. Simply register to download immediately. This is a short demo that auto-runs. Drawing the graph of a quadratic function? 8 7 LoS y Drawing quadratic graphs of the form y = ax2 + bx + c Equation of Line of symmetry is x = 1 6 5 Example 1. 4 y = x2 - 2x - 8 3 2 x 1 -3 -2 -1 0 -1 -2 -3 -4 1 2 3 4 5 x -3 -2 -1 0 1 -7 -8 -9 3 4 5 x2 9 4 1 0 1 4 9 16 25 -2x 6 4 2 0 -2 -4 -6 -8 -10 -8 -8 -8 -8 -8 -8 -8 -8 -8 -8 y 7 0 -5 -8 -9 -8 -5 0 7 -5 -6 2 Minimum point at (1, -9) Look at graphs of some trig functions? The Trigonometric Ratios for any angle 90o -360 -270 180o 0o -360 -180 -90 -270 -180 -90 -450o -360o -270o -180o -90o 0 0 0o 90 180 270 360 90 180 270 90o 180o 270o 360o 450o 270o 1 sinx + circle 0o -1 90o 180o 270o 360 360o 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 -1 -2 90 180 270 360 Introducing addition of fractions with different denominators? 2 3 1 4 + 8 12 = + 3 12 11 12 Multiples of 3 and 4 3 4 6 8 9 12 12 16 15 20 12 is the LCM Probability for Dependent Events 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. 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 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) Enlarge on object? Enlargements from a Given Point Centre of Enlargement To enlarge the kite by scale factor x3 from the point shown. B A Object C B/ 1. Draw the ray lines through vertices. D C/ A/ Image No Grid 2 D/ 2. Mark off x3 distances along lines from C of E. 3. Draw and label image. Do some Loci? 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. Scale:1cm = 3m Wall 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. Investigate some Properties of Pascal’s Triangle Pascal’s Triangle 1 1 1 1 Blaisé Pascal (1623-1662) 1 1 1 1 1 1 1 1 1. Complete the rest of the triangle. 1 7 8 9 6 2 3 4 5 1 1 6 10 10 Counting/Natural Numbers 1 3 4 1 5 15 20 15 1 6 21 35 35 21 Triangular Numbers 1 1 7 28 56 70 56 28 36 84 126 126 84 36 1 8 9 Tetrahedral Numbers 1 10 45 120 210 252 210 120 45 10 1 Pascal’s 1 66 220 495Triangle 792 924 792 495 220 66 12 1 11 55 165 330 462 462 330 165 55 11 12 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 Pyramid Numbers (square base) Add the numbers shown along each of the shallow diagonals to find Leonardo of Pisa another well known - 1250 sequence1180 of numbers. The Fibonacci Sequence 1 1 1 1 1 2 3 1 1 3 1 2 3 5 1 1 1 The sequence first appears as a recreational maths problem about the growth in population of rabbits in book 3 of his famous work, Liber – abaci (the book of the calculator). 10 45 120 210 252 210 120 45 10 1 Fibonacci 1 66 220 495 792 924 792 495 220 66 12 1 Sequence 11 55 165 330 462 462 330 165 55 11 12 13 21 34 55 89 144 233 377 Fibonacci travelled 1 4 6 4 1 extensively throughout 1 5 10 10 5 1 the Middle East and elsewhere. He strongly 1 6 15 20 15 6 1 recommended that 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 1 9 36 84 126 126 84 36 9 1 zero “zephirum” 1 8 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 National Lottery Jackpot? Row 0 12 7 31 49C 6 39 16 4 9 46 34 2 29 42 5 There are 13 983 816 ways of choosing 6 balls from a set of 49. So buying a single ticket means that the probability of a win is 1/13 983 816 10 28 25 1 13 43 36 24 3 45 15 33 38 30 17 20 11 49 21 49 balls choose 6 Choose 6 35 14 37 19 22 44 40 32 47 8 26 23 6 41 18 27 Row 49 13 983 816 48 The Theorem of Pythagoras? A 3rd Pythagorean Triple In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. 49 625 7, 24, 25 25 7 24 576 72+ 242= 252 49 + 576 = 625 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 Look at one of the 6 proofs of the Theorem from the Pythagorean Treasury. 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 Area of trapezium Draw line:The boundary shape is a trapezium = ½ (a + b)(a + b) = ½ (a2 +2ab + b2) Area of trapezium is also equal to the areas of the 3 right-angled triangles. yo = ½ ab + ½ ab + ½ c2 c xo a b c ½ (a2 +2ab + b2) = ½ ab + ½ ab + ½ c2 a2 +2ab + b2 = 2ab + c2 yo b So a2 + b2 = c2 xo QED a Take 1 identical copy of this right-angled triangle and arrange like so. Sample some material from the Golden section presentation. THE GOLDEN SECTION Constructing a Golden Rectangle. 1. Construct a square and the perpendicular bisector of a side to find its midpoint p. 2. Extend the sides as shown. M L Q 3. Set compass to length PM and draw an arc as shown. 1 O P LQRO is a Golden Rectangle. N R 4. Construct a perpendicular QR. 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 Or just simply ride your bike! Wheels in Motion The Cycloid It’s true! The point at the bottom of a moving wheel is not moving! Wheel Choose Order Forms/New for V5 to view latest material and other catalogues. Whiteboardmaths.com Stand SW 100 © 2004 - 2008 All rights reserved Click when ready Don’t forget to pick up your 8 free presentations.