The concept of decimals are presented in grid format to enhance the understanding of decimals. The subtle difference between 0.3 and 0.30 is made approachable to younger pupils of primary school age. The idea of money is used to explaining that. For example, while the former represents 3 ten pence coins, the latter represents 30 one pence coins. In more advanced way, one may explain it in the following way: While in 0.3, one is not quite sure whether the digit 3 is the result of rounding up or down, whereas in 0.30, one will have the confidence that it is not. Another way of looking at it is, that, if these figures were the results of measurements, then, 0.30 is more accurately measured than 0.3

Unit circle is very useful to find the values of sine, cosine and tangent of angles such as 0, 90º, 180º, 270º or 360º degrees without a calculator. One can also use equilateral triangle to find the values of sine, cosine and tangent of 30º 60º . The sine, cosine and tangent of 45º can be found using isosceles right angle triangle. Proof of some trigonometric identities are also provided. Examples on how to use the unit circle and exercises on its applications and solutions to the exercises are also included.

A line segment is divided into to parts. The division of the segment is carried out in different ratios. The line segment need not be horizontal or vertical. Three coordinate, P(a,b), Q(x, y) and R(c, d) that are on a line segment are related in such a way that PQ:QR = m:n After introduction, three different examples are provided. Given P(a, b) and R(c, d) and PQ:QR, the coordinates of Q(x, y) is be found. All three coordinates are given and the ratio of PQ:QR is sought. The intermediate coordinate, Q(x, y), the ratio PQ:QR and one other coordinate is given to find the third coordinate. This qualifies for full lesson. Additional exercises with answers provided.

The exact relationship between degrees Celsius and Fahrenheit is to be established. There are two parts of the lesson. One is to establish the relationship and the other is to compare the established formula with an estimate. The exact formula is F = 1.8C +32, whereas the estimate formula is F = 2C + 30 ( in words, double and add 30). Straight line graphs are to be plotted on the graph paper provided. The steps of the derivation are given, but require explanations as to why each step follows the previous step, until one arrives at the required exact relationship. Note: While the ice (freezing) and the steam (boiling) points of pure water are 0 and 100 degrees Celsius respectively, the corresponding points for the Fahrenheit scale are 32 and 212 degrees. Points of discussion: At 10 degrees Celsius, both the exact and the estimate formula produce the sam value of degrees Fahrenheit - 50 degrees Fahrenheit Impurities will lower the freezing point of water and raise the boing point. Work out the one particular temperature at which both scales read the same value.

An interactive mortgage calculation using Excel. The inputs are: the value of the property, current interest rate (enter as decimal), the number of years the mortgage lasts (years) and the period (12 monthly payments in the year). The output is the monthly payment of the mortgage (PMT in red = monthly payment of the mortgage). Proof of the mortgage formula is also included. Note: this exercise has motivational value that each student may consider themselves as mortgage consultants. Students may be asked to investigate by how much the monthly payment of the mortgage change when the interest rate becomes 15% instead of say 3% or 1.6% etc. (historically this has once happened during Mrs Thacher Premiership ) and the likely consequence of it - perhaps re-possession.

The aim is to establish, experimentally, the value of ‘pi’. The relationship between the circumference and the diameter of a circle is to be investigated. Pupils are encouraged to measure the circumference and diameter of several round objects. They record their measures in the table provided. A plot of circumference against diameter is to be done and gradient (slope) to be measured , which is the approximate value of pi. The instructions how to carry out the experiment is given in one of the sheets. There is also small exercise on other worksheet. Points of discussion: Is the mean better summary value than individual measures? Discussion on measurement errors To try to find the mean of all the means of whole class

It may be of interest of find the area of a triangle given three sides. This methods does not require to know the base and the height of the triangle (the basic definition of finding the area of a triangle), nor an angle in order to use trigonometry to find the area of a triangle. The type of triangle need not be right angled or any other special type of triangle - it can be of any type. Here though, for simplicity, we used right angled triangles, in the exercises given, to make it easier to use the traditional way of finding the area of a triangle - 1/2 x base x height in order to confirm the Heron’s formula works. The proof of Heron’s formula is given in a separate worksheet.

