method and procedure of the study -...
TRANSCRIPT
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CHAPTER- III
METHOD AND PROCEDURE OF THE STUDYThe methods and procedure of conducting a research study are, by and large,
determined by the design of the study and realization of its objectives, stipulated
purposes and testing the variables involved. This would imply sub-heads like:
3.1 Method of research
3.2 Design of the study
3.3 Population and sample
3.4 Procedure followed
3.5 Statistical analysis
3.6 Precautions observed
3.7 Constraints and difficulties
3.1 METHOD OF RESEARCH
According to Hillway (1964, p.138) ----- ‘to describe in detail the specific
method being used, incidentally, constitutes a very good way of determining whether
the method chosen has been worked out properly and is likely to prove effective. If
the scholar cannot describe his method, the chances are that it is too vague and
general to yield him satisfactory results’.
Broudy (1963) stated that “Method refers to the formal structure of the
sequence of acts commonly denoted by instruction. The term method covers both
strategy and tactics of teaching and involves the choice of what is to be taught and the
order in which it is to be taught.
George J. Mouly has classified research methods into three basic types:
survey, historical and experimental methods.
Methods to conduct a research study differ in their nature and intent. Choice of
the methods of research is determined by the nature of the problem.
The present study is an attempt to study the effect of ICT on the students’
academic achievements. It is obvious that the effect of ICT cannot be studied through
survey or historical method. It needs an experimental setting. Keeping this in mind,
the investigator used pre-test, post-test experimental method to conduct this study .
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3.2 DESIGN OF THE STUDY
A design is used to structure the research, to show how all the major parts of
the research projects the sample or groups, measures, treatments or programmes and
method of assignments work together to try to address the central research question.
Winer (1971) compared the design of an experiment to an architect’s plan for
the structure of a building. The designer of experiments performs a role similar to that
of the architect. The prospective owner of a building gives his basic requirements to
the architect, who then exercising his ingenuity prepares a plan or a blueprint
outlining the final shape of the structure. Similarly, the designer of the experiment has
to do the planning of the experiment so that the experiment on completion fulfils the
objectives of research.
In the present study, students in the ICT group were taught using a power
point programme saved to CD-ROM. The power point presentation included animated
pictures, video clips. Students in the traditional group were taught using a chalkboard,
textbooks, models and charts. Experimental classes housed a ceiling-mounted LCD
projector that was connected to a computer and classroom projector projected onto a
interactive whiteboard. The power point presentation was presented on a ceiling-
mounted LCD projector. The presentation expanded each lesson by providing extra
examples and examples from the homework. Students were able to solve example
problems and then instantly see the answer on a large screen in the classroom. This
presented them immediate feedback.
In the present study, pre-test post-test control group quasi experimental,
design was employed with a purposive sample in the form of intact sections of class
IX of the same school.
The study included a control group (34 students) and an experimental group
(34 students). The experimental group was taught through ICT used teaching and the
control group through traditional method.
The intact sections were equated on intelligence and socio-economic status. A
figurative representation of the design is given in Table 3.1.
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Table 3.1 Design of the Study
Groups Pre-Test Independent variable Post-test
Experimental Y1 ICT-used teaching Y2
control Y1 Traditional teaching Y2
The study involved three operational stages as identification stage, treatment
stage and post-testing stage. The first stage involved pre-testing of all the students of
both groups on intelligence, socio-economic status, and achievement and confidence
level in mathematics. The second stage involved the experimental treatment, which
consisted of ten subunits of IX grade mathematics taught through ICT used teaching
and through traditional teaching to control group. The third stage dealt with post
testing of the control and experimental group using the achievement test in
mathematics. A schematic view of the phases of experiment is presented in Table3.2.
Table 3.2 Phases of the study
Stage Control Group Experimental Group
I Pre-testing 1.Measurement of intelligence of
pupils
2.Measurement of socio-economic
status of pupils
3. Measurement of achievement in
mathematics
4. Measurement of confidence
level of pupils
1.Measurement of intelligence of
pupils
2.Measurement of socio-
economic status of pupils
3. Measurement of achievement
in mathematics
4. Measurement of confidence
level of pupils
II. Treatment Teaching mathematics through
conventional method
Teaching mathematics through
ICT used method
III.Post-testing 1. Measurement of achievement in
mathematics.
2. Measurement of confidence
level in answering test questions of
pupils
1.Measurement of achievement
in mathematics
2. Measurement of confidence
level in answering test questions
of pupils
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Variables under Study
In an experimental research, the relationship between two types of variables,
namely independent and dependent variables, is studied. Independent variables are the
causes, while dependent ones are effects. Another category of variables, which is
equally important, is of the intervening variables. The three kinds of variables,
identified for the study are:
- Independent Variables
These variables are manipulated in order to see their effect on the learning
outcome of students. In this study ‘Treatments’ acted as an independent variable. The
treatments involved the two approaches of teaching viz., ICT-used teaching and
traditional teaching. The experimental group was taught through ICT- used teaching
and the control group was taught through the traditional teaching. Thus, ICT-used
teaching and traditional teachings were the two independent variables for the study.
- Dependent Variables
Achievement in mathematics and confidence level were taken as dependent
variables, which were measured twice during the course of the study. First, before
beginning the experimental treatment, i.e., at the pre-test stage and then, after
completing the experimental treatment, i.e., at the post-test stage.
- Intervening Variables
There are certain variables known as intervening variables which have their
effect on the learning outcomes, and influence both independent and dependent
variables. Intervening variables such as nature of school, grade level, subject to be
taught, intelligence of pupils, socio-economic status of pupils, previous knowledge of
pupils etc. were successfully controlled experimentally.
Control Employed
It is necessary to control all those variables that may significantly affect the
dependent variables. Hence, such intervening variables were controlled by employing
suitable controls.
1. Nature of school
The sample was selected from a single English medium public school (M.M.
Public School, Assandh) situated in Karnal district.
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2. Grade Level
IXclass students were selected for the study and grade level was thus kept
constant during the study.
3. Teacher behaviour
The investigator herself taught the content to both experimental and control
group i.e., inter-teacher variation was eliminated. She herself outlined the entry level
behaviour, prepared achievement test, lesson plans etc. Hence there was equal
familiarity with all the treatments.
4. Subject
The two groups were taught same units of mathematics.
5. Socio-Economic Status
The experimental group and the control group were given S.E.S. Test. t-test
was applied to find out the difference between S.E.S. test scores of the two groups.
The results are given in the Table 3.3
Table 3.3‘t’-Value of S.E.S. test scores of Experimental groupand Control Group
Group N Mean SD t-Value
Control Group
Experimental
Group
34
34
19.265
19.853
4.166
3.823
.607(NS)
(NS) Not significant at 0.01 level
Table3.3. shows that the t-value between the groups is (0.607) which is not
significant at 0.01 levels. It means that no significant differences existed between
the S.E.S. of the two groups, indicating that they belonged almost to the same kind of
a socio -economic milieu.
6. Intelligence of Pupils:
To eliminate the initial variability of the pupils statistically in the two groups,
they were measured on general mental ability, through Cattell,S Culture Fair
Intelligence Test.General mental ability is an index of intelligence which might have
effected the independent variables.
t-value was computed to analyse the difference between intelligence test scores of the
two groups. The results are given in Table 3.4.
