Maths practical activities class 9.pdf

The Purpose of the Mathematics Laboratory
A mathematics laboratory can foster mathematical awareness, skill building,
positive attitudes and learning by doing experiences in different branches of
mathematics such asAlgebra, Geometry, Mensuration, Trigonometry, Coordinate
Geometry, Statistics and Probability etc. It is the place where students can learn
certain concepts using concrete objects and verify many mathematical facts
and properties using models, measurements and other activities. It will also
provide an opportunity to the students to do certain calculations using tables,
calculators, etc., and also to listen or view certain audio-video cassettes, remedial
instructions, enrichment materials, etc., of his/her own choice on a computer.
Thus, it will act as an individualised learning centre for a student. It provides
opportunities for discovering, remedial instruction, reinforcement and
enrichment. Mathematics laboratory will also provide an opportunity for the
teacher to explain and demonstrate many mathematical concepts, facts and
properties using concrete materials, models, charts, etc. The teacher may also
encourage students to prepare similar models and charts using materials like
thermocol, cardboard, etc., in the laboratory. The laboratory will also act as a
forum for the teachers to discuss and deliberate on some important mathematical
issues and problems of the day. It may also act as a place for teachers and the
students to perform a number of mathematical celebrations and recreational
activities. Thus, the purpose of a mathematics laboratory is to enable:
• A student to learn mathematics with the help of concrete objects and to
exhibit the relatedness of mathematics with everyday life.
• A student to verify or discover some geometric properties using models,
measurements, paper cutting, paper folding, etc.
• A student to use different tables and ready reckoners in solving some
problems.
• A student to draw graphs and do certain calculations using computers and
calculators.
• The students to do some field work like surveying, finding heights, making
badminton courts, etc., using instruments kept in the laboratory.
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• The students and teachers to organise mathematics club activities including
celebration of birthdays of famous mathematicians.
• The students to listen or view certain audio or video cassettes. CDs relating
to different mathematical concepts/topics.
• A student to see a certain programme on a computer as a part of remedial
instruction or enrichment under the proper guidance of the teacher.
• The students to perform certain experiments, which can be easily evaluated
by the teacher.
• The students to do certain projects under the proper guidance of the teacher.
• The students to perform certain recreational activities in mathematics.
• A teacher to visually explain some abstract concepts by using three-
dimensional models.
• A teacher to demonstrate certain concepts and patterns using charts and
models.
• A teacher to demonstrate and reinforce the truth of certain algebraic
identities using different models.
• A teacher to demonstrate the truth of various formulae for areas and volumes
of different plane and solid figures using models.
• A teacher to explain certain concepts using computers and calculators.
• The teachers and students to consult good reference mathematics books,
journals, etc., kept in the laboratory.
• The teachers to meet and discuss important issues relating to mathematics
from time to time.
• A teacher explain certain concepts, data, graphs, etc., using slides.
• A teacher to generate different sets of parallel tests using a computer for
testing the achievement of students.
• The budding mathematicians to take inspiration from the lives, works and
anecodetes relating to great mathematicians.
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Role of Mathematics Laboratory inTeaching-Learning
Mathematics is a compulsory subject at the Secondary Stage. Access to quality
mathematics education is the right of every child. Mathematics engages children
to use abstractions to establish precise relationships, to see structures, to reason
out things, to find truth or falsity of statements (NCF - 2005). Therefore,
mathematics teaching in schools must be planned in such away that they should
nurture the ability to explore and seek solutions to problems of not only the
academic areas but also of daily life. In order to do this, access to laboratory is an
essential requirement which our system has not been able to provide so far. It is
proposed to fill this gap by providing a mathematics laboratory at the secondary
stage to all the schools. This facility will bring about a renewed thrust in our schools
so far teaching-learning of mathematics is concerned.
The teaching-learning of mathematics needs to be characterised by focussed
emphasis on processes such as activity-based learning, making observations,
collection of data, classification, analysis, making hypothesis, drawing inferences
and arriving at a conclusion for establishing the objective truth.
The main goal of mathematics education is to ‘develop the child’s resources
to think and reason mathematically, to pursue assumptions to their logical
conclusion and to handle abstractions’ (NCF-2005). To achieve this, a variety of
methods and skills have to be adopted in the teaching-learning situation. The
basic arithmetical skills offered in the first eight years of schooling will stand
in good stead to achieve the higher goals visualised at the secondary stage. A
stronger emphasis is to be laid on problem solving and acquisition of analytical
skills in order to prepare children to tackle a wide variety of life situations.
Abstraction, quantification, analogy, case analysis, guesses and verification
exercises are useful in many problem-solving situations (NCF-2005). Another
area of concern which teachers will have to address is of the perceived ‘stand-
alone’ status that mathematics has vis-a-vis other subject areas in the school
curriculum.
One of the biggest challenges of a mathematics teacher is to create and sustain
interest in his students. There is a general feeling that mathematics is all about
formulas and mechanical procedures. Under these circumstances, a mathematics
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laboratory will help teachers to reorient their strategies and make mathematics
also an activity-oriented programme in schools.
In the huddled and bundled classroom situations it is indeed difficult to make
complex theoretical concepts very clear to all the students. Developing the habit
of critical thinking and logical reasoning which are most important in
mathematics learning also suffer under such claustrophobic classroom situations.
A mathematics corner in the lower classes and a mathematics laboratory with
appropriate tools at the secondary stage will enable children to translate
abstractions into specific figures, shapes and patterns that will provide
opportunities to visualise abstractions with greater ease. To promote interest in
the subject, mathematics laboratory has become a reality at many places and is
considered as an established strategy for mathematics teaching-learning. Since
a practical exercise takes a longer time than a theoretical solution might require,
it gives the student additional time for better assimilation leading to stronger
retention. For students in whom aptitude for mathematics is limited, practical
activities besides overcoming drudgery, boredom and indifference, may help
create positive attitude and a new thirst for knowledge.
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Management and Maintenance of Laboratory
There is no second opinion that for effective teaching and learning ‘Learning by
Doing’ is of great importance as the experiences gained remains permanently
affixed in the mind of the child. Exploring what mathematics is about and arriving
at truth provides for pleasure of doing, understanding, developing positive attitude
and learning processes of mathematics and above all the great feeling of
attachment with the teacher as facilitator. It is said, ‘a bad teacher teaches the
truth but a good teacher teaches how to arrive at the truth’.
A principle or a concept learnt as a conclusion through activities under the
guidance of the teacher stands above all other methods of learning and the theory
built upon it, can not be forgotten. On the contrary, a concept stated in the
classroom and verified later on in the laboratory doesn’t provide for any great
experience nor make child’s curiosity to know any good nor provides for any
sense of achievement.
A laboratory is equipped with instruments, apparatus, equipments, models
apart from facilities like water, electricity, etc. Non availability of a single
material or facility out of these may hinder the performance of any experiment
activity in the laboratory. Therefore, the laboratory must be well managed and
well maintained.
A laboratory is managed and maintained by persons and the material
required. Therefore, management and maintenance of a laboratory may be
classified into two categorise namely the personal management and
maintenance and the material management and maintenance.
(A) PERSONAL MANAGEMENT AND MAINTENANCE
The persons who manage and maintain laboratories are generally called
laboratory assistant and laboratory attendant. Collectively they are known as
laboratory staff. Teaching staff also helps in managing and maintenance of
the laboratory whenever and wherever it is required.
In personal management and maintenance following points are considered:
1. Cleanliness
A laboratory should always be neat and clean. When students perform
experiment activities during the day, it certainly becomes dirty and