Material suitable for a complete lesson in sequences are provided in two worksheets. One deals with a differentiated lesson and the other, the more involved one, is to find the nth term of a triangular sequence. The steps arriving at the nth term of the triangular sequence are given, but the pupils are expected to answer as to why each step is true, until one arrives at the last step. Because of the inductive nature of the process, it is good practice to check if the generalised expression of the sequence works for some other situations of the sequence number.

A bundle of theorems are included here. They are: angle in a semicircle, angle in a major and minor sector, cyclic quadrilateral, angle subtended by an arc, tangent to circle property and alternate segment theorem. These materials are good for discussion and exercises can always be found in text books. The focus here is on concepts. Also, there are applications of perpendicular and angle bisectors as applied to the sides and angles of triangles to achieve circumscribed and inscribed circles. This can enhance and perfect the skills required to carry out perpendicular and angle bisectors.

A pair of compasses and a ruler only is to be used. After pupils get familiar in carrying out angle bisectors and perpendicular bisectors, they are challenged to apply those skills in finding the following specific angles: 60, 30, 45. The most challenging exercise is to construct a perpendicular line to a given line and the perpendicular line must pass through a given point. For example: a perpendicular line on line a segment AB, which is not horizontal, and from a point P above the line segment AB, is required to be constructed. The following knowledge is required: The angle bisector of an isosceles triangle can also become perpendicular bisector. A line from the centre of circle to the midpoint of a chord forms 90 degrees. The radius of a circle to a tangent to the circle will form 90 degrees at the point of contact.

Spreadsheets (excel) is used to solve a cubic equation, x^3 - 5x + 3 = 0; the equation is solved iteratively by Newton Raphson method. A plot of line chart is carried out and the cart is the visualization of six convergent series, which provide the three approximate roots of the equation. The detail is to found on the worksheet. The three approximate solutions are: -2.49086, 1.834243, 0.65662

This sheet enhances the concept of rotation. The rigid body to be rotated is a triangle. Rotating the three corners of the triangle is sufficient to rotate the whole of the object (the triangle). The distance from the centre of rotation to the points of interest (the corners of the triangle) of the object and the corresponding point of the image are equal, and rotating it through an angle of (90 degrees in this case) will locate the position of the image, and since it is a rigid body all the regions bounded by the corners of the triangle will have been moved from the object position to the image position. The above note is for the teacher but the worksheet is simply to enhance the understanding of the concept of rotation and it is left for the pupil to summarise their observations.

There are three worksheets in which pupils are challenged to find the areas of the shaded parts. The challenging bit is to justify why, in one the figures, the diameter of the square and that of the circle. The answer lies in the application of the angle in a semicircle. Knowledge required: How to find area of a triangle How to find area of a circle Use of Pythagoras theorem

Interactive puzzle that is engaging. It gives pupils the opportunity to solve a problem that requires flexible thinking. The problem is simple. Numbers form 1 to 9 is used, but each number is used only once. The goal to achieve is the sum of each row and column, and diagonally, should be 15. There are four different ways to achieve that. Note: for 3 x3 magic square 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 45 /3 = 15 ( 3 rows or columns share the 45) For 5 x 5 magic square the sum of each row, column or diagonal is 65. For 7 x 7 the sum is 175. The rule: Enter 1 in the middle of the first row. Go through right corner (small arrow line through the corner will help). Once outside the boundary of the main square fill the next number at end of that row or column. Keep on going through right corner and if you find empty square fill the next number there, but if you find it is already occupied, fill the next number in the square immediately below. Exception: since there there is no row or column after filling top right corner square, fill the next number in the square immediately under it. The general formula for the row sum is: Last x (Last + 1) /( 2*R) where R = Number of rows Example, for 3 x 3 magic square (9 x 10 )/(2 x 3) =15 for 5 x 5 magic square (25 x 26)/(2 x 5) = 65 for 7 x 7 magic square (49 x 50)/(2 x 7) = 175 Hint: One could use the 5 x 5 and 7 x7 magic squares to practice mental arithmetic, using the idea of number bonds.