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Table 3.4‘t’-VALUE OF INTELLIGENCE TEST SCORES OF
EXPERIMENTAL GROUP AND CONTROL GROUP
Group N Mean SD t-Value
Control Group
Experimental
Group
34
34
101.294
101.764
11.68
7.61
0..21(NS)
(NS) Not significant at 0.01 level
Table3.4. shows that the t-value between the two groups is (0.21) which is
not significant at 0.01 level. It means that no significant difference existed between
the intelligence of the two groups.
Initially, general mental ability was thought to be controlled statistically
through covariance but since the two groups selected did not differ on general mental
ability at the pre-test stage, there was no need to control covariate.
The independent variables, dependent variables, control variables and the kind of
control employed in the study are summarized in Table 3.5.
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Table3.5 Control Employed to Variables
Independent
Variables
Dependent
Variables
Control
Variables
Control
Employed
Method
of
Teaching
1. Achievement
In Mathematics
2.Confidence level of
pupils in answering
test questions
1. Nature
of school
2. Grade Level
3. Teacher
4. Subject to be
taught
5. Duration of
treatment
6. Pupils’ socio-
economic status
7. Pupils’
intelligence
1. Administrative
(Single School)
2. Administrative
(Only IX class chosen
as sample and taught)
3. Both the groups
were taught by
the same teacher
(investigator hereby)
4. Administrative
(Same units of
Mathematics taught in
both groups)
5. The two groups
taught for 33 days, 40
minutes each period
daily.
6. Belonged to the
same milieu.
7. No need.
Specific events and factors like anxiety, home environment, adjustment, and
social maturity could have only a marginal effect upon the experiment, so these
factors were not taken into account.
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3.3 POPULATION AND SAMPLE
The term ‘Population’ is used in research to describe any group of individuals,
events or observations in which the researcher is interested. In the present study, the
term population refers to class IX students studying in English Medium Public School
of Karnal district.
A sample is a finite part of a statistical population whose properties are studied
to gain information about the whole (Webster, 1985). When dealing with people, it
can be defined as the set of respondents (people) selected from a larger population for
the purpose of a survey. In majority of the studies, it is just not feasible to collect data
from each and every subject. In addition, to work on a sample saves time, labour and
money.
Sampling makes it possible to draw valid generalisation by studying a
relatively small proportion of the population selected for observation and analysis. In
the present investigation, Karnal district of Haryana was the field of study. The
sample of the study comprised 34 pupils each studying in two sections of the IX class
of M.M. public School, Assandh situated in Karnal district. One section formed the
control group and the other section formed the experimental group, as shown in
Table3.6.
TABLE 3.6 SAMLE OF STUDY
S. NO. Groups Total No. of Students
1 Experimental group 34
2 Control Group 34
Total 68
No doubt, the sample is small for the result of the study to be generalized, an
experimental study is normally more suitable on a small sample, as is evident from
earlier investigations conducted through experimental design, which used small
samples only.Krulger (1999) and Angrist and Lavy (2004) provide evidence in favour
of positive and significant effect of small classes in experimental studies.
Arias and Walker (2004) conducted an experiment to test the relationship
between class size and student performance. They controlled variation in instruction,
lecture material, and topic coverage by using the same instructors. Their results were
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statistically significant showing that small class size had a positive impact on student
performance.
3.4 PROCEDURE FOLLOWED
Procedure of the experiment comprised of two main stages, that is, selection
of the sample and conducting the experiment.
Stage1: Selection of the sample
The sample of the study comprised of 68 students of class IX (34 as control
group and 34 as experimental group) studying in M.M. Public School, Assandh distt.
Karnal.
Selection of Experimental Group:
For the experimental group, a total of 34 learners studying in IX standard,
section A was chosen from M.M. Public School Assandh distt. Karnal.
Selection of Control Group:
The control group consisted of 34 learners studying in IX standard, section B
of the same school. The group was exposed to traditional method of instruction. No
novel treatment was given to the control group of students.
Stage2: Conducting the experiment
The experiment was conducted in three phases:
Phase I: Administration of the Pre-test
Before the start of the experiment, the sample subjects were contacted and
rapport was established with them. They were oriented about the tests to be used.
Three pre-tests i.e., S.E.S., Intelligence, Achievement Test were administered to the
students of two groups by the researcher herself. Cooperation of the class teacher was
sought for administering the tests properly. The instructions pertaining to the tests
were explained verbally in clear terms to the students before administering the test.
The administration of the tests was carried out as per norms and instructions
contained in respective test manuals.
After this, the students of both the groups were provided orientation and
instructions about the treatment to be allotted to them to get over the anxiety and
curiosity of the students. The students of the experimental group were given a trial of
their respective materials, which helped them in getting over the curiosity and anxiety
around via the electronic system being applied in the classroom setting. The students
of the control group were also made familiar about the objectives, etc, of the tests to
elicit their cooperation in the conduct of the study.
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Phase II: Conducting the Instructional programme
The second phase of the experiment was addressed to the real execution of the
experiment. In this phase, the experimental group students were taught by ICT-used
teaching and the control group students were taught by traditional method of teaching.
The instructional treatment was given about 40 days to the experimental
group, where as the control group was taught by the traditional method. Same content
was taught to both the groups.
Phase III: Administration of Post-test
Immediately after the instructional treatment was over, the researcher tested
the subjects of experimental group and control group on the dependent variables
(Academic Achievement and Confidence level ).
Date Schedule of the Instructional Phase for both the groups:
Phase 1: Pre-test Stage
9th November 2009-Administration of Achievement Test in Mathematics
Phase2: Instructional Programme
Table3.7 Date Schedule of the Instructional Phase for both the groups
S.NO. Name of Sub-unit Date Schedule of the
Instructional Phase
1 Surface Area of cuboids 10th-13th NOV.2009
2 Surface Area of cube 14th-18thNOV.2009
3 Surface Area of right circular cylinder 19th-23rd NOV.2009
4 Surface Area of right circular cone 24th-27th NOV.2009
5 Surface Area of Sphere 28NOV.-2nd DEC.2009
6. Volume of cuboids 3rd.-5th DEC.2009
7 Volume of cube 7th.-10th DEC.2009
8. Volume of right circular cylinder 11th-16th DEC.2009
9. volume of a right circular cone 17th-19th DEC.2009
10 Volume of a Sphere 21st-24thDEC.2009
25th December2009- Achievement Test in Mathematics
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As many as 354 slides were prepared (Appendix-I) and demonstrated before
the students in the experimental classroom situation to explain and elaborate the
teaching of different topics listed above to enrich the teaching-learning process and to
improve the quality of schooling. Their details are briefly summarized lesson –wise as
follows as part of the delivery of demonstration and teaching topic-wise and day-wise
to facilitate understanding the merit of the use of ICT in the classroom.
Day-wise details of the Instructional Programme
S.NO Name of Sub-unit Date Schedule of the
Instructional Phase
1 Surface Area of cuboids 10th-13th NOV.2009
(1) Topic- ‘Surface Area of cuboids’ (10 th Nov.2009)
Material used:
Power-point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them (Slide-1).
After that announced the topic ‘Surface Area of cuboids’ and showed it on
screen (slide2-3).
Step 2: Explained about its Geometrical Representation and also told
them that in cuboids there were 8 vertexes, 12 edges and 6 rectangular faces
(slide4-5).
