Mathematics Tasks - Math Problem Example

Running Head: ICT MATHS Discuss your own work on three of the mathematics tasks in Chapters 2, 3 and 4, using no more than one task from any one chapter Name: University: Course: Date: Tutor: Date: Working on mathematical tasks involving geometric thinking, as experienced in various tasks, falls under different levels of difficult from learner to learner. For example, there are problems those mere calculations in order to arrive at conclusions, while there are problems that need applications of mathematical concepts and problem solving techniques in order to arrive at a solution. These latter problems may be tricky and hard to conceptualize but with a clear basis and knowledge of different types of problems and techniques to apply, they can be solved easily. Invariance in relation to the tasks worked on before can be said to be concepts or ideas that remain constant and do not change. Qn 1. Similarities and Ratio According to Chapter 2, task 2.2.1 on similarities and ratio, so many mathematical tasks can be developed and solved. A mathematical task like the one below entails many more mathematical objects apart from geometry because algebra can also be involved. Fig. 1.1 D B C 52 13 G A F E In fig. 1.1 above AB = 13 cm, DE = 52cm and DF = 48cm. AB is parallel to DE. To find the perimeter of CDE, and that of ABC, similarity must be applied. Noting that ABC is similar to CDE, then the ratio between DE and AB is 52/13 = 4. This is the linear scale factor that has been applied on triangle ABC to enlarge it to triangle CDE. Therefore line BG then is calculated by dividing line DF by the L.D.F. i.e. 48/4 = 12cm To get the length of BC, Pythagoras theorem is applicable which can be based on structural algebra (Cundy, Rollett,1989). This verifies that the length of AC = 14cm, BC = 15cm while that of AG is 132 = AG2 +122 169 - 144 = AG2 25 = AG2 5 = AG The length of CG 52 = CG2 + 122 225 - 144 = CG2 181 = CG2 9 = CG The perimeter of ABC then becomes 13+14+15 = 42cm To get the length of CD, AC is multiplied with the L.S.F. which is 4. Then we have 14 x 4 -= 56cm Similarly, CD = 15 x 4 = 60 cm The perimeter of CDE then becomes 60+52+56 = 168cm Measurement as a mathematical object is also very necessary in this mathematical task where the measurement units are given in centimeters. Mathematical process like connection where mathematic ideas interconnect and build on one another to produce a coherent whole has highly been applied on this task. Mathematical processes like representation is also very important as it helps the learner to organize, record, and communicate mathematical ideas. Geometric reasoning. (ii) Chapter 3 is about geometric reasoning. Basing on task 3.3.3 the following mathematical task has been formulated. B A C To construct a circle inside the triangle to touch the side of the triangle which is generally termed as an inscribed circle, then there is a lot of geometry involved (Bates, 1979). This can be done as follows: - (a) Bisection of angle BAC and extending the bisecting line to a point, p. (b) Bisecting the angle ACB and extending the bisector to a point, m. Where the two bisecting lines meet, say point O, becomes the centre of the circle. With a pair of compasses, you pin at the meeting point of the two lines and draw a circle that touches the vertices of the triangle ABC. This is what is called an inscribed circle. This can be shown in fig,1.2 below: - B o A C Measurement as a mathematical tool is highly used here. The length of the sides of the triangle can be measured as well as the radius of the inscribed circle. Reasoning and proof are mathematical processes that are also applied on this mathematical task. It is only through concrete reasoning and proofing what you reason by for example bisecting the angles that one comes to see that the circle touches the sides of the triangle but not the vertices. Use of mathematical tools like a pair of compasses, calculators and rulers has also been highly applied to solve this mathematical task. Mathematics thinking and language is highly exercised and mathematics ideas expressed precisely. Seeing Solids as revolutionary. (iii) In chapter 4 which is on seeing solids as revolutionary and example of a mathematical task formulated is that of calculation of the volume of a cone which can be created or produced by rotating a symmetric trapezium around its axis of symmetry. The cone can be represented as shown in figure 1.3 below: - Source: http://www.math.com/school/subject3/lessons/S3U4L4DP.ht V = Bh, where B is the base area Since the base of the cone is a circle, then we solve the base using the area of a circle which is A = r2 The volume of a cone then becomes; V = r2h For example, suppose the cone had a height of 9cm and a radius of 2cm, then the volume will be V = (3.14) (2)2 x 9cm = 37.68 Cubic centimeters Measurement is a very important mathematical object which should be kept in mind while tackling this mathematical task. This is evidenced by different units in measurements of the area and volume which are square centimeters and cubic centimeters respectively. Q2. (a) Picking on Q1, the mathematical task on similarities and ratio, for the first time it was learnt that similarity has a correlation with enlargement. A constant value notably linear scale factor or area scale factor is used to relate the two. In this case, L.S.F. was found to be 4. About the mathematical task on geometric reasoning, it has been learnt for the first time that bisection of two of the angles of the triangle and extending the bisecting lines to meet at a certain point, then treating the point as the centre of the circle gives an inscribed circle, and not through bisecting the lines of the triangle which leads to escribed circles. This is a circle that touches the vertices of the triangle as follows: - B A C (b) In the task for similarity and ratio, ICT was used to recognize the figure1.1 and realize that it is made of two triangles. ICT has also helped to know the relationship between the two triangles and their behavior. It has also helped in making generalizations that are based on mathematical vocabulary. It has also helped in development of vocabulary and language such as enlargement, similarity and linear scale factor. In this mathematical task, ICT has also been helpful in that it has enabled in accessing, selecting and interpreting of the information given in the instructions. Invariance and Change Invariance and change are expressed in that; a constant factor has been used to multiply the lengths of the small triangle ABC in order to obtain the larger triangle CDE. Hence there is no change between the two except in lengths. In other words, the triangles ABC and CDE are similar. ICT has influenced a lot in determining this invariance and change because it has helped in acquiring of information and interpreting it. (c) Geometry as a course construct is used to help the learner be able to analyse characteristics and properties of geometrical shapes for example two or three dimensional geometric shapes. It is also helpful in enabling the learner to develop mathematics arguments about geometry relationships, that is, to develop logical thinking abilities (Freudenthal, 1991). The learner is also able to specify locations and describe spatial relationships for example using co-ordinate geometry and other representational systems. An example of geometry is Euclidean geometry which the learner may use to be able to study the properties within plane geometry (this involves two dimensional geometry). These properties include angle properties, relationships and similarities. REFERENCES Bates, T. (1979) 'Triangles and squares', Mathematics Teaching, 88: 38-41. Cundy, H. and Rollett, A. (1989) Mathematical Models, 3rd ed. Stradbroke, England: Tarquin Publishers Freudenthal, H. (1991) Revisiting Mathematics Education: China Lectures. Dordrecht: Kluwer Academic. http://www.math.com/school/subject3/lessons/S3U4L4DP.ht Accessed on 6 November 2006. Read More