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6 Laboratory Manual
things are scattered. So, it is the duty of the lab staff to clean the
laboratory when the day’s work is over and also place the things at their
proper places if these are lying scattered.
2. Checking and arranging materials for the day’s work
Lab staff should know that what activities are going to be performed on
a particular day. The material required for the day’s activities must be
arranged one day before.
The materials and instruments should be arranged on tables before the
class comes to perform an activity or the teacher brings the class for a
demonstration.
3. The facilities like water, electricity, etc. must be checked and made
available at the time of experiments.
4. It is better if a list of materials and equipments is pasted on the wall of
the laboratory.
5. Many safety measures are required while working in laboratory. A list
of such measures may be pasted on a wall of the laboratory.
6. While selecting the laboratory staff, the school authority must see that
the persons should have their education with mathematics background.
7. A training of 7 to 10 days may be arranged for the newly selected
laboratory staff with the help of mathematics teachers of the school or
some resource persons outside the school.
8. A first aid kit may be kept in the laboratory.
(B) MANAGEMENT AND MAINTENANCE OF MATERIALS
A laboratory requires a variety of materials to run it properly. The quantity
of materials however depends upon the number of students in the school.
To manage and maintain materials for a laboratory following points must be
considered:
1. A list of instruments, apparatus, activities and material may be prepared
according to the experiments included in the syllabus of mathematics.
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Mathematics 7
2. A group of mathematics teachers may visit the agencies or shops to
check the quality of the materials and compare the rates. This will help
to acquire the material of good quality at appropriate rates.
3. The materials required for the laboratory must be checked from time
to time. If some materials or other consumable things are exhausted,
orders may be placed for the same.
4. The instruments, equipments and apparatus should also be checked
regularly by the laboratory staff. If any repair is required it should be
done immediately. If any part is to be replaced, it should be ordered and
replaced.
5. All the instruments, equipments, apparatus, etc. must be stored in the
almirahs and cupboards in the laboratory or in a separate store room.
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EquipmentsforMathematicsLaboratoryatSecondaryStage
As the students will be involved in a lot of model making activities under the
guidance of the teacher, the smooth running of the mathematics laboratory will
depend upon the supply of oddments such as strings and threads, cellotape, white
cardboard, hardboard, needles and pins, drawing pins, sandpaper, pliers, screw-
drivers, rubber bands of different colours, gummed papers and labels, squared
papers, plywood, scissors, saw, paint, soldering, solder wire, steel wire, cotton
wool, tin and plastic sheets, glazed papers, etc. Besides these, some models,
charts, slides, etc., made up of a good durable material should also be there for
the teacher to demonstrate some mathematical concepts, facts and properties
before the students. Different tables, ready reckner should also be there (in the
laminated form) so that these can be used by the students for different purposes.
Further, for performing activities such as measuring, drawing and calculating,
consulting reference books, etc., there should be equipments like mathematical
instruments, calculators, computers, books, journals mathematical dictionaries
etc., in the laboratory. In view of the above, following is the list of suggested
instruments/models for the laboratory:
EQUIPMENTS
Mathematical instrument set (Wooden Geometry Box for demonstration
containing rulers, set-squares, divider, protractor and compasses), some geometry
boxes, metre scales of 100 cm, 50 cm and 30 cm, measuring tape, diagonal scale,
clinometer, calculators, computers including related software etc.
MODELS
• Number line
• Geoboards - rectangular, circular and isometric
• Models for verifying the following identities:
(i) (a + b)2
= a2
+2ab + b2
(ii) (a – b)2
= a2
– 2ab + b2
(iii) a2
– b2
= (a – b) (a + b) (iv) k (a + b + c) = ka+ kb + kc
(v) (a + b)3
= a3
+ b3
+ 3a2
b + 3ab2
(vi) (a – b)3
= a3
– b3
– 3a2
b + 3ab2
(vii) a3
+ b3
= (a + b)3
– 3ab (a + b) (viii) a3
– b3
= (a – b)3
+ 3ab (a – b)
• Concrete models of the following:
Equilateral triangle, Isosceles triangle, Scalene triangle, Right triangle.
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Different types of quadrilaterals such as square, parallelogram, kite,
rhombus, rectangle etc., regular pentagon, regular hexagon, regular octagon,
circle, sphere, hemisphere, cuboid, cube, right circular cylinder, cone,
frustum of a cone, tetrahedron, hexahedron, regular octahedron,
dodecahedron, icosahedron.
Model for finding the centre of a circle. Models illustrating the following
concepts/properties:
Height and slant height of a cone, various criteria of congruency of triangles
(SSS,ASA, SAS, RHS),Angles in a semi-circle. Major and Minor segments
of a circle,
Models for verifying Pythagoras theorem by different methods.
• Model for deriving formula for area of a circle sliced into sectors.
• Hinged models for demonstrating the symmetry of a square, rectangle
isosceles triangle, equilateral triangle, circle.
• Overhead projector along with slides.
• CDs and films regarding teaching of mathematics, specially on some
selected topics.
• Calculators
• Computers
• Reference Books and Journals
• Photographs of mathematicians alongwith their brief life histories and
contributions in mathematics.
STATIONERY AND ODDMENTS
Rubber-bands of different colours, Marbles of different colours, a pack of
playing cards, graph paper/squared paper, dotted paper, drawing pins, erasers,
pencils, sketch pens, cellotapes, threads of different colours, glazed papers,
kite papers, tracing papers, adhesive, pins, scissors and cutters, hammers, saw,
thermocol sheets, sand paper, nails and screws of different sizes, screw drivers,
drill machine with bit set, and pliers.
Mathematics 9
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The basic principles of learning mathematics are :
(a) learning should be related to each child individually
(b) the need for mathematics should develop from an
intimate acquaintance with the environment (c) the child
should be active and interested (d) concrete material and
wide variety of illustrations are needed to aid the learning
process (e) understanding should be encouraged at each
stage of acquiring a particular skill (f) content should be
broadly based with adequate appreciation of the links
between the various branches of mathematics (g) correct
mathematical usage should be encouraged at all stages.
– Ronwill
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Activities for
Class IX
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12 Laboratory Manual
Mathematics is one of the most important cultural
components of every modern society. Its influence an other
cultural elements has been so fundamental and wide-spread
as to warrant the statement that her “most modern” ways
of life would hardly have been possilbly without mathematics.
Appeal to such obvious examples as electronics radio,
television, computing machines, and space travel, to
substantiate this statement is unnecessary : the elementary
art of calculating is evidence enough. Imagine trying to get
through three day without using numbers in some fashion
or other!
– R.L. Wilder
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Mathematics 13
METHOD OF CONSTRUCTION
1. Take a piece of plywood with dimensions 30 cm × 30 cm.
2. Taking 2 cm = 1 unit, draw a line segment AB of length one unit.
3. Construct a perpendicular BX at the line segment AB using set squares (or
compasses).
4. From BX, cut off BC = 1 unit. Join AC.
5. Using blue coloured thread (of length equal to AC) and adhesive, fix the
thread along AC.
6. WithAC as base and using set squares (or compasses), draw CYperpendicular
to AC.
7. From CY, cut-off CD = 1 unit and join AD.
OBJECTIVE MATERIAL REQUIRED
To construct a square-root spiral. Coloured threads, adhesive,
drawing pins, nails, geometry box,
sketch pens, marker, a piece of
plywood.
Fig. 1
Activity 1
Mathematics 13
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8. Fix orange coloured thread (of length equal to AD) along AD with adhesive.
9. With AD as base and using set squares (or compasses), draw DZ
perpendicular to AD.
10. From DZ, cut off DE = 1 unit and join AE.
11. Fix green coloured thread (of length equal to AE) along AE with adhesive
[see Fig. 1].
Repeat the above process for a sufficient number of times. This is called “a
square root spiral”.
DEMONSTRATION
1. From the figure, AC2
= AB2
+ BC2
= 12 + 12 = 2 or AC = 2 .
AD2
= AC2
+ CD2
= 2 + 1 = 3 or AD = 3 .
2. Similarly, we get the other lengths AE, AF, AG, ... as 4 or 2, 5 , 6 ....
OBSERVATION
On actual measurement
AC = ..... , AD = ...... , AE =...... , AF =......., AG = ......
2 = AC = ............... (approx.),
3 = AD = ............... (approx.),
4 = AE = ............... (approx.),
5 = AF = ............... (approx.)
APPLICATION
Through this activity, existence of irrational numbers can be illustrated.
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Mathematics 15
1. Make a straight slit on the top of one of the wooden strips. Fix another
wooden strip on the slit perpendicular to the former strip with a screw at
the bottom so that it can move freely along the slit [see Fig.1].