Concept of a cuboid was made clear by showing the examples related to daily
life, i.e.,picture of a match box and picture of a book (slide6-8)
Step 3: Explained the procedure about deriving the formula for finding the
‘.Surface Area of cuboids’ (slide 10).
\To explain it, followed the following steps:
Explained that pair of opposite rectangular faces is of same area which were of
different colors (slide9-12).
With the help of MS Animation Clips explained that to find the ‘Surface
Area of cuboids’ we have to add the surface areas of all its 6 faces (slide13-
15).
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Step 4: Also explained the steps for solution of a problem based on the
‘Surface Area of cuboids’ (slide16-17).
Step 5: Gave the students one sum based on the above topic to solve in their
note books (slide-18)
and also check their note books.
Step 6: Gave the students one sum based on the above topic to solve in their
note books at their (slide-19).
(2) Topic- ‘Lateral Surface Area of cuboids’ (11 th Nov.2009)
Material used: -
Power point CD with LCD projector
Method:
Step 1: Asked some questions from the students to know their previous
knowledge and motivate them and then announced the topic ‘Lateral
Surface Area of cuboids’ and showed it on screen.(slide-20)
Step 2: Explained the procedure about deriving the formula for finding the
‘Lateral Surface Area of cuboids’. To explain it, the researcher followed the
following steps
Explained that pair of opposite rectangular faces is of same areas which were
of different colors .With the help of MS Animation Clips, explained that to
find the ‘Lateral Surface Area of cuboids’ we have to add the surface areas
of all its 4faces leaving the bottom and top faces. (slide21-23).
Step 3: Also explained the steps for solution of a problem based on the
‘Lateral Surface Area of cuboids’ (slide24-25).
Step 4: Gave the students one sum based on the above topic to solve in their
note books (slide-26)
and also checked their note books.
Step 5: Gave the students two sums based on the above topic to solve in their
note books at their homes (slide-27).
3) Topic- ‘Solutions of the Problem based on ‘Surface Area of cuboids’
(12 Th Nov.2009)
Material used:
Power - point CD with LCD projector
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Method:
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them.
Step 2: Explained the steps for solution of a problem based on the ‘Surface
Area of cuboids’ (slide-28).
Step 3: In Question No.1 (slide-30).
Also told the formula to calculate the area of sheet required for making the
box as well as the steps to be followed for its calculation.
Step 4: In Question No.2 (slide-31).
- Explained the formula for calculating the area of four walls of the
rooms as well as the steps for its calculation.
- Also told them to add the areas of four walls of the rooms and area of
the ceiling.
- Explained that to calculate total cost of white washing we have to
multiply the Total Area by Cost per meter square.
Step 5: In Question No.3 (slide-32).
- Explained the formula for calculating the area of four walls of the
rooms as well as the steps for its calculation.
- Explained the formula for calculating the Perimeter of the hall as well
as the steps for its calculation.
- Also told them the procedure of calculating the height of the hall.
Step 6: Gave the students one sum based on the above topic to solve in their
note books (slide-33) and also check their note books.
Step 7: Gave the students two sums based on the above topic to solve in their
note books at their homes (slide-34).
1. Surface Area of cube 13th-18th NOV.2009
(1) Topic- ‘Surface Area of cube’ (13 th Nov.2009)
Material used:
Power point CD with LCD projector
Method :
Step1: Asked some questions from the students to know their previous
knowledge and motivate them.
After that announced the topic ‘Surface Area of a cube’ and showed it on
screen. (Slide35-37)
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Step 2: Explained about its Geometrical Representation and also told them
that in cuboids there were 8 vertexes, 12 edges and 6squre faces (slide38-40).
Concept of a cube was made clear by showing the examples related to daily
life i.e. Picture of a Ruby cube and picture of ice cubes (slide45-46)
Step 3: Explained the procedure about deriving the formula for finding the
‘.Surface Area of cuboids’ (slide41- 42).
To explain it , followed the following steps:-
- Explained that pair of opposite squared faces are of same area which
were of same colors (slide44).
- With the help of MS Animation Clips explained that to find the
‘Surface Area of a cube’ we have to add the surface areas of all its 6 faces
(slide45-47).
Step 4: Also explained the steps for solution of a problem based on the
‘Surface Area of a cube’’ (slide-49).
Step 5: Gave to the students one sum based on the above topic to solve in
their note books (slide-50) and also checked their note books.
Step 6: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-51).
(2) Topic- ‘Lateral Surface Area of a cube’ (14 th Nov.2009)
Material used:
Power - point CD with LCD projector
Method:
Step 1: Aked some questions from the students to know their previous
knowledge and to motivate them. After that, announced the topic ‘Lateral
Surface Area of a cube’ and showed it on screen.(slide-52)
Step 2: Explained the procedure about deriving the formula for finding the
‘Lateral Surface Area of a cube’. To explain it, followed the following steps
- Explained that pair of opposite square faces is of same areas which
were of same colors.
- With the help of MS Animations Clips, explained that to find the ‘Lateral
Surface Area ofa cube’ we have to add the surface areas of all its 4faces
leaving the bottom and top faces. (slide53-55).
\Step 3: Also explained the steps for solution of a problem based on the
‘Lateral Surface Area of a cube’ (slide56-57).
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Step4: Gave to the students one sum based on the above topic to solve in
their note books (slide-58)
and also checked their note books.
Step 5: Gave to the students two sums based on the above topic to solve in
their note books at their home (slide-59 ).
(3) Topic- ‘Solutions of the Problem based (16 Th Nov.2009)
on ‘Surface Area of cube’
Material used:
Power - point CD with LCD projector
Method:
Step 1: Asked some questions to the students to know their previous
knowledge and to motivate them.
Step 2: Also explained the steps for solution of a problem based on the
‘Surface Area of cube’ (slide-60).
Step 3: Question No.4 (slide-61).
The researcher explained the formula to calculate the area of a brick and also
told them the procedure of finding the nos. of bricks required when the total
surface area as well as surface area of a brick was given.
Step 4: In Question No.5 (slide62-63).
- Explained the formula for calculating the Lateral Surface Area of a
cube and of a cubical box.
- Also told them to subtract the Lateral Surface Areas of a cubical box
from a cube.
- Explained the formula for calculating the Total Surface Area of a cube
and of a cubical box.
- Also told them to subs tract the Total Surface Areas of a cubical box
from a cube.
Step 5: Gave to the students two sums based on the above topic to solve in
their note books and also checked their note books (slide64-65).
Step 6: Gave to the students two sums based on the above topic to solve in
their note books at their home (slide-66).
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(4) Topic- ‘Solutions of the Problem based on‘Surface Area of cuboids’
(17 Th Nov.2009)
Material used:
Power - point CD with LCD projector
Method:
Step 1 Asked some questions to the students to knowtheir previous
knowledge and to motivate them.
Step 2: Also explained the steps for solution of a problem based on the
‘Surface Area of cuboids’ (slide-67).
Step 3: In Question No.6 (slide-68).
- Explained the students that the surface area of a glass can be calculated
by using the surface areas of cuboids.
- The researcher also explained the students that length of tape can be
calculated by calculating the length of 12 edges.
Step 4: Gave to the students one sum based on the above topic to solve in
their note books and also checked their note books. (slide-69).
Step 5: Gave to the students two sums based on the above topic to solve in
their note books at their homes (slide-70).