Reversing the direction of the input and output in a number machine, the operator changes ( + into - and x into ÷ and vice versa). A number of exercises are provided. The lesson is suitable for KS2 pupils. The commutative property comes handy when solving a problem that seems to defy common sense; that is, when dividing a number by another the usual expectation is the answer to being smaller in magnitude than the number to be divided. Similarly, when multiplying two numbers one expects the answer to be larger in magnitude than both numbers that are to be multiplied. This puzzling situation arises because of the fact that dividing or multiplying by a number between -1 and 1 produces that effect. A challenging situation is when one asks, what is 3 divided by a number and the answer is 4. Using number machines idea and the commutative property one can get the answer. This is explained in one of the worksheets, and the interesting thing is, it is not beyond the level of KS2 pupils to understand it. It is suggested that a calculator should be used first to explore the answer - in the interest of enhancing problem solving culture.

After multiplying or dividing by 10, 100 or 1000 a pattern is observed and pupils are expected to summarise their observations. A skill is then developed to aim to change the denominator, if possible, to make 10, 100 or 1000 from which one will find it easy to write down the corresponding decimals. For example, 1/2 = 5/10 = 0.5; 1/4 = 25/100 =0.25 and if this is difficult, try, 1/4 = 0.5/2 = 2.5/10 = 0.25; or 1/8 = 0.5/4 = 0.25/2 = (.25 x 5)/(2 x 5) =1.25/10 = 0.125 Note, one could move in a convenient direction to achieve the denominator to become 10, 100 or 1000. More explanations and answers are provided.

The purpose of the investigation is to establish, through observation of patterns, the exterior angle of a regular polygon is equal to 360 degrees divided by the number of sides of the regular polygon (i.e. ext = 360/n). Then applying their knowledge of angle at a point is equal to 360 degrees, students will be challenged to find the number of sides of a larger regular polygon, given one or two other regular polygons in a tessellation.

This visual proof of Pythagoras theorem is an adaptation of an excellent animation by Alan Kitching. Search “YouTube” under the title “Pythagoras in 60 seconds”, it is great fun to see.

Theoretical out come of the sum of two dice is first carried out. This is then plotted to observe the binomial distribution characteristics - reflecting in the limit, the normal distribution shape. This is achieved by using all 36 probable outcomes. However, when a simulation is carried out (experiment), the binomial distribution shape is achieved only if the sample size is increased. This experiment reinforces the importance of sample size when probability investigation is carried. All the probable outcome of the throw of two dice is: (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2.1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) Experimentally (simulation), 36 throws will unlikely to achieve the the theoretical outcome probabilities, unless the sample size is increased a great deal. Again what the optimum sample size would be, will be a question of compromise, as the sample size is increased, the more probable it becomes to achieving the shape of the theoretical distribution. But, one has to keep in mind the cost attached to unnecessarily large data size.

A sequence right angled triangles are presented. The smallest sides of the triangles are represented by odd integers - the smallest in this case chosen to be 3. The challenge is when only one side (the smallest) is given and then finding the other two sides. It is expected the pupils to notice the other two lengths differ by 1. Then Pythagoras theorem is used to find the other two lengths. For example, 4th pattern in the sequence, n is 4, so the smallest side is 9. Then, to find the other two lengths: 9^2 + x^2 = (x + 1)^2 81 + x^2 = x^2 + 2x + 1 81 = 2x + 1 2x = 80 x = 40 Filling column 2 is easy, but filling 4 and noticing the pattern of the sequence is a little challenging. In column 3 the following sequence should be found and factorised to notice triangular number sequence, whose generalisation is another challenge. The sequence in column 4 is: 4, 12, 24, 40, 60, 84 etc. This sequence can the be rewritten as: 4 x 1, 4 x 3, 4 x 6, 4 x 10, 4 x 15, 4 x 21 … So now, it becomes obvious the sequence summarises to 4 x triangular number sequence, i.e., 4 x n(n+1)/2 = 2n(n + 1). The derivation of the triangular number sequence formula maybe left as exercise for pupils to establish.

The properties of parallel lines in which, what corresponding, alternating and vertically opposite angles mean. Students are given the opportunity to measure these angles and familiarize themselves with these terms. Then, these properties are used to prove that the sum of the angles of a triangle add up to 180 degrees.