Class 08 Project 02

VALUE OF

BACKGROUND

In a circle, the ratio of the circumference to its diameter is a constant and is

denoted by Greek lettern. For calculation purpose, value of is generally

taken as or 3.14.

OBJECTIVE

In this project, an attempt has been made the value of using simple different met NCERT

DESCRIPTION

Method-1

Take a circular disc. Mark a point on the edge of the disc. Take a cardboard of convenient size and paste a white paper on it. Draw a line on the paper. Mark a point P on the line. Place the card on the line such that the point marked on it touches the point marked on the line. (Fig. 1). Now roll the card along the line till its marked point again touches the line at some point Q (Fig. 2).

Mathematics

PQ is the circumference of the disc. Denote it by c. Measure the diameter of

the circular disc and denote it by d. Find the ratio which is equal to t

293

Thus n=

On actual measurement:

c=21.9 cm

d= 7.0 cm

So, x=

Method-2:

A Take a squared paper of dimensions say 20 cm x 20 cm. of radius say 8 em on it as shown in Fig. 3. draw a circle

21.9

= 3.13 approx.

50/5

1:00 1

4ill 57% 4.57%

рах

LTE B

← Activities(...

Count the number of complete squares inside the circle and denote it

by a

Count the number of squares through which the circle passes and denote

it by h Find +

In this case, a = 14 x 6+12x4+ 10 x 2+6x2=164

b= 60

b So a+ 164+ 60 = 164 +30= 194

2

That is area of the circle 194 sq. units approx.

Divide a+ by (radius)", to get approximate value of r

2

Fig. 4

295

Mathematics

194 1.e. 3.03125 (approx.).

194 64

8 Again draw a circle of radius 10 cm on the squared paper and repeat

B

the above process to find value of .

In this case,

a = 18 x8 + 16 x 4 + 14 x 2 + 12 x2+8x2 = 276

b= 68

68 So a+ =276+ 310

Thus = 310 310 3.1 approx.

10 100 As radius r increases, the value of it comes closer to 3.1

Method-3:

Take a squared paper of dimension say 20 cm

cm. L

Divide a by r" to get value of it.

radius say 10 cm on it as shown in Fig. 4. Count the number of vertices of the in the circles. Let it be

Un this case. fillos increases, the value of comes closer to 3.14.

= = 3.03125 (approx.

i.e. 194 194

8- 64

).

Again draw a circle of radius 10 cm on the squared paper and repeat

the above process to find value of a.

In this case.

a = 18x8 +16x4+14x2+12x2+8x2=276

b= 68

So a+=276+ 68 = 310

Thus

310

10

310

100

3.1 approx.

As radius r increases, the value of it comes closer to 3 

Method-3:

Take a squared paper of dimension 20 cm

radius say 10 cm on it as shown in Fig. 4.

Count the number of vertices of the s

Divide a by to get value of n

In this case, m rellished 3.17

317 317 1001

10

As radius r increases, the value of comes closer to 3.14.

CONCLUSION 

The ratio of the circumference and diameter of a circle is always constant

and is denoted by . Its approximate value is 3.14.

Class 08 Project 1 [PERIMETERS AND AREAS OF RECTANGLES]

Project 1 [PERIMETERS AND AREAS OF RECTANGLES]

BACKGROUND

The most common problems that people come across in daily life are to fence their fields and to determine the size of the field to grow a particular type of crop. These two problems are closely related with the mathematical concepts - perimeter and area, respectively. There is an impression that if the perimeter of a figure is increased, then its area will also increase and vice-versa. However, this impression is not always correct. This project tries to explain this idea by taking rectangle as a basic figure.

OBJECTIVE

To explore the changes in behaviours of perimeters and areas of rectangles with respect to each other.

DESCRIPTION

(A) Rectangles with equal perimeters

Let us consider the rectangles of the following lengths and breadths:

(1) Length = 10 cm : Breadth = 6 cm

Length = 11 cm: Breadth = 5 cm 

Length = 8 cm: Breadth = 8 cm

(IV) Length = 9.5 cm: Breadth = 6.5 cm

(V) Length = 8.5 cm : Breadth = 7.5 cm

We may note that perimeter of each of the above rectangles is the same.

namely. 32 cm. Let us calculate the areas of these rectangles:

[1] Area of the rectangle of length 10 cm and breadth 6 em

= 10 cm x 6 cm = 60 cm³.

(ii) Area of the rectangle of length 11 cm and breadth 5 cm

= 11 cm x 5 cm = 55 cm

(iii) Area of the rectangle of length 8 cm and breadth 8 cm

=8 cm x 8 cm = 64 cm² (Maximum)

(iv) Area of the rectangle of length 9.5 cm and breadth 6.5 cm

9.5 cm x 6.5 cm = 61.75 cm²

(v) Area of the rectangle of length 8.5 cm and breadth 7.5 cm

8.5 cm x 7.5 cm = 63.75 cm

From the above, it can be observed that in case (u), the rectangle has t maximum area. Also, in this case, the rectangle is a square.

(1)

(B) Rectangles with equal areas

Let us consider the rectangles with following lengths and breadths

Length = 16 cm: Breadth = 4 cm 

Length = 8 cm: Breadth = 8 cm

Length 10 cm: Breadth=6,4 cm

length-22 cm Breadth - 2 cm

length=64 cm: Breadth-1cm 

We may note that area of each of the above rectangles is the same namely

 64 cm². Let us calculate the perimeter of each of these rectangles: 

(1) Perimeter of rectangle of length 16 cm and breadth 4 cm

=2 (16+4) cm 40 cm

(1) Perimeter of rectangle of length 8 cm and breadth 8 cm

= 2 (8+8) cm = 32 cm (Minimum) (iii) Perimeter of rectangle of length 10 cm and breadth 6.4 cm

= 2 (10 +6.4) cm = 32.8 cm

(iv) Perimeter of rectangle of length 32 cm and breadth 2 cm = 2 (32 + 2) cm = 68 cm

(v) Perimeter of rectangle of length 64 cm and breadth 1 cm

= 2 (64+ 1) cm = 130 cm

 From the above, it can be observed that in case (ii), the rectangle has the minimum perimeter. Further, in this case, the rectangle is a square.