(5) Topic- ‘Solutions of the Problem based on (18 Th Nov.2009)
‘Surface Area of cuboids’
Material used:
Power - point CD with LCD projector
Method:
Step 1: Asked some questions to the students to know their previous
knowledge and to motivate them.
Step 2: Also explained the steps for solution of a problem based on the
‘Surface Area of cuboids’ .
Step 3: In Question No.7 (slide72-75).
- Explained to the students that the surface area of a bigger box can be
calculated by using the surface areas of cuboids.
- Explained to the students that the surface area of a 250 bigger boxes
can be calculated by multiplying the surface area of a bigger box by 250.
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- Also explained t0 the students that the surface area of a smaller box
can be calculated by using the surface areas of cuboids.
- Explained to the students that the surface area of a 250 smaller boxes
can be calculated by multiplying the surface area of a smaller box by 250.
- Told the students to add the areas of both type of boxes
- Explained to the students about the area of sheet required for making
250 boxes of including extra required area of 5% for overlaps.
- Explained to the students about total cost of sheet at the rate of Rs.4 for
1000cm2.
Step4: Question No.8 (slide76-77).
- Explained to the students that the area of trapallin required to cover
four sides and the roof can be calculated by the formula i.e. lxb+2(l+b) x h in
which L=4m, b=3m, h=2.5m.
Step 5: Gave to the students one sum based on the above topic to solve in
their note books and also checked their note books. (Slide-78).
Step 6 Gave to the students two sums based on the above topic to solve in
their note books at their home (slide-79).
3. Surface Area of right circular cylinder 19th-23rd NOV.2009
(1) Topic- ‘Surface Area of right circular cylinder’ (19 th Nov.2009)
Material used:
Power -point CD with LCD projector
Method :
Step 1: Asked some questions to the students to know their previous
knowledge and to motivate them. After that announced the topic ‘Surface
Area of a right circular cylinder’ and showed it on screen.(slide-80)
Step 2: With the help of MS Animation Clips, explained that in a right
circular cylinder the angle between radius and height is of 900.(slide81-82)
Step 3: Explained about its Geometrical Representation (slide-84) and also
told them that in a right circular cylinder there were 3 surfaces (slide-83) and
also explained them the shapes of these three surfaces (slide-84).
Step 4 Also explained the definition of a right circular cylinder (slide-85).
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Step 5: Explained the procedure about the deriving the formula for finding
the‘.Surface Area of a right circular cylinder (slide 86-94).
To explain it, followed the following steps
- Explained that in a right circular cylinder areas of top and bottom
circles are of same area.
With the help of MS Animation Clips explained that to find the ‘Surface Area
of right circular cylinder’ we have to add the surface areas of all its 3
surfaces.
Step 6: Also explained the steps for finding the area of middle part which is
of curved shape. With the help of MS Animation Clips, explained the
students that area of curved surface is equal to the area of a rectangle of
length 2Π r and breadth h.
Step 7: Also told them to calculate the total surface area of a right circular
cylinder.
Step 8: Also explained the steps for solution of a problem based on the
‘Surface Area of a right circular cylinder’ (slide95).
Step 9: Gave to the students one sum based on the above topic to solve in
their note books (slide-96) and also checked their note books.
Step 10: Gave the students two sums based on the above topic to solve in their
note books at their home (slide97-98).
(2) Topic- ‘Solutions of the Problem based on (20 Th Nov.2009)
‘Surface Area of right circular cylinder’
Material used:
Power - point CD with LCD projector
Method:
Step 1: Asked some questions from the students to know the previous
knowledge and to motivate them.
Step 2: Also explained the steps for solution of a problem based on the
‘Surface Area of right circular cylinder.
Step 3: Question No.2 (slide100).
-Explained the students that the surface area of a cylindrical tank can be
calculated by using the formula of surface areas of right circular cylinder.
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Step 4: Question No.3(slide101-102)
- Explained to the students that the inner curved surface area of a metal
pipe can be calculated by using the formula 2Πrh in which ‘r’ is 2 cm and ‘h’
is 77 cm.
- Also explained the students that the outer curved surface area of a pipe
can be calculated by using the surface area of right circular cylinder in which
‘R’ is 2.2 cm and height ‘77’ cm.
- Also told the students that the total surface area can be calculated by
adding the curved surface area of a hollow pipe and area of inner and outer
curved surface areas.
Step 5: Question No.4 (slide103)
-Explained the students that the area covered by one revolution can be
calculated by using the formula of curved surface area of a right circular
cylinder in which height is 120 cm and radius is 42 cm.
- Told the students to multiply the area covered by one revolution by
500 to find area of the playground.
Step 6: Gave to the students one sum based on the above topic to solve in
their note books (slide-104) and also check their note books.
Step 7: Gave to the students two sums based on the above topic to solve in
their note books at their home (slide-105-106).
(3) Topic- ‘Solutions of the Problem based on (21st Nov.2009)
‘Surface Area of right circular cylinder’
Material used: -
Power point CD with LCD projector
Method:
Step 1: Asked some questions from to the students to know the previous
knowledge and to motivate them.
Step 2: Also explained the steps for solution of a problem based on the
‘Surface Area of right circular cylinder.
Step 3: Question No.5 (slide108).
- Explained the students that the curved surface area of a cylindrical pillar can
be calculated by using the formula of surface areas of right circular cylinder in
which ‘r’ is 25 cm and ‘h’ is 3.5 cm.
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- Explained them that total cost of painting the curved surface of the pillar can
be calculated by multiplying its curved surface area by cost of paint per m2.
Step 4: Question No.6 (slide-109)
-Explained the students that the inner curved surface area of a well can be
calculated by using the formula 2Πrh in which ‘2r’ is 3.5 m and ‘h’ is10m.
Step 5: Question No.7 (slide-110)
-Also told the students that the total cost of plastering this curved surface area
can be calculated by multiplying its curved surface area by cost of plastering
per m2.
Step 6: Question No.8 (slide-111)
-Explained the students that the total radiating surface area can be calculated
by using the curved surface area of a cylinder in which r=5/2 cm and h= 28 m.
-Gave to the students two sums based on the above topic to solve in their note
books at their home (slide-112).
Step 7 Gave to the students one sum based on the above topic to solve in their
note books at their home (slide-113).
(4) Topic- ‘Solutions of the Problem based on (23rd Nov.2009)
‘Surface Area of right circular cylinder’
Material used:
Power point CD with LCD projector
Method:
Step 1: Asked some questions to the students to know the previous
knowledge and to motivate them. The researcher also explained the steps for
solution of a problem based on the ‘Lateral Surface Area of right circular
cylinder.
Step 2: Question No.9 (slide115).
-Explained to the students that the lateral surface area of a closed cylindrical
tank can be calculated by using the formula of surface areas of right circular
cylinder in which ‘2r’ is 4.2 m and ‘h’ is 4.5 m.
-Explained to the students that the total surface area of a closed cylindrical
tank can be calculated by using the formula of surface areas of right circular
cylinder in which ‘2r’ is 4.2 m and ‘h’ is 4.5 m.
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Step 3: Question No.10 (slide-116)
-Explained to the students that the total height of a frame of a Lampshade is
35cm, i.e., 30+2.5+2.5 and the length of the cloth required for covering the
lampshade can be calculated by using the formula 2Πrh in which ‘r’ is 10cm
and ‘h’ is 35 cm.
Step 4: Question No.11 (slide117-118)
- Explained the students the cardboard required for one penholder can be
calculated by using the formula (r2+2Πrh) in which ‘r’ is 3 cm and ‘h’ 10.5
cm.
-Explained the students the cardboard required for 35 penholders can be
calculated by multiplying cardboard required for one penholder by 35.
Step 5: Gave to the students two sums based on the above topic to solve in
their note books at their home (slide-119).
Step 6: Gave to the students two sums based on the above topic to solve in
their note books at their home (slide-120).
4 Surface Area of right circular cone 24th-27th NOV.2009
(1) Topic- ‘Surface Area of right circular cone’ (24 th Nov.2009)
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions to the students to know their previous
knowledge and to motivate them.
After that Announced the topic ‘Surface Area of a right circular cone’ and
showed it on screen.(slide-121-122)
Step 2: Also explained the definition of a right circular
cone (slide-123).
Step 3: Explained about its Geometrical
Representation (slide124-126) and also told them about ‘vertex’
‘height ‘radius’ and ‘slant height’.
Step 4: With the help of MS Animation Clips, explained that there are two
surfaces in a right circular cylinder .one is flat base circular in shape and
second is a lateral surface curved in shape (slide-127).
Step 5: Explained the procedure about deriving the formula for finding the
‘.Surface Area of a right circular cone (slide 128-140).
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To explain it, followed the following steps
- Explained that in a right circular cone area of base is Πr2.
- With the help of MS Animation Clips, explained that to find the
‘Surface Area of right circular cone’ we have to add the surface areas of all
its 2 surfaces.
Step 6: Also explained the steps for finding the area of curved surface. With
the help of MS Animation Clips, explained to the students that area of curved
surface is equal to half of the length of entire curved
boundary.
Step 7: Also told them to calculate the total surface area of a right circular
cone.
Step 8: Also explained the steps for solution of a
problem based on the ‘Surface Area of a right circular cone’ (slide-141) .
Step 9: Gave to the students one sum based on the above topic to solve in
their note books (slide-142) and also check their note books.
Step 10: Gave to the students two sums based on the above topic to solve in
their note books at their home (slide-143-144).
2) Topic- Solutions of the Problem based on (25th Nov.2009)
‘Surface Area of right circular cone’
Material used:
Power - point CD with LCD projector
Method:
Step 1: Asked some questions to the students to know
the previous knowledge and to motivate them.
Step 2: Also explained the steps for solution of a problem
based on the ‘Surface Area of right circular cone.
Step 3: Question No. 2 (slide146).
- Explained to the students that the total surface area of a cone can
be calculated by using the formula of surface areas of right circular cylinder in
which ‘r’ is 12 cm and ‘l’ is 21 cm.
Step 4: Question No.3 (slide147-148)
- Explained to the students that the radius of a cone can be obtained by
dividing its curved surface area with Π times its slant height.
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- Also told the students that the total surface area can be calculated by
substituting the value of ‘r’ and ‘h’.
Step 5: Question No.4 (slide149-150)
- Explained, the students that the slant height of a right circular cone
can be calculated by using the Pythagoras theorem.
- Also told the students that area of canvas can be calculated by curved
surface area of a right circular cone.
Step 6: Also told the students that cost of canvas can be calculated by
multiplying its curved surface area by Rs.70.
Step 7: Question No.5 (slide151)
- Explained to the students that the slant height of a right circular cone
can be calculated by using the Pythagoras theorem.
- Also told the students that area of canvas can be calculated by curved
surface area of a right circular cone.
Step 8: Also told the students that cost of canvas can be calculated by
multiplying its curved surface area by Rs.70.
Step 9: Explained to the students that the length of canvas
required can be calculated by adding its curved surface area and
stitching margins.
Step10: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-152).
Step 11 Gave to the students two sums based on the above topic to solve in
their note books at their home (slide-153).
3) Topic- Solutions of the Problem based on (26th Nov.2009)
‘Surface Area of right circular cone’
Material used:
Power - point CD with LCD projector
Method:
Step 1: Asked some questions to the students to know
the previous knowledge and to motivate them.
Step 2: Question No.6 (slide155)
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- Explained to the students that the curved surface area of a cone can be
calculated by using its formula in which ‘r’ are 7m and ‘l’ is 25 cm.
- Also explained to the students that the cost of white
washing can be calculated by multiplying its curved surface by Rs.210 per 100
m2.
Step 3: Question No.7 (slide156)
- Explained to the students that the area of the sheet required to make
one can be calculated by using curved surface area of a right circular cone.
- Also told the students that sheet required for 10 caps can be calculated
multiplying curved surface area of a right circular cone by 10.
Step 4: Question No.8 (slide157)
- Explained to the students that the slant height of a right circular cone
can be calculated by using the Pythagoras theorem.
- Explained to the students that the curved surface area of 50 cones can
be calculated multiplying curved surface area of a right circular cone by 50.
- Also told the students that cost of painting all these cones can be
calculated multiplying curved surface area of a right circular cone by Rs.12
per m2 Step 5: Gave to the students one sum based on the above topic
to solve in their note books at their home (slide-158).
Step 6: Gave the students one sum based on the above
topic to solve in their note books at their home (slide-159).
4 Surface Area of Sphere 28 th NOV. _ 2nd DEC.2009
(1) Topic- ‘Surface Area of Sphere’ (28 th NOV.2009)
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate After that, announced the topic ‘Surface Area of
a Sphere’ and showed it on screen (slide-160) .
Step 2: Also explained the definition of a right circular cone (slide-162).
Step 3: Explained about its Geometrical
Representation and also told them about its’ radius’ and ‘centre’
(slide163).
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Step 4: Concept of a Sphere was made clear by showin examples related to
daily life, i.e. , Marble Ball,Ball,footbal globe.(slide164-168)
Step 5: Explained the procedure about the deriving the formula for finding
the ‘.Surface Area of a Sphere. (slide 169-17
To explained it, followed the following steps
- With the help of MS Animation Clips, explained that the ‘Surface
Area of Sphere is equal to the four times area of a circle having radius ‘r’.
Step 5: Also explained the steps for solution of a problem based on the
‘Surface Area of a Sphere’.
Step 6: Question No.1 (slide178)
Explained the students that the surface area of a Sphere cone can be calculated
by substituting the value ‘r’ in the formula (4Π r2).
Step 7: Question No.2 (slide179)
Explained to the students that the surface area of aSphere cone can be
calculated by substituting the value ‘2r’ in the formula a (4Π r2).
Step 8: In Question No.4 (slide180)
Explained to the students that the ratio of two surface areas of spherical
balloon can be calculated by dividing their surface area.
Step 9: Gave the students one sum based on the above topic to solve in their
note books at their home (slide-181).
Step 10: Gave the students one sum based on the above topic to solve in their
note books at their home (slide-182).
(2) Topic- ‘Curved Surface Area of a Hemisphere’ (30th NOV.2009)
- With the help of MS Animation Clips, explained that the ‘ Total
Surface Area of a Hemisphere is sum of the areas of both surfaces (area of
Base and area of curved surface).
Step 4: Also explained the steps for solution of a problem based on the
‘Surface Area of a Hemisphere’.
Step 5: Question No.3 (slide190)
Explained to the students that the surface area of a Hemisphere cone can be
calculated by substituting the value ‘r’ in the formula (3Π r2).