2. Paste one photocopy of the scale on each of these two strips as shown in
Fig. 1.
3. Fix nails at a distance of 1 unit each, starting from 0, on both the strips as
shown in the figure.
4. Tie a thread at the nail at 0 on the horizontal strip.
To represent some irrational numbers
on the number line.
Two cuboidal wooden strips,
thread, nails, hammer, two photo
copies of a scale, a screw with nut,
glue, cutter.
Activity 2
Fig. 1
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DEMONSTRATION
1. Take 1 unit on the horizontal scale and fix the perpendicular wooden strip
at 1 by the screw at the bottom.
2. Tie the other end of the thread to unit ‘1’ on the perpendicular strip.
3. Remove the thread from unit ‘1’ on the perpendicular strip and place it on
the horizontal strip to represent 2 on the horizontal strip [see Fig. 1].
Similarly, to represent 3 , fix the perpendicular wooden strip at 2 and
repeat the process as above. To represent a , a > 1, fix the perpendicular
scale at –1
a and proceed as above to get a .
OBSERVATION
On actual measurement:
a – 1 = ........... a = ...........
APPLICATION
The activity may help in representing some
irrational numbers such as 2 , 3 , 4 ,
5 , 6 , 7 , .... on the number line.
NOTE
You may also find a such as
13 by fixing the perpendicular
strip at 3 on the horizontal strip
and tying the other end of thread
at 2 on the vertical strip.
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Mathematics 17
1. Cut out a square of side length a units from a drawing sheet/cardboard and
name it as square ABCD [see Fig. 1].
2. Cut out another square of length b units from a drawing sheet/cardboard and
name it as square CHGF [see Fig. 2].
To verify the algebraic identity :
(a + b)2
= a2
+ 2ab + b2
Drawing sheet, cardboard, cello-
tape, coloured papers, cutter and
ruler.
Activity 3
Fig. 1 Fig. 2
3. Cut out a rectangle of length a units and breadth b units from a drawing
sheet/cardbaord and name it as a rectangle DCFE [see Fig. 3].
4. Cut out another rectangle of length b units and breadth a units from a drawing
sheet/cardboard and name it as a rectangle BIHC [see Fig. 4].
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5. Total area of these four cut-out figures
= Area of square ABCD + Area of square CHGF + Area of rectangle DCFE
+ Area of rectangle BIHC
= a2
+ b2
+ ab + ba = a2
+ b2
+ 2ab.
6. Join the four quadrilaterals using cello-tape as shown in Fig. 5.
Fig. 3 Fig. 4
Fig. 5
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Mathematics 19
Clearly, AIGE is a square of side (a + b). Therefore, its area is (a + b)2
. The
combined area of the constituent units = a2
+ b2
+ ab + ab = a2
+ b2
+ 2ab.
Hence, the algebraic identity (a + b)2
= a2
+ 2ab + b2
Here, area is in square units.
OBSERVATION
a = .............., b = .............. (a+b) = ..............,
So, a2
= .............. b2
= .............., ab = ..............
(a+b)2
= .............., 2ab = ..............
Therefore, (a+b)2
= a2
+ 2ab + b2
.
The identity may be verified by taking different values of a and b.
APPLICATION
The identity may be used for
1. calculating the square of a number expressed as the sum of two convenient
numbers.
2. simplifications/factorisation of some algebraic expressions.
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1. Cut out a squareABCD of side a units from a drawing sheet/cardboard [see Fig. 1].
2. Cut out a square EBHI of side b units (b < a) from a drawing sheet/cardboard
[see Fig. 2].
3. Cut out a rectangle GDCJ of length a units and breadth b units from a drawing
sheet/cardboard [see Fig. 3].
4. Cut out a rectangle IFJH of length a units and breadth b units from a drawing
sheet/cardboard [see Fig. 4].
(a – b)2
= a2
– 2ab + b2
Drawing sheets, cardboard,
coloured papers, scissors, ruler and
adhesive.
Activity 4
Fig. 1
Fig. 2
Fig. 3
Fig. 4
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Mathematics 21
5. Arrange these cut outs as shown in Fig. 5.
DEMONSTRATION
According to figure 1, 2, 3, and 4, Area of
square ABCD = a2
, Area of square EBHI = b2
Area of rectangle GDCJ = ab, Area of
rectangle IFJH = ab
From Fig. 5, area of square AGFE =AG × GF
= (a – b) (a – b) = (a – b)2
Now, area of square AGFE = Area of square
ABCD + Area of square EBHI
– Area of rectangle IFJH – Area of rectangle
GDCJ
= a2
+ b2
– ab – ab
= a2
– 2ab + b2
OBSERVATION
a = .............., b = .............., (a – b) = ..............,
So, a2
= .............., b2
= .............., (a – b)2
= ..............,
ab = .............., 2ab = ..............
Therefore, (a – b)2
= a2
– 2ab + b2
APPLICATION
1. calculating the square of a number expressed as a difference of two
convenient numbers.
2. simplifying/factorisation of some algebraic expressions.
Fig. 5
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1. Take a cardboard of a convenient size and paste a coloured paper on it.
2. Cut out one square ABCD of side a units from a drawing sheet [see Fig. 1].
3. Cut out one square AEFG of side b units (b < a) from another drawing sheet
[see Fig. 2].
a2
– b2
= (a + b)(a – b)
Drawing sheets, cardboard,
coloured papers, scissors, sketch
pen, ruler, transparent sheet and
adhesive.
Activity 5
Fig. 1 Fig. 2
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Mathematics 23
4. Arrange these squares as shown in Fig. 3.
5. Join F to C using sketch pen. Cut out trapeziums congruent to EBCF and
GFCD using a transparent sheet and name them as EBCF and GFCD,
respectively [see Fig. 4 and Fig. 5].
Fig. 3 Fig. 4
Fig. 5
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6. Arrange these trapeziums as shown in
Fig. 6.
DEMONSTRATION
Area of square ABCD = a2
Area of square AEFG = b2
In Fig. 3,
Area of square ABCD – Area of square
AEFG
= Area of trapezium EBCF + Area of
trapezium GFCD
= Area of rectangle EBGD [Fig. 6].
= ED × DG
Thus, a2
– b2
= (a+b) (a–b)
OBSERVATION
a = .............., b = .............., (a+b) = ..............,
So, a2
= .............., b2
= .............., (a–b) = ..............,
a2
–b2
= ..............,(a+b) (a–b) = ..............,
Therefore, a2
–b2
= (a+b) (a–b)
APPLICATION
1. difference of two squares
2. some products involving two numbers
3. simplification and factorisation of algebraic expressions.
Fig. 6
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Mathematics 25
1. Take a hardboard of a convenient size and paste a white paper on it.
2. Cut out a square of side a units from a coloured paper [see Fig. 1].
3. Cut out a square of side b units from a coloured paper [see Fig. 2].
4. Cut out a square of side c units from a coloured paper [see Fig. 3].
5. Cut out two rectangles of dimensions a× b, two rectangles of dimensions
b × c and two rectangles of dimensions c × a square units from a coloured
paper [see Fig. 4].
(a+b+c)2
= a2
+ b2
+ c2
+ 2ab + 2bc + 2ca
Hardboard, adhesive, coloured
papers, white paper.
Activity 6
Fig. 1
Fig. 4
Fig. 2 Fig. 3
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6. Arrange the squares and rectangles on
the hardboard as shown in Fig. 5.
DEMONSTRATION
From the arrangement of squares and
rectangles in Fig. 5, a square ABCD is
obtained whose side is (a+b+c) units.
Area of square ABCD = (a+b+c)2
.
Therefore, (a+b+c)2
= sum of all the
squares and rectangles shown in Fig. 1 to
Fig. 4.
= a2
+ ab + ac + ab + b2
+ bc + ac + bc + c2
= a2
+ b2
+ c2
+ 2ab + 2bc + 2ca
OBSERVATION
a = .............., b = .............., c = ..............,
So, a2
= .............., b2
= ..............,c2
= .............., ab= ..............,
bc= .............., ca = ..............,2ab = .............., 2bc = ..............,
2ca= .............., a+b+c = .............., (a+b+c)2
= ..............,
Therefore, (a+b+c)2
= a2
+ b2
+c2
+2ab + 2bc + 2ca
APPLICATION
1. simiplification/factorisation of algebraic expressions
2. calculating the square of a number expressed as a sum of three convenient
numbers.
Fig. 5
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Mathematics 27
1. Make a cube of side a units and one more cube of side b units (b < a), using
acrylic sheet and cello-tape/adhesive [see Fig. 1 and Fig. 2].
2. Similarly, make three cuboids of dimensions a×a×b and three cuboids of
dimensions a×b×b [see Fig. 3 and Fig. 4].
(a+b)3
= a3
+ b3
+ 3a2
b + 3ab2
Acrylic sheet, coloured papers,
glazed papers, saw, sketch pen,
adhesive, Cello-tape.
Activity 7
Fig. 1
Fig. 2
Fig. 3
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3. Arrange the cubes and cuboids as shown in Fig. 5.
Fig. 4
Fig. 5
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Mathematics 29
DEMONSTRATION
Volume of the cube of side a = a×a×a = a3
, volume of the cube of side b = b3
Volume of the cuboid of dimensions a×a×b = a2
b, volume of three such cuboids
= 3a2
b
Volume of the cuboid of dimensions a×b×b = ab2
, volume of three such cuboids
= 3ab2
Solid figure obtained in Fig. 5 is a cube of side (a + b)
Its volume = (a + b)3
Therefore, (a+b)3
= a3
+ b3
+ 3a2
b + 3ab2
Here, volume is in cubic units.
OBSERVATION
a = .............., b = ............., a3
= ..............,
So, a3
= .............., b3
= ............., a2
b = .............., 3a2
b= ..............,
ab2
= .............., 3ab2
= .............., (a+b)3
= ..............,
Therefore, (a+b)3
= a3
+ b3
+3a2
b + 3ab2
APPLICATION
1. calculating cube of a number expressed as the sum of two convenient
numbers
2. simplification and factorisation of algebraic expressions.
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1. Make a cube of side (a – b) units (a > b)using acrylic sheet and cellotape/
adhesive [see Fig. 1].
2. Make three cuboids each of dimensions (a–b) × a × b and one cube of side
b units using acrylic sheet and cellotape [see Fig. 2 and Fig. 3].
3. Arrange the cubes and cuboids as shown in Fig. 4.
To verify the algebraic identity
(a – b)3
= a3
– b3
– 3(a – b)ab
Acrylic sheet, coloured papers,
saw, sketch pens, adhesive, Cello-
tape.
Activity 8
Fig. 1
Fig. 2
14/04/18