CONCLUSION

The impression that when the perimeter of a rectangle is increased, its area is also increased and vice-versa is not correct. In fact,

(i) of all the rectangles with equal perimeters, the square has the

maximum area. (ii) of all the rectangles with equal areas, the square has t inimum perimeter.

APPLICATION

1. This project is useful in estimating a field of maximum area within a given fencing (perimeter) and also a fencing (or boundary) of minimum length enclosed in a given area. T

2. A square is a regular quadrilateral. Increasing the number of sides of a regular polygon, we can arrive at the result in Description (A) above, that with a given perimeter, a circle has the maximum area.

3. If the measures of length and breadth are considered only in terms of natural numbers, then in Description (B) above, it can be seen that with a given area, the perimeter is maximum, when one of the sides is of unit length [case (v) of BJ. However, if this condition is not there, then there is no limit for the perimeter to be maximum. That is, for a given area, a rectangle can have infinitely large perimeter. nom Is it not surprising that you can make a belt of infinite length having a given area?

 Suggested  List  of Projects

Project 01

Class 08 Project 1 [PERIMETERS AND AREAS OF RECTANGLES]

Class 08 Project 2 Method of calculate value of pi

Suggested List of Projects
1. About an Indian mathematician and his/her contributions to mathematics.
2. Verification of Pythagoras theorem in different ways.
3. Magic squares: 3 x 3, 4 x 4, and 5 x 5.
4. Congruent shapes.
5. Exploring Pythagorean Triplets.
6. Drawing map of your school/locality, 
7. Collection of data and its pictorial representation in different ways.
8.Decimal system versus other number systems with base 5, 8 and 2.
9. Divisibility Tests with special reference to 7, 11 and 13.
10. Verification of Euler's formula for different 3 D shapes (polyhedra).
11. Application of direct and inverse proportions in day to day life. 
12.Use of double bar graph in different situations.
13. Hardy Ramanujan Numbers.
14. Use of algebraic identities in solving problems.
15. Areas of different polygons.
16. Graphs in day to day life.

LIST OF PROJECTS

LIST OF PROJECTS

1.       To develop Heron's formulae for area of a triangle.

2.       Story of p.

3.       Development of Number Systems with their needs.

4.       Chronology of Indian Mathematicians with their contributions.

5.       Chronological development of solution of a quadratic equations.

6.       Development of Formula for the area of a cyclic quadrilateral.

7.       Pythagoras Theorem-Proofs other than given in the present textbook.

8.       Extensions of Pythagoras Theorem.

9.       With rectangle of given perimeter finding the one with a maximum area and with rectangle of given area, finding the one with least perimeter.

10.      Knowledge and classification of solid figures with respect to surface areas and volumes.

11.      Sum of the exterior angles of a polygon taken in an order.

12.      Generation of Pythagorean triplets.

13.      Magic squares.

14.      With cuboids of given surface area, finding the one with maximum volume and with cuboids of given volumes finding one with least surface area.

15.      Mathematical designs and patterns.

16.      Indian Mathematicians and their contibutions.

SCHEME OF EVALUATION

The following weightage are assigned for evaluation at Secondary Stage in mathematics: Theory Examination                              :           80 marks

Internal Assessment                             :           20 marks

l. Internal assessment of 20 marks, based on school based examination will have following break-up:

Year-end assessment of activities          :           12 marks Assessment of Project Work                 :           5 marks

Viva-voce                                             :           3 marks

•        Assessment of Activity Work

(a)     Every student will be asked to perform two given activities during the allotted time.

(b)     The assessment may be carried out by a team of two mathematics teachers, including the teacher who is taking practical classes.

(c)     The break-up of 12 marks for assessment for a single activity may be as under:

•     Statement of objective of the activity :          1 mark

•     Material required                               :           1 mark

•     Preparation for the activity                :           3 marks

•     Conduct of the activity                      :           3 marks

•     Observation and analysis                   :           3 marks

•     Results and Conclusion                     :           1 mark

Total              :           12 marks

(d)     The marks for two activities may be added first and then marks calculated out of 12.

(e)     Full record of activities may be kept by each student.

•       Evaluation of Project Work

(a)     Every student will be asked to do at least one project based on the concepts learnt in the classroom.

(b)     The project may be carried out individually (or in a group of two or three students).

(c)     The weightage of 5 marks for the project may be as under :

•     Identification and statement of the project     :           1 mark

•     Planning the project                                      :           1 mark

•     Procedure adopted                                        :           1 mark

•     Observations from data collected                   :           1 mark

•     Interpretation and application of result           :           1 mark

Total Score out of 20 : The marks obtained in year-end assessment of activities and project work be added to the marks in viva-voce to get the total score out of 20.

Note : Every student should be asked to perform at least twenty activities in one academic year.

17.      To prepare a list of quotations on mathematics.

18.      Ramanujan number (1729)

19.      Mathematical Crosswords

20.      Application of Geometry in day-to-day life

21.      Application of Algebra in day-to-day life.

22.      Application of Mensuration in day-to-day life.