Step 6: Question No.5(slide191)
- Explained to the students that the inner surface area of a Bowl can be
calculated by substituting the value ‘2r’ in the formula (2Π r2).
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- Also explained to the students that the cost of tin painting can be
calculated multiplying its inner surface area by Rs.16 per 100cm2.
Step 7: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-192).
Step 8: Gave to the students two sums based on the above topic to solve
in their note books at their home (slide-193).
(2) Topic- Solutions of the Problem based on 1st DEC.2009
‘Surface Area of Sphere’
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them.
Step 2: Also explained the steps for solution of a problem based on the
Surface Area of a Sphere’.
Step 3: Question No.6 (slide195)
Explained to the students that the radius ’r’ of a Sphere can be calculated by
substituting the values in the formula (4Π r2).
Step 4: Question No.7 (slide196)
Explained to the students that the ratio of the surface area of moon and earth
can be calculated by dividing their surface areas.
Step 5 : Question No.8 (slide197)
- Explained to the students that the radius of outer radius can be calculated
by adding radius of inner radius and thickness of steel sheet.
- Also explained the students that the outer curved surface area of a
spherical bowl can be calculated by using formula 2ΠR2.
Step 6: Question No.9 (slide198-199)
-Explained to the students that the radius of a right circular cylinder is equal to
that of a Sphere (r).and height of a cylinder is twice that the height of a
Sphere.
-Explained to the students that the ratio of the surface area of a Sphere to that
of curved surface area can be calculated by dividing them.
Step 7: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-200).
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Step 8: Gave to the students two sums based on the above topic to solve in
their note books at their home (slide-201).
6 . Volume of cube 3 rd DEC. - 5th DEC.2009
(1) Topic- ‘Volume of cube’ (3 rd DEC.2009)
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them. After that , announced the topic
‘Volume of cube’ and showed it on screen. (slide-203)
Step 2: Explained the procedure about the deriving the formula for finding
the ‘Volume of cube’ (slide 204-205).
Step 3: Asked objective type questions related to the
topic to know whether they understood or not (slide 206).
Step 4: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-207).
(2) Topic-Solution of problem based (4 th DEC.2009)
on ‘Volume of cube’
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to knowtheir previous
knowledge and to motivate them.
Step 2: Also explained the steps for solution of problems based on the
‘Volume of cube’.
Step 3: Question No. 8 (slide-209)
Explained to the students that the length of the side of the
new cube can be calculated by equating the volume of eight cubes
with the volume of a cube whose edge is ‘a’.
Step 4: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-210).
Step 5: Gave the students one sum based on the above topic to solve in their
note books at their home (slide-211).
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(3) Topic-Solution of problem based (5 th DEC.2009)
On ‘Volume of cube’
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them.
Also explained the steps for solution of a problem based on the ‘Volume of
cube’.
Step 2: Question (slide-213)
Explained to the students that the edge cube can be calculated by equating the
volume of cube with the (edge)3.
Step 3: Gave the students one sum based on the above topic to solve in their
note books at their home (slide-214).
Step 4: Gave the students one sum based on the above topic to solve in their
note books at their home (slide-215).
7. Volume of cuboids 7 th DEC. - 10th DEC.2009
1) Topic- ‘Volume of a cuboids’ (7 th DEC.2009)
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them.
After that , announced the topic ‘Volume of a cuboids’ and showed it on
screen (slide-218) .
Step 2: Explained the procedure about the deriving the formula for
finding the ‘Volume of a cuboids’ (slide 219-224).
To explain it, followed the following steps
-With the help of MS Animation Clips, explained that to find the ‘Volume of
a cuboids’ fill the bottom of the cuboids with a rectangle until the cuboids is
completely filled.
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- With the help of MS Animation Clips, Also explained that the volume of a
cuboids can be calculated multiplying the area of the plane region occupied by
each rectangle by height.
- Also told them to calculate the volume of a cuboids.
- Also explained the steps for solution of a problem based on the
‘Volume of a cuboids’.
Step 3: Question No.1 (slide-225)
- Explained to the students that the volume of a packet containing
one matchbox be calculated by using the volume of a cube.
- Explained to the students that the volume of 12 packets
can be calculated by multiplying the volume of one matchbox by 12.
Step 4: Question No.2 (slide-226)
Explained to the students that the capacity of water tank
calculated by using the volume of a cube.
Step 5: Gave to the students one sum based on the above
topic to solve in their note books at their home (slide-227).
Step 6: Gave to the students one sum based on the above
topic to solve in their note books at their home (slide-228).
2) Topic-Solution of problems based on (8 th DEC.2009)
Volume of a cuboids’
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them.
Step 2: Also explained the steps for solution of a problem
based on the ‘Volume of a cuboids’.
Step 3: Question No.3 (slide-230)
Explained to the students that the high of a cubical vessel can
be calculated by using the volume of cuboids having its length 10 m and
wide 8 m.
Step 4: Question No. 4 (slide-231)
Explained to the students that the internal space of the pit can
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be calculated by using the volume of a cuboids in which ‘l’ is 8m,’b’
is 6m and ‘h’ is3m.
Step 5: Explained to the students that the cost of digging a
cubical pit can be calculated its volume by Rs.30.
Step 6: Question No.5 (slide-232)
Explained to the students that the breadth can be calculated
dividing its volume by the product of length and depth.
Step 7: Gave to the students one sum based on the above
topic to solve in their note books. (slide-233).
Step 8: Gave to the students one sum based on the above
topic to solve in their note books at their home (slide-234).
3) Topic-Solution of problems based on (9 th DEC.2009)
‘Volume of a cuboids’
Material used:
Power- point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know
their previous knowledge and to motivate them.
Step 2: Also explained the steps for solution of a problem based on the
‘Volume of a cuboids’.
Step 3: Question No.6 (slide-236)
Explained to the students that the number water of days can be calculated by
dividing the capacity of tank by quantity of water per day.
Step 4: Question No.7 (slide-237)
Explained to the students that the maximum number Of wooden crates can be
calculated by dividing the capacity of godown by volume of one wood crate.
Step 5: Question No.8 (slide-238)
Explained to the students that the side of a new cube can be calculated by
equating the volumes of eight cubes with the volume of a
cube having edge ‘12’.
Step 6: Gave to the students one sum based on the above
topic to solve in their note books. (Slide-239).
Step 7: Gave to the students one sum based on the above
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topic to solve in their note books at their home (slide-240).
8 . Volume of right circular cylinder 11 th DEC. - 16th DEC.2009
1) Topic- ‘Volume of right circular cylinder (11 th DEC.2009)
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them.
After that, announced the topic ‘Volume of right circular cylinder ’ and
showed it on screen (slide-244) .
Step 2: Explained the procedure about deriving the formula for finding the.
‘Volume of right circular cylinder’ (slide 245-253).
To explain it, followed the following steps
- With the help of MS Animation Clips, explained that to find the
‘Volume of right circular cylinder’ fill the bottom of the cylinder with a
circle of radius ‘r’until the cylinder is completely filled.
- With the help of MS Animation Clips, also explained that the volume
of a cylinder can be calculated multiplying the Base Area by height.
- Also told them to calculate the volume of right circular cylinder.
- Also explained the steps for solution of a problem based on the
‘Volume of right circular cylinder’.