Mathematics 31
Fig. 3 Fig. 4
DEMONSTRATION
Volume of the cube of side (a – b) units in Fig. 1 = (a– b)3
Volume of a cuboid in Fig. 2 = (a–b) ab
Volume of three cuboids in Fig. 2 = 3 (a–b) ab
Volume of the cube of side b in Fig. 3 = b3
Volume of the solid in Fig. 4 = (a–b)3
+ (a–b) ab + (a–b) ab + (a – b) ab + b3
= (a–b)3
+ 3(a–b) ab + b3
(1)
Also, the solid obtained in Fig. 4 is a cube of side a
Therefore, its volume = a3
(2)
From (1) and (2),
(a–b)3
+ 3(a–b) ab + b3
= a3
or (a–b)3
= a3
– b3
– 3ab (a–b).
Here, volume is in cubic units.
14/04/18

OBSERVATION
a = .............., b = .............., a–b = ..............,
So, a3
= .............., ab = ..............,
b3
= .............., ab(a–b) = ..............,
3ab (a–b) = .............., (a–b)3
= ..............,
Therefore, (a–b)3
= a3
– b3
– 3ab(a–b)
APPLICATION
1. calculating cube of a number
expressed as a difference of two
convenient numbers
2. simplification and factorisation of
algebraic expressions.
NOTE
This identity can also be
expressed as :
(a – b)3
= a3
– 3a2
b + 3ab2
– b3
.
14/04/18

Mathematics 33
1. Make a cube of side a units and another cube of side b units as shown in
Fig. 1 and Fig. 2 by using acrylic sheet and cellotape/adhesive.
2. Make a cuboid of dimensions a × a × b [see Fig. 3].
3. Make a cuboid of dimensions a × b × b [see Fig. 4].
4. Arrange these cubes and cuboids as shown in Fig. 5.
a3
+ b3
= (a + b) (a2
– ab + b2
)
Acrylic sheet, glazed papers, saw,
adhesive, cellotape, coloured
papers, sketch pen, etc.
Activity 9
Fig. 1
Fig. 2
Fig. 3
Fig. 4 Fig. 5
14/04/18

DEMONSTRATION
Volume of cube in Fig. 1 = a3
Volume of cube in Fig. 2 = b3
Volume of cuboid in Fig. 3 = a2
b
Volume of cuboid in Fig. 4 = ab2
Volume of solid in Fig. 5 = a3
+b3
+ a2
b + ab2
= (a+b) (a2
+ b2
)
Removing cuboids of volumes a2
b and ab2
, i.e.,
ab (a + b) from solid obtained in Fig. 5, we
get the solid in Fig. 6.
+ b3
.
Therefore, a3
+ b3
= (a+b) (a2
+ b2
) – ab (a + b)
= (a+b) (a2
+ b2
– ab)
Here, volumes are in cubic units.
OBSERVATION
a = .............., b = ..............,
So, a3
= .............., b3
= .............., (a+b) = .............., (a+b)a2
= ..............,
(a+b) b2
= .............., a2
b = .............., ab2
= ..............,
ab (a+b) = ..............,
Therefore, a3
+ b3
= (a + b) (a2
+ b2
– ab).
APPLICATION
The identity may be used in simplification and factorisation of algebraic
expressions.
Fig. 6
14/04/18

Mathematics 35
1. Make a cuboid of dimensions (a–b) × a × a (b < a), using acrylic sheet and
cellotape/adhesive as shown in Fig. 1.
2. Make another cuboid of dimensions (a–b) × a × b, using acrylic sheet and
cellotape/adhesive as shown in Fig. 2.
3. Make one more cuboid of dimensions (a–b) × b × b as shown in Fig. 3.
4. Make a cube of dimensions b × b × b using acrylic sheet as shown in Fig. 4.
a3
– b3
= (a – b)(a2
+ ab + b2
)
Acrylic sheet, sketch pen, glazed
papers, scissors, adhesive, cello-
tape, coloured papers, cutter.
Activity 10
Fig. 1
Fig. 2
Fig. 3
Fig. 4
14/04/18

5. Arrange the cubes and cuboids made above in Steps (1), (2), (3) and (4) to
obtain a solid as shown in Fig. 5, which is a cube of volume a3
cubic units.
DEMONSTRATION
Volume of cuboid in Fig. 1 = (a–b) × a × a cubic units.
Volume of cuboid in Fig. 2 = (a–b) × a × b cubic units.
Volume of cuboid in Fig. 3 = (a–b) × b × b cubic units.
Volume of cube in Fig. 4 = b3
cubic units.
cubic units.
Removing a cube of size b3
cubic units from the solid in Fig. 5, we obtain a solid
as shown in Fig. 6.
Volume of solid in Fig. 6 = (a–b) a2
+ (a–b) ab + (a–b) b2
= (a–b) (a2
+ ab + b2
)
Therefore, a3
– b3
= (a – b)(a2
+ ab + b2
)
Fig. 5
Fig. 6
14/04/18

Mathematics 37
OBSERVATION
a = .............., b = ..............,
So, a3
= .............., b3
= .............., (a–b) = .............., ab = ..............,
a2
= .............., b2
= ..............,
Therefore, a3
– b3
= (a – b) (a2
+ ab + b2
).
APPLICATION
The identity may be used in simplification/factorisation of algebraic expressions.
14/04/18

1. Take a cardboard of a convenient size and paste a white paper on it.
2. Paste the given graph paper alongwith various points drawn on it [see Fig. 1].
3. Look at the graph paper and the points whose abcissae and ordinates are to
be found.
DEMONSTRATION
To find abscissa and ordinate of a point, say A, draw perpendiculars AM and AN
from A to x-axis and y-axis, respectively. Then abscissa of A is OM and ordinate
of A is ON. Here, OM = 2 and AM = ON = 9. The point A is in first quadrant.
Coordinates of A are (2, 9).
OBSERVATION
To find the values of abscissae and
ordinates of various points given in a
cartesian plane.
Cardboard, white paper, graph paper
with various given points, geometry
box, pen/pencil.
Activity 11
Point Abscissa Ordinate Quadrant Coordinates
B
C
...
...
...
...
14/04/18

Mathematics 39
APPLICATION
This activity is helpful in locating the position
of a particular city/place or country on map.
PRECAUTION
The students should be careful
while reading the coordinates,
otherwise the location of the
object will differ.
Fig. 1
14/04/18

2. Take a graph paper and paste it on the white paper.
3. Draw two rectangular axes X′OX and Y′OY as shown in Fig. 1.
4. Plot the points A, B, C, ... with given coordinates (a, b), (c, d), (e, f), ...,
respectively as shown in Fig. 2.
5. Join the points in a given order say A→B→C→D→.....→A [see Fig. 3].
To find a hidden picture by plotting
and joining the various points with
given coordinates in a plane.
Cardboard, white paper, cutter,
adhesive, graph paper/squared
paper, geometry box, pencil.
Activity 12
Fig. 1
Fig. 2
14/04/18

Mathematics 41
DEMONSTRATION
By joining the points as per given instructions, a ‘hidden’ picture of an ‘aeroplane’
is formed.
OBSERVATION
In Fig. 3:
Coordinates of points A, B, C, D, .......................
are ........, ........, ........, ........, ........, ........, ........
Hidden picture is of ______________.
APPLICATION
This activity is useful in understanding the plotting of points in a cartesian plane
which in turn may be useful in preparing the road maps, seating plan in the
classroom, etc.
Fig. 3
14/04/18