Step 3: Question No.1 (slide-254)
- Explained to the students that the value of radius ‘r’ of a cylindrical
vessel can be calculated dividing the diameter by 2.
vessel can be calculated by substituting the values 0f ‘r’ and ‘h’ in volume
formula.
Step 4: Gave to the students one sum based on the above
topic to solve in their note books. (Slide-255).
Step 5: Gave to the students one sum based on the above
topic to solve in their note books at their home (slide-256).
2) Topic-Solution of problems based on (12 th DEC.2009)
‘Volume of a cylinder’
Material used:
Power - point CD with LCD projector
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Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them.
Also explained the steps for solution of a problem based on the ‘Volume of a
cylinder’.
Step 2: Question No.2 (slide-258)
- Explained to the students that the volume of wood can be calculated by
using the formula Π (R2-r2)x h.
- Also explained to the students that the mass of the pipe can be obtained
multiplying the volume of wood by 0.6.
Step 3: Question No.3 (slide-259)
- Explained to the students that the volume of a tin with a rectangular
base can be calculated by using the volume formula of cuboid.
- Also explained to the students that the volume of a plastic cylinder
with circular base can be calculated by using the volume formula of cylinder.
- Also told the students to compare their volumes to find out the
container having the greater capacity and how much.
Step 4: Question No.4 (slide-260)
- Explained to the students that the radius of a cylinder can be calculated
by substituting the values of ‘volume’ and height ‘h’ in the formula and
volume can be calculated by using the values of ‘r’ and height’s’.
Step 5: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-261).
Step 6: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-262).
3) Topic-Solution of problems based on (14 th DEC.2009)
‘Volume of a cylinder’
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them.
Also explained the steps for solution of a problem based on the ‘Volume of a
cylinder’.
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Step 2: Question No.5 (slide263-264)
- Explained to the students that the inner curved surface area of a
cylindrical vessel can be obtained dividing the total cost by cost of painting
per m2.
- Also told that the radius ‘r’ of the cylindrical vessel can be calculated
by substituting the values of ‘h’ in the curved surface area.
- Also explained that the capacity of the vessel can be obtained by
substituting the value of ‘r’ and ‘h’.
Step 3: Question No.6 (slide-265)
- Explained that the radius ‘r’ of the cylindrical vessel can be calculated
by substituting the values of ‘h’ in its volume formula.
- Gave to the students one sum based on the above topic to solve in their
note books at their home (slide-266).
- Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-267).
4) Topic-Solution of problems based on (15 th DEC.2009)
‘Volume of a cylinder’
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them. Also explained the steps for solution of a
problem based on the ‘Volume of a cylinder’.
Step 2: Question No.7 (slide-269)
-Explained to the students that the volume of Graphite can be obtained by using
the volume formula of cylinder.
-Also explained the method for the calculation of the volume of wood.
Step 3: Question No.8 (slide-270)
Explained that the capacity of one bowl can be obtained by using the volume
formula of a cylinder.
Step 4: Gave to the students one sum based on the above
topic to solve in their note books at their home (slide-271).
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Step 5: Gave to the students one sum based on the above topic to solve in their
note books at their home (slide-272).
8 . Volume of right circular cylinder 17 th DEC. - 19th DEC.2009
1) Topic- ‘Volume of right circular cone (17 th DEC.2009)
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them.
After that, announced the topic ‘Volume of right circular cone’ can be
determined from the volume formula for a cylinder and showed it on screen.
(slide274)
Step 2: Explained the procedure about the deriving the formula for finding
the. ‘Volume of right circular cylinder’ (slide275-281).
To explain it, followed the following steps
- With the help of MS Animation Clips, explained that to find the
‘Volume of right circular cone’ take a cylinder and a cone having the equal
heights(h) and radii(r) and fill the cone up to the brim with sand once.
- With the help of MS Animation Clips, also explained them to empty
the sand into the cylinder. It fills up only a part of the cylinder and when the
cone is filled up for the third time, and emptied into the cylinder, it can be seen
that the cylinder is also full to the brim.
- With the help of MS Animation Clips, also explained that the volume
of a cone.
- Also told them to calculate the volume of right circular cylinder.
Step 3: Also explained the steps for solution of a problem based on the
‘Volume of right circular cylinder’.
Step 4: Question No.1 (slide-282)
Explained to the students that the volume of the right circular cone can be
calculated by substituting the value of radius ‘r’ and its height.
Step 5: Question No.2 (slide283-284)
Explained to the students that the capacity of a conical vessel can be
calculated by substituting the values 0f ‘r’ and ‘h’ in volume formula.
Step 6: Gave to the students one sum based on the
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above topic to solve in their note books at their home (slide-285).
Step 7: Gave to the students one sum based on the above topic to solve in their
note books at their home (slide-286).
2) Topic-Solution of problems based on (18 th DEC.2009)
‘Volume of a cone’
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them.
Also explained the steps for solution of a problem based on the ‘Volume of a
cone’.
Step 2: Question No.3 (slide-288)
Explained to the students that the radius of the base of a cone can be
calculated by using the formula of a cone having ‘h’ as its height.
Step 3: Question No.4 (slide-289)
Explained the relationship of radius and diameter and also explained them
that the radius of the base of a cone can be calculated by using the formula of
a cone having ‘h’ as its height.
Step 4: Question No.5 (slide-290)
Explained to the students that the capacity of a conical vessel can be
calculated by substituting the values of ‘r’ and ‘h’ in volume formula.
Step 5: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-291).
Step 6: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-292).
3) Topic-Solution of problems based on (19 th DEC.2009)
‘Volume of a cone’
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previou
knowledge and to motivate them. Also explained the steps for solution o
a problem based on the ‘Volume of a cone’.
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Step 2: Question No.6 (slide293-294)
- Explained to the students that the height of a cone can be calculated by
using the formula of a cone having ‘r’as radius.
- Explained to the students that the curved surface area can be calculated by
substituting the value of ‘r’ and ‘h’ in the formula.
Step 3: Question No.7 (slide-295)
- Explained to the students that revolution of a right triangle ABC about
the side 12 cm, will take the shape of a cone and volume of this solid cone can
be calculated by substituting the value of ‘r’ and ‘h’ in the volume formula.
Step 4: Question No.8 (slide-296)
- Explained to the students that revolution of a right triangle ABC about
the side 5 cm it will take the shape of a cone and volume of this solid cone can
be calculated by substituting the value of ‘r’ and ‘h’ in the volume formula.
- Also told them to find the ratio of these volumes.
Step 5: Question No.9 (slide-297)
- Explained to the students that volume of a cone can
be calculated by substituting the value of’r’ and ‘h’ in the formula.
- Explained to the students that the area of canvas can be calculated by
using the curved surface formula of a cone. volume
Step 6: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-298).
Step 7: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-299).
10) Volume of a Sphere 21st DEC. - 24th DEC.2009
1) Topic- ‘Volume of a Sphere (21st th DEC.2009)
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them. After that, announced the topic ‘Volume
of a Sphere’ that can be determined from the volume formula for a cylinder
and showed it on screen.(slide303)
Step 2: Explained the procedure about the deriving the formula for finding
the. ‘Volume of a Sphere’ (slide 304-309).
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To explain it, followed the following steps
- With the help of MS Animation Clips, explained that to find
the‘Volume of a Sphere ’ take a cylinder and a sphere having the equal and
radii(r) and height of a cylinder is equal to the Diameter of a sphere and fill the
sphere up to the brim with sand once.