2. Paste a full protractor (0° to 360º) on the cardboard, as shown in Fig. 1.
3. Mark the centre of the protractor as O.
4. Make a hole in the middle of each transparent strip containing two
intersecting lines.
5. Now fix both the strips at O by putting a nail as shown in Fig. 1.
To verify experimentally that if two lines
intersect, then
(i) the vertically opposite angles are
equal
(ii) thesumoftwoadjacentanglesis180º
(iii) the sum of all the four angles is 360º.
Two transparent strips marked as
AB and CD, a full protractor, a nail,
cardboard, white paper, etc.
Activity 13
Fig. 1
14/04/18

Mathematics 43
DEMONSTRATION
1. Observe the adjacent angles and the vertically opposite angles formed in
different positions of the strips.
2. Compare vertically opposite angles formed by the two lines in the strips in
different positions.
3. Check the relationship between the vertically opposite angles.
4. Check that the vertically opposite angles ∠AOD, ∠COB, ∠COA and ∠BOD
are equal.
5. Compare the pairs of adjacent angles and check that ∠COA + ∠DOA= 180º,
etc.
6. Find the sum of all the four angles formed at the point O and see that the
sum is equal to 360º.
OBSERVATION
On actual measurement of angles in one position of the strips :
1. ∠AOD = ................., ∠AOC = ...................
∠COB = ................., ∠BOD = .................
Therefore, ∠AOD = ∠COB and ∠AOC = ............ (vertically opposite angles).
2. ∠AOC + ∠AOD = ............., ∠AOC + ∠BOC = ...................,
∠COB + ∠BOD = ...................
∠AOD + ∠BOD = ................... (Linear pairs).
3. ∠AOD + ∠AOC + ∠COB + ∠BOD = .................... (angles formed at a point).
APPLICATION
These properties are used in solving many geometrical problems.
14/04/18

2. Make a pair of trianglesABC and DEF in which AB = DE, BC = EF, AC = DF
on a glazed paper and cut them out [see Fig. 1].
3. Make a pair of triangles GHI, JKL in which GH = JK, GI = JL, ∠G = ∠J on
a glazed paper and cut them out [see Fig. 2].
To verify experimentally the different
criteria for congruency of triangles using
triangle cut-outs.
Cardboard, scissors, cutter, white
paper, geometry box, pencil/sketch
pens, coloured glazed papers.
Activity 14
Fig. 1
Fig. 2
14/04/18

Mathematics 45
4. Make a pair of triangles PQR, STU in which QR = TU, ∠Q = ∠T, ∠R = ∠U
on a glazed paper and cut them out [see Fig. 3].
5. Make two right triangles XYZ, LMN in which hypotenuse YZ = hypotenuse
MN and XZ = LN on a glazed paper and cut them out [see Fig. 4].
Fig. 3
Fig. 4
DEMONSTRATION
1. Superpose DABC on DDEF and see whether one triangle covers the other
triangle or not by suitable arrangement. See that ∆ABC covers ∆DEF
completely only under the correspondence A↔D, B↔E, C→F. So, ∆ABC
≅ ∆DEF, if AB = DE, BC = EF and AC = DF.
This is SSS criterion for congruency.
14/04/18

2. Similarly, establish ∆GHI ≅ ∆JKL if GH = JK. ∠G = ∠J and GI = JL. This is
SAS criterion for congruency.
3. Establish ∆PQR ≅ ∆STU, if QR = TU, ∠Q = ∠T and ∠R = ∠U.
This is ASA criterion for congruency.
4. In the same way, ∆STU ≅ ∆LMN, if hypotenuse YZ = hypotenuse MN and
XZ = LN.
This is RHS criterion for right triangles.
OBSERVATION
On actual measurement :
In ∆ABC and ∆DEF,
AB = DE = ..................., BC = EF = ...................,
AC = DF = ..................., ∠A = ...................,
∠D = ..................., ∠B = ..................., ∠E = ...................,
∠C = ..................., ∠F = ....................
Therefore, ∆ABC ≅ ∆DEF.
2. In ∆GHI and ∆JKL,
GH = JK = ..................., GI = JL = ...................., HI = ...................,
KL= ..................., ∠G = ..................., ∠J = ...................,
∠H = ..................., ∠K = ..................., ∠I = ...................,
∠L = ....................
Therefore, ∆GHI ≅ ∆JKL.
3. In ∆PQR and ∆STU,
QR = TU = ..................., PQ = ..................., ST = ...................,
PR = ..................., SU = .................... ∠S = ...................,
∠Q = ∠T = ..................., ∠R = ∠U = ..................., ∠P = ....................
Therefore, ∆PQR ≅ ∆STU.
14/04/18

Mathematics 47
4. In ∆XYZ and ∆LMN, hypotenuse YZ = hypotenuse MN = .............
XZ = LN = ..................., XY = ...................,
LM = ..................., ∠X = ∠L = 90°
∠Y = ..................., ∠M = ..................., ∠Z = ...................,
∠N = ...................,
Therefore, ∆XYZ ≅ ∆LMN.
APPLICATION
These criteria are useful in solving a number of problems in geometry.
These criteria are also useful in solving some practical problems such as finding
width of a river without crossing it.
14/04/18

1. Take a hardboard sheet of a convenient size and paste a white paper on it.
2. Cut out a triangle from a drawing sheet, and paste it on the hardboard and
name it as ∆ABC.
3. Mark its three angles as shown in Fig. 1
4. Cut out the angles respectively equal to ∠A, ∠B and ∠C from a drawing
sheet using tracing paper [see Fig. 2].
To verify that the sum of the angles of
a triangle is 180º.
Hardboard sheet, glazed papers,
sketch pens/pencils, adhesive,
cutter, tracing paper, drawing sheet,
geometry box.
Activity 15
Fig. 1
Fig. 2
14/04/18

Mathematics 49
5. Draw a line on the hardboard and arrange the cut-outs of three angles at a
point O as shown in Fig. 3.
DEMONSTRATION
The three cut-outs of the three angles A, B and C placed adjacent to each other
at a point form a line forming a straight angle = 180°. It shows that sum of the
three angles of a triangle is 180º. Therefore, ∠A + ∠B + ∠C = 180°.
OBSERVATION
Measure of ∠A = -------------------.
Measure of ∠B = -------------------.
Measure of ∠C = -------------------.
Sum (∠A + ∠B + ∠C) = -------------------.
APPLICATION
This result may be used in a number of geometrical problems such as to find the
sum of the angles of a quadrilateral, pentagon, etc.
Fig. 3
14/04/18

1. Take a hardboard sheet of a convenient size and paste a white paper on it.
2. Cut out a triangle from a drawing sheet/glazed paper and name it as ∆ABC
and paste it on the hardboard, as shown in Fig. 1.
3. Produce the side BC of the triangle to a point D as shown in Fig. 2.
To verify exterior angle property of a
triangle.
Hardboard sheet, adhesive, glazed
papers, sketch pens/pencils,
drawing sheet, geometry box,
tracing paper, cutter, etc.
Activity 16
Fig. 1
Fig. 2
14/04/18

Mathematics 51
4. Cut out the angles from the drawing sheet equal to ∠A and ∠B using a tracing
paper [see Fig. 3].
5. Arrange the two cutout angles as shown in Fig. 4.
DEMONSTRATION
∠ACD is an exterior angle.
∠A and ∠B are its two interior opposite angles.
∠A and ∠B in Fig. 4 are adjacent angles.
From the Fig. 4, ∠ACD = ∠A + ∠B.
OBSERVATION
Measure of ∠A= __________, Measure of ∠B = __________,
Sum (∠A + ∠B) = ________, Measure of ∠ACD = _______.
Therefore, ∠ACD = ∠A + ∠B.
APPLICATION
This property is useful in solving many geometrical problems.
Fig. 3
Fig. 4
14/04/18

1. Take a rectangular cardboard piece of a convenient size and paste a white
paper on it.
2. Cut out a quadrilateral ABCD from a drawing sheet and paste it on the
cardboard [see Fig. 1].
3. Make cut-outs of all the four angles of the quadrilateral with the help of a
tracing paper [see Fig. 2]
To verify experimentally that the sum
of the angles of a quadrilateral is 360º.
Cardboard, white paper, coloured
drawing sheet, cutter, adhesive,
geometry box, sketch pens, tracing
paper.
Activity 17
Fig. 1
14/04/18