- With the help of MS Animation Clips, also explained to them to empty
the sand into the cylinder. It fills up only a part of the cylinder.
- With the help of MS Animation Clips, also explained that the volume of
a sphere is equal to 2/3 volume of a cylinder.
- Also told them to calculate the volume of a sphere.
- Also explained the steps for solution of a problem based on the
‘Volume of a sphere’.
Step 3: Question No.1 (slide-310)
Explained to the students that the volume of a sphere can be calculated by
substituting the value of radius ‘r’ in its volume formula.
Step 4: Question No.2 (slide311)
Explained to the students that the amount of water displaced can be calculated
by substituting the values of ‘r’ and ‘h’ in volume formula of a sphere.
Step 5: Question No.7 (slide312)
Explained to the students that the value of the radius ‘r’of a sphere can be
calculated by using the surface area.
Step 6: Also explained to the students that the volume of a sphere can
be calculated by substituting the value of radius ‘r’ in its volume formula.
Step 7: Gave to the students one sum based on the
above topic to solve in their note books at their home (slide-313).
Step 8: Gave to the students one sum based on the
above topic to solve in their note books at their home (slide-314).
2) Topic-Solution of problems based on (22 nd DEC.2009)
‘Volume of a sphere’
Material used:
Power - point CD with LCD projector
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Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them. Also explained the steps for solution of a
problem based on the ‘Volume of a Sphere’.
Step 2: Question No.3 (slide-316)
- Explained to the students that the volume of a metallic ball can be calculated
by using the volume formula of a sphere having ‘r’ as radius.
- Explained to the students that the mass of the ball can be calculated by
multiplying volume of a sphere by density of the metal.
Step 3: Question No.4 (slide-317)
- Explained to the students that the shape of both moon and earth are of
sphere and their can be calculated by using the volume formula of a sphere.
- Also told the students to compare the volumes of both moon and earth.
Step 4: Question No.5 (slide-318)
Explained to the students the capacity of the bowl can be
calculated by using the volume formula of a hemisphere, i.e.,(1/2 volume of a
sphere).
Step 5: Question No.6 (slide-319)
- Explained to the students that Radius of outer hemispherical bowl can
be calculated by adding the value of the inner radius and 1/100 m.
- Also explained to the students that volume of outer hemispherical
bowl can be calculated by using its Volume formula.
Step 6: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-320).
Step 7: Gave to the students one sum based on the above topic to solve in
their note books at their home (slide-321).
3) Topic-Solution of problems based on (23 rd DEC.2009)
‘Volume of a sphere’
Material used:
Power - point CD with LCD projector
Method :
Step 1: Asked some questions from the students to know their previous
knowledge and to motivate them.
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Also explained the steps for solution of a problem based on the ‘Volume of a
sphere’.
Step 2: Question No.7 (slide-323)
- Explained to the students that radius of a sphere can be calculated by
using its surface area formula.
- Also explained to the students that volume of sphere can be calculated
by using its Volume formula.
Step 5: Question No.8 (slide-324)
- Explained to the students that the inner surface area of hemispherical
dome can be calculated dividing the total cost of white washed by per square
meter.
- Also told the students to calculate the inner radius of hemispherical
dome using surface area of hemisphere.
- Explained to the students that the volume of inside of the dome can be
calculated by using Volume formula of a sphere.
Step 6: Question No.9 (slide-325)
- Explained to the students that the radius of new sphere can be
calculated by equating the volume of new sphere with the Volume of twenty
seven iron sphere.
- Explained to the students to calculate the surface area of both spheres
surface.
- Explained to the students to compare their surface areas.
Step 7: Question No.9 (slide-326)
- Explained to the students that the radius (r’) of new sphere can be
calculated by equating the volume of new sphere with the Volume of twenty
seven iron spheres having radius ’r’.
- Also explained to the students to find surface areas (s) and(s’).
- Explained to the students to find out the ratio of s and s’.
Step 8: Question No.10 (slide-327)
- Explained to the students that capacity of the capsule can be calculated
by using the formula’4/3Π r3, where ‘r’ is 3.5/2 m.
- Gave the students one sum based on the above topic to solve in their
note books at their home (slide-328).
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- Gave the students one sum based on the above topic to solve in their
note books at their home (slide-329).
3.5 STATISTICAL ANALYSIS
To achieve objectives of the study, the data collected was statistically analysed
using the following techniques:
1. Descriptive statistics such as mean and S.D worked out on the score of
achievement and confidence level in Mathematics.
2. ‘t’ value was computed in order to adjudge pupil’s intelligence and socio-
economic status.
3. ‘t’ test was employed for testing the significance of difference between the
means of pupils’ achievement in mathematics on pre - test, post -test and gain
scores.
4. t’ test was employed for testing the significance of difference between the
means of pupils’ confidence level in answering the test questions in
mathematics on pre - test, post - test and gain scores. The value of‘ ‘t’ was
computed with the help of the following formula:
M1 - M2
t = √σ1 + σ2
NI N2
WhereM1=Mean of first group
M2 = Mean of second group
σ1 = Variance of first group
σ2 = Variance of first group
NI = Number of cases in first group
N2 = Number of cases in second group
Mean scores in respect of achievement in mathematics and confidence level in
answering the test questions were pictorially presented in the form of histograms.
Histograms were drawn in respect of pre-test, post-test and gain scores of
experimental group and control group.
5. Pearson Product moment correlation coefficient was computed to adjudge the
relationship between students’ achievement and their confidence level in answering
the test items on both groups of students
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3.6 PRECAUTIONS OBSERVED
Following precautions were observed during the course of
experiment (Pre-test –treatment-Post-test) for ensuring effectiveness and high
precision in experimental condition which may have contributed to the results.
No undue stress or control of any kind was imposed on the subjects at any
time during the study and the experiment was conducted in a relaxed natural
sitting.
Both the experimental and control groups were taught by the investigator
herself to avoid any variation.
The effectiveness of the experimental treatment was ensured by establishing
rapport with students and teachers, maintaining natural setting, harmonious
atmosphere, providing sufficient time for various activities in the
experimentation and the like.
It was ensured that the topics on contents of treatment had not been previously
taught to the students in both the experimental and the control groups.
Care was taken to keep importance of content matter during the course of
treatment and it was not underplayed while fitting into the instructional
treatment.
Teaching periods of 40 minutes duration was utilized fully for treatment and
time was not wasted during experimentation.
3.7 CONSTRAINTS AND DIFFICULTIES FACED DURING
EXPERIMENT
It may not be out of place to mention some of the difficulties faced or constraints
of the experiment that needed to taken note of. These, as sorted out by the
researcher were :
Power failure
Infrastructural lapses.
Time-table related difficulties.
Efforts were needed to convince the teachers and the principal about the
experiment and to seek their co-operation in the conduct of the experiment
with in the frame work of the school schedule. The researcher contacted the
school authorities and convinced them about the programme and its usefulness
to ensure that the treatment be fully provided to every student and that the
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sample groups regularly attended the school during that period. The
experiment was accordingly adjusted as per the time table in vogue for
pursuing a regular course of studies, Making some minor changes in the
regular time-table in consultation with the time-table in charge. The students’
motivation and the helpful attitude of the school authorities encouraged the
researcher to carry out the experiment with full enthusiasm and very smoothly
to study the effectiveness of ICT on students’ achievement in Mathematics.