Mathematics 53
Fig. 2
4. Arrange the four cut-out angles at a point O as shown in Fig. 3.
DEMONSTRATION
1. The vertex of each cut-out angle
coincides at the point O.
2. Such arrangement of cut-outs
shows that the sum of the angles
of a quadrilateral forms a
complete angle and hence is
equal to 360º.
OBSERVATION
Measure of ∠A = ----------.
Measure of ∠B = ----------. Measure of ∠C = ----------.
Measure of ∠D = ----------. Sum [ ∠A+ ∠B+ ∠C+ ∠D] = -------------.
APPLICATION
This property can be used in solving problems relating to special types of
quadrilaterals, such as trapeziums, parallelograms, rhombuses, etc.
Fig. 3
14/04/18

1. Take a piece of cardboard of a convenient size and paste a white paper on it.
2. Cut out a ∆ABC from a coloured paper and paste it on the cardboard [see
Fig. 1].
3. Measure the lengths of the sides of ∆ABC.
4. Colour all the angles of the triangle ABC as shown in Fig. 2.
5. Make the cut-out of the angle opposite to the longest side using a tracing
paper [see Fig. 3].
To verify experimentally that in a
triangle, the longer side has the greater
angle opposite to it.
Coloured paper, scissors, tracing
paper, geometry box, cardboard
sheet, sketch pens.
Activity 18
Fig. 1
Fig. 2
Fig. 3
14/04/18

Mathematics 55
DEMONSTRATION
Take the cut-out angle and compare it with other two angles as shown in Fig. 4.
∠A is greater than both ∠B and ∠C.
i.e., the angle opposite the longer side is greater than the angle opposite the
other side.
OBSERVATION
Length of side AB = .......................
Length of side BC = .......................
Length of side CA = .......................
Measure of the angle opposite to longest side = .......................
Measure of the other two angles = ...................... and .......................
The angle opposite the ...................... side is ...................... than either of the other
two angles.
APPLICATION
The result may be used in solving different geometrical problems.
Fig. 4
14/04/18

1. Take a rectangular piece of plywood of convenient size and paste a graph
paper on it.
2. Fix two horizontal wooden strips on it parallel to each other [see Fig. 1].
To verify experimentally that the
parallelograms on the same base and
between same parallels are equal in area.
A piece of plywood, two wooden
strips, nails, elastic strings, graph
paper.
Activity 19
3. Fix two nails A1
and A2
on one of the strips [see Fig. 1].
4. Fix nails at equal distances on the other strip as shown in the figure.
DEMONSTRATION
1. Put a string along A1
, A2
, B8
, B2
which forms a parallelogram A1
A2
B8
B2
. By
counting number of squares, find the area of this parallelogram.
Fig. 1
14/04/18

Mathematics 57
2. Keeping same base A1
A2
, make another parallelogram A1
A2
B9
B3
and find
the area of this parallelogram by counting the squares.
3. Area of parallelogram in Step 1 = Area of parallelogram in Step 2.
OBSERVATION
Number of squares in 1st parallelogram = --------------.
Number of squares in 2nd parallelogram = -------------------.
Number of squares in 1st parallelogram = Number of squares in 2nd
parallelogram.
Area of 1st parallelogram = --------- of 2nd parallelogram
APPLICATION
This result helps in solving various
geometrical problems. It also helps in
deriving the formula for the area of a
paralleogram.
NOTE
In finding the area of a
parallelogram, by counting
squares, find the number of
complete squares, half squares,
more than half squares. Less than
half squares may be ignored.
14/04/18

1. Cut a rectangular plywood of a convenient size.
2. Paste a graph paper on it.
3. Fix any two horizontal wooden strips on it which are parallel to each other.
4. Fix two points A and B on the paper along the first strip (base strip).
5. Fix a pin at a point, say at C, on the second strip.
6. Join C to A and B as shown in Fig. 1.
To verify that the triangles on the same
base and between the same parallels are
equal in area.
Apiece of plywood, graph paper, pair
of wooden strips, colour box , scissors,
cutter, adhesive, geometry box.
Activity 20
Fig. 1
7. Take any other two points on the second strip say C′
and C′′
[see Fig. 2].
8. Join C′
A, C′
B, C′′
A and C′′
B to form two more triangles.
14/04/18

Mathematics 59
DEMONSTRATION
1. Count the number of squares contained in each of the above triangles, taking
half square as
1
2
and more than half as 1 square, leaving those squares which
contain less than
1
2
squares.
2. See that the area of all these triangles is the same. This shows that triangles
on the same base and between the same parallels are equal in area.
OBSERVATION
1. The number of squares in triangle ABC =.........., Area of ∆ABC = ........ units
2. The number of squares in triangle ABC′
=......., Area of D ABC′
= ........ units
3. The number of squares in triangle ABC′′
=....... ,Area of D ABC′′
= ........ units
Therefore, area (∆ABC) = ar(ABC′
) = ar(ABC′′
).
APPLICATION
This result helps in solving various geometric problems. It also helps in finding
the formula for area of a triangle.
Fig. 2
14/04/18

1. Take a rectangular plywood sheet.
2. Paste a graph paper on it.
3. Take any pair of wooden strips or wooden scale and fix these two horizontally
so that they are parallel.
4. Fix any two points A and B on the base strip (say Strip I) and take any two
points C and D on the second strip (say Strip II) such that AB = CD.
5. Take any point P on the second strip and join it to A and B [see Fig. 1].
To verify that the ratio of the areas of a
parallelogram and a triangle on the same
baseandbetweenthesameparallelsis2:1.
Plywood sheet of convenient size,
graph paper, colour box, a pair of
wooden strips, scissors, cutter,
adhesive, geometry box.
Activity 21
Fig. 1
14/04/18

Mathematics 61
DEMONSTRATION
1. AB is parallel to CD and P is any point on CD.
2. Triangle PAB and parallelogramABCD are on the same baseAB and between
the same parallels.
3. Count the number of squares contained in each of the above triangle and
parallelograms, keeping half square as
1
2
and more than half as 1 square,
leaving those squares which contain less than half square.
4. See that area of the triangle PAB is half of the area of parallelogramsABCD.
OBSERVATION
1. The number of squares in triangle PAB =...............
2. The number of squares in parallelogram ABCD =............... .
So, the area of parallelogram ABCD = 2 [Area of triangle PAB]
Thus, area of parallelogram ABCD : area of DPAB = ........ : ...........
Fig. 2
14/04/18

NOTE
You may take different triangles
PAB by taking different positions
of point P and the two parallel
strips as shown in Fig. 2.
APPLICATION
This activity is useful in deriving formula
for the area of a triangle and also in solving
problems on mensuration.
One should study Mathematics because it is only through
Mathematics that nature can be conceived in hormonious
form.
– Birkhoff
14/04/18

Mathematics 63
1. Take a rectangular cardboard of a
convenient size and paste a white paper
on it.
2. Cut out a circle of suitable radius on a
coloured drawing sheet and paste on the
cardboard.
3. Take two points B and C on the circle to
obtain the arc BC [see Fig. 1].
4. Join the points B and C to the centre O to
obtain an angle subtended by the arc BC
at the centre O.
5. Take any point A on the remaining part of
the circle. Join it to B and C to get ∠BAC
subtended by the arc BC on any point A
on the remaining part of the circle
[see Fig. 1].
6. Make a cut-out of ∠BOC and two cut-
outs of angle BAC, using transparent sheet
[see Fig. 2].
To verify that the angle subtended by an
arc of a circle at the centre is double the
angle subtended by it at any point on the
remaining part of the circle.
Cardboard, coloured drawing
sheets, scissors, sketch pens,
adhesive, geometry box, transparent
sheet.
Activity 22
Fig. 1
Fig. 2
14/04/18

DEMONSTRATION
Place the two cut-outs of ∠BAC on the cut-out of angle BOC, adjacent to each
as shown in the Fig. 3. Clearly, 2 ∠BAC = ∠BOC, i.e., the angle subtended by an
arc at the centre is double the angle subtended by it at any point on the remaining
part of the circle.
OBSERVATION
Measure of ∠BOC = .........................
Measure of ∠BAC = .........................
Therefore, ∠BOC = 2 ........................
APPLICATION
This property is used in proving many other important results such as angles in
the same segment of a circle are equal, opposite angles of a cyclic quadrilateral
are supplementary, etc.
Fig. 3
14/04/18

Mathematics 65
To verify that the angles in the same
segment of a circle are equal.
Geometry box, coloured glazed
papers, scissors, cardboard, white
paper and adhesive.
Activity 23
1. Take a cardboard of suitable size and paste a white paper on it.
2. Take a sheet of glazed paper and draw a circle of radius a units on it [see Fig. 1].
3. Make a cut-out of the circle and paste it on the cardboard.
4. Take two points A and B on the circle and join them to form chord AB
[see Fig. 2].
5. Now take two points C and D on the circle in the same segment and join AC,
BC, AD and BD [see Fig. 3].
6. Take replicas of the angles ∠ACB and ∠ADB.
Fig. 1 Fig. 2
14/04/18

DEMONSTRATION
Put the cut-outs of ∠ACB and ∠ADB on each other such that vertex C falls on
vertex D [see Fig. 4]. In Fig. 4, ∠ACB covers ∠ADB completely. So,
∠ACB = ∠ADB.
OBSERVATION
∠ACB = ---------------, ∠ADB = ---------------
So, ∠ACB = ∠ADB. Thus, angles in the same segment are ---------.
APPLICATION
This result may be used in proving other theorems/riders of geometry related to
circles.
Fig. 3 Fig. 4
14/04/18

Mathematics 67
To verify that the opposite angles of a
cyclic quadrilateral are supplementary.
Chart paper, geometry box,
scissors, sketch pens, adhesive,
transparent sheet.
Activity 24
1. Take a chart paper and draw a circle of
radius on it.
2. In the circle, draw a quadrilateral so that
all the four vertices of the quadrilateral
lie on the circle. Name the angles and
colour them as shown in Fig. 1.
3. Make the cut-outs of the angles as shown
in Fig. 2.
Fig. 2
Fig. 3
Fig. 1
14/04/18

DEMONSTRATION
Paste cut-outs of the opposite angles ∠1 and ∠3, ∠2 and ∠4 to make straight
angles as shown in Fig. 3. Thus ∠1 + ∠3 = 180° and ∠2 + ∠4 = 180°.
OBSERVATION
∠1 = ................; ∠2 = ................; ∠3 = ................; ∠4 = .................
So, ∠1 + ∠3 = ..........; ∠2 + ∠4 = ..........;
Therefore, sum of each pair of the opposite angles of a cyclic quadrilateral is
........................ .
APPLICATION
The concept may be used in solving various problems in geometry.
14/04/18

Mathematics 69
To find the formula for the area of a
trapezium experimentally.
Hardboard, thermocol, coloured
glazed papers, adhesive, scissors.
Activity 25
1. Take a piece of hardboard for the base of the model.
2. Cut two congruent trapeziums of parallel sides a and b units [see Fig. 1].
3. Place them on the hardboard as shown in Fig. 2.
Fig. 2
Fig. 1
14/04/18

DEMONSTRATION
1. Figure formed by the two trapeziums [see Fig. 2] is a parallelogram ABCD.
2. Side AB of the parallelogram = (a + b) units and its corresponding altitude
= h units.
3. Area of each trapezium
1
2
= (area of parallelogram) ( )
1
2
a b h
= + ×
Therefore, area of trapezium ( )
1
2
a b h
= + ×
1
2
= (sum of parallel sides) × perpendicular distance.
OBSERVATION
Lengths of parallel sides of the trapezium = -------,-------.
Length of altitude of the parallelogram = --------.
Area of parallelogram = ---------------.
Area of the trapezium
1
2
= (Sum of ---------- sides) × ---------.
APPLICATION
This concept is used for finding the formula for area of a triangle in coordinate
geometry. This may also be used in finding the area of a field which can be split
into different trapeziums and right triangles.
14/04/18

Mathematics 71
To form a cube and find the formula for
its surface area experimentally.
Cardboard, ruler, cutter, cellotape,
sketch pen/pencil.
Activity 26
1. Make six identical squares each of side a units, using cardboard and join
them as shown in Fig. 1 using a cellotape.
2. Fold the squares along the dotted markings to form a cube [see Fig. 2].
Fig. 1
14/04/18

DEMONSTRATION
1. Each face of the cube so obtained is a square of side a units. Therefore, area
of one face of the cube is a2
square units.
2. Thus, the surface area of the cube with side a units = 6a2
square units.
OBSERVATION
Length of side a = ..................
Area of one square / one face = a2
= ............... .
So, sum of the areas of all the squares = ..........+............+..........+ ..........+ ..........
+ ............
Therefore, surface area of the cube = 6a2
APPLICATION
This result is useful in estimating
materials required for making cubical
boxes needed for packing.
Fig. 2
NOTE
Instead of making six squares
separately as done in the activity,
a net of a cube be directly
prepared on the cardboard itself.
14/04/18

Mathematics 73
To form a cuboid and find the formula
for its surface area experimentally.
Cardboard, cellotape, cutter, ruler,
sketch pen/pencil.
Activity 27
1. Make two identical rectangles of dimensions a units × b units, two identical
rectangles of dimensions b units × c units and two identical rectangles of
dimensions c units × a units, using a cardboard and cut them out.
2. Arrange these six rectangles as shown in Fig. 1 to obtain a net for the cuboid
to be made.
3. Fold the rectangles along the dotted markings using cello-tape to form a
cuboid [see Fig. 2].
Fig. 1
Fig. 2
14/04/18

DEMONSTRATION
Area of a rectangle of dimensions ( a units × b units) = ab square units.
Area of a rectangle of dimensions ( b units × c units) = bc square units.
Area of a rectangle of dimensions ( c units × a units) = ca square units.
Surface area of the cuboid so formed
= (2 × ab + 2 × bc + 2 × ca) square units = 2 (ab + bc + ca) square units.
OBSERVATION
a = ....................., b = ....................., c = .....................,
So, ab = ....................., bc = ....................., ca = .....................,
2ab = ....................., 2bc = ....................., 2ca = .....................
Sum of areas of all the six rectangles = ..............
Therefore, surface area of the cuboid = 2 (ab+bc+ca)
APPLICATION
This result is useful in estimating
materials required for making cuboidal
boxes/almirahs, etc.
NOTE
Instead of making six rectangles
separately, as done in the activity,
a net of a cuboid be directly
prepared on the cardboard itself.
14/04/18

Mathematics 75
To form a cone from a sector of a circle
and to find the formula for its curved
surface area.
Wooden hardboard, acrylic sheets,
cellotape, glazed papers, sketch
pens, white paper, nails, marker.
Activity 28
1. Take a wooden hardboard of a convenient size and paste a white paper on it.
2. Cut out a circle of radius l from a acrylic sheet [see Fig. 1].
3. Cut out a sector of angle q degrees from the circle [see Fig. 2].
4. Bring together both the radii of the sector to form a cone and paste the ends
using a cellotape and fix it on the hardboard [see Fig. 3].
Fig. 2
Fig. 1
Fig. 3
14/04/18

DEMONSTRATION
1. Slant height of the cone = radius of the circle = l.
2. Radius of the base of the cone = r.
3. Circumference of the base of the cone = Arc length of the sector = 2πr.
4. Curved surface area of the cone = Area of the sector
Arc length
= Area of the circle
Circumference of the circle
×
2
2
  .
2
= × =
r
l rl
l
OBSERVATION
The slant height l of the cone = -----------------------, r = ------------------------
So, arc length l = ----------------------,
Area of the sector = ----------------, Curved surface area of the cone = ---------
------------------
Therefore, curved surface area of the cone = Area of the sector.
APPLICATION
The result is useful in
1. estimating canvas required to make a conical tent
2. estimating material required to make Joker’s cap, ice cream cone, etc.
14/04/18

Mathematics 77
To find the relationship among the
volumes of a right circular cone, a
hemisphere and a right circular cylinder
of equal radii and equal heights.
Cardboard, acrylic sheet, cutter, a
hollow ball, adhesive, marker, sand
or salt.
Activity 29
1. Take a hollow ball of radius, say, a units and cut this ball into two halves
[see Fig. 1].
2. Make a cone of radius a and height a by cutting a sector of a circle of
suitable radius using acrylic sheet and place it on the cardboard [see Fig. 2].
3. Make a cylinder of radius a and height a, by cutting a rectangular sheet of a
suitable size. Stick it on the cardboard [see Fig. 3].
Fig. 2
Fig. 1
Fig. 3
14/04/18

DEMONSTRATION
1. Fill the cone with sand (or salt) and pour it twice into the hemisphere. The
hemisphere is completely filled with sand.
Therefore, volume of cone
1
2
= volume of hemisphere.
2. Fill the cone with sand (or salt ) and pour it thrice into the cylinder. The
cylinder is completely filled with sand.
Therefore, volume of cone
1
3
= volume of cylinder.
3. Volume of cone : Volume of hemisphere : Volume of cylinder = 1:2:3
OBSERVATION
Radius of cone = Height of the cone = ----------.
Volume of cone
1
2
= Volume of ---------------.
Volume of cone
1
3
= Volume of ---------------.
Volume of cone : Volume of a hemisphere = -------- : ----------
Volume of cone : Volume of a cylinder = -------- : ----------
Volume of cone : Volume of hemisphere : Volume of cylinder = -------- :
---------- : ----------
APPLICATION
1. This relationship is useful in obtaining the formula for the volume of a cone
and that of a hemisphere/sphere from the formula of volume of a cylinder.
2. This relationship among the volumes can be used in making packages of the
same material in containers of different shapes such as cone, hemisphere,
cylinder.
14/04/18

Mathematics 79
To find a formula for the curved surface
area of a right circular cylinder,
experimentally.
Coloured chart paper, cellotape,
ruler.
Activity 30
1. Take a rectangular chart paper of length l units and breadth b units [see Fig. 1].
2. Fold this paper along its breadth and join the two ends by using cellotape
and obtain a cylinder as shown in Fig. 2.
Fig. 2
Fig. 1
14/04/18

DEMONSTRATION
1. Length of the rectangular paper = l = circumference of the base of the
cylinder = 2πr, where r is the radius of the cylinder.
2. Breadth of the rectangular paper = b = height (h) of the cylinder.
3. The curved surface area of the cylinder is equal to the area of the rectangle
= l × b = 2πr × h = 2πrh square units.
OBSERVATION
l = ...................., b = .....................,
2π r = l = ...................., h = b = ....................,
Area of the rectangular paper = l × b = .................
Therefore, curved surface area of the cylinder = 2πrh.
APPLICATION
This result can be used in finding the material used in making cylindrical
containers, i.e., powder tins, drums, oil tanks used in industrial units, overhead
water tanks, etc.
14/04/18

Mathematics 81
To obtain the formula for the surface area
of a sphere.
A ball, cardboard/wooden strips,
thick sheet of paper, ruler, cutter,
string, measuring tape, adhesive.
Activity 31
1. Take a spherical ball and find its diameter by placing it between two vertical
boards (or wooden strips) [see Fig. 1]. Denote the diameter as d.
2. Mark the topmost part of ball and fix a pin [see Fig. 2].
3. Taking support of pin, wrap the ball (spirally) with string completely, so that
on the ball no space is left uncovered [see Fig. 2].
4. Mark the starting and finishing points on the string, measure the length
between these two marks and denote it by l. Slowly, unwind the string from
the surface of ball.
5. On the thick sheet of paper, draw 4 circles of radius ‘r’ (radius equal to the
radius of ball).
Fig. 2
Fig. 1
14/04/18

6. Start filling the circles [see Fig. 3] one by one with string that you have
wound around the ball.
DEMONSTRATION
Let the length of string which covers a circle (radius r) be denoted by a.
The string which had completely covered the surface area of ball has been used
completely to fill the region of four circles (all of the same radius as of ball or
sphere).
This suggests:
Length of string needed to cover sphere of radius r = 4 × length of string
needed to cover one circle
i.e., l = 4a
or, surface area of sphere = 4 × area of a circle of radius r
So, surface area of a sphere = 4πr2
Fig. 3
14/04/18

Mathematics 83
OBSERVATION
Diameter d of the spherical ball =................ units
radius r =................ units
Length l of string used to cover ball = ................ units
Length a of string used to cover one circle =............... units
So l = 4 × ____
Surface area of a sphere of radius r = 4 × Area of a circle of radius _____ = 4πr2
.
APPLICATION
This result is useful in finding the cost
of painting, repairing, constructing
spherical and hemispherical objects.
PRECAUTIONS
• Measure diameter of ball
carefully.
• Wrap the ball completely so
that no space is left uncovered.
• Thinner the string more is the
accuracy.
14/04/18

1. Collect data from day to day life such as weights of students in a class and
make a frequency distribution table.
Case I : For classes of equal widths
To draw histograms for classes of equal
widths and varying widths.
Graph paper, geometry box, sketch
pens, scissors, adhesive, cardboard.
Activity 32
Case II : For classes of varying widths
Here : d – f = 2 (a – b)
Class a-b b-c c-d d-f
(width x) (width x) (width x) (width 2x)
Frequency f1
f2
f3
f4
Modified frequency f1
f2
f3
4
F =
2
f
′
Class a-b b-c c-d d-e e-f
Frequency f1
f2
f3
f4
f5
2. Take a graph paper ( 20 cm × 20 cm) and paste it on a cardboard.
3. Draw two perpendicular axes X′OX and YOY′ on the graph paper.
4. Mark classes on x-axis and frequencies on y-axis at equal distances as shown
in Fig. 1.
14/04/18

Mathematics 85
Fig. 1
Fig. 2
14/04/18

5. On intervals (a-b), (b-c), (c-d), (d-e), (e- f), draw rectangles of equal widths
and of heights f1
, f2
, f3
, f4
and f5
, respectively, as shown in Fig. 2.
6. On intervals (a-b), (b-c), (c-d), (d-f), draw rectangles of heights f1
, f2
, f3
, and
F ' as shown in Fig. 3.
DEMONSTRATION
1. Different numerical values can be taken for a, b, c, d, e and f.
2. With these numerical values, histograms of equal widths and varying widths
can be drawn.
OBSERVATION
Case I
1. The intervals are
a-b = ................., b-c = ................., c-d = .................,
Fig. 3
14/04/18

Mathematics 87
d-e = ................., e– f = .................
2. f1
= ................., f2
= ................., f3
= .................,
f4
= ................., f5
= .................
Case II
1. a-b = ................., b-c = ................., c-d = .................,
d-f = .................,
2. f1
= ................., f2
= ................., f3
= .................,
f4
= ................., 4
F =
2
f
′ = .................
APPLICATION
Histograms are used in presenting large data in a concise form pictorially.
14/04/18

1. Take a telephone directory and select a page at random.
2. Count the number of telephone numbers on the selected page. Let it be ‘N’.
3. Unit place of a telephone number can be occupied by any one of the digits
0, 1, ..., 9.
4. Prepare a frequency distribution table for the digits, at unit’s place using
tally marks.
5. Write the frequency of each of the digits 0, 1, 2, ...8, 9 from the table.
6. Find the probability of each digit using the formula for experimental
probability.
DEMONSTRATION
1. Prepare a frequency distribution table (using tally marks) for digits 0, 1, ...,
8, 9 as shown below:
To find experimental probability of
unit’s digits of telephone numbers listed
on a page selected at random of a
telephone directory.
Telephone directory, note book,
pen, ruler.
Activity 33
Digit 0 1 2 3 4 5 6 7 8 9
Tally marks
Frequency n0
n1
n2
n3
n4
n5
n6
n7
n8
n9
14/04/18

Mathematics 89
2. Note down frequency of each digit (0, 1, 2, 3,...,9) from the table.
Digits 0, 1, 2, 3, ..., 9 are occuring respectively n0
, n1
, n2
, n3
, ..., n9
times.
3. Calculate probability of each digit considering it as an event ‘E’ using the
formula
( )
Numberof trials in which the event occured
P E
Total number of trials
=
4. Therefore, respective experimental probability of occurence of 0, 1, 2, ...,
9 is given by
( ) ( ) ( ) ( )
0 9
1 2
P 0 , P 1 , P 2 ,..., P 9
N N N N
n n
n n
= = = = .
OBSERVATION
Total number of telephone numbers on a page (N) = .......................... .
Number of times 0 occurring at unit’s place (n0
) = ........................ .
) = ........................ .
) = ........................ .
---------------------3 -------------------------- (n3
) = ......................... .
--------------------- 4 ------------------------ (n4
) = .........................
)= ........................
Therefore, experimental probability of occurence of 0 ( ) 0
P 0
N
n
= = = .............,
Experimental probability of occurence of 1 ( ) 1
P 1
N
n
= = = ............. .
( ) 2
P 2
N
n
= = ............., ...,
( ) 9
P 9
N
n
= = ............. .
.
.
.
14/04/18

APPLICATION
Concept of experimental probability is used for deciding premium tables by
insurance companies, by metreological department to forecast weather, for
forecasting the performance of a company in stock market.
The mathematics experience of the students is incomplete
if he never had the opportunity to solve a problem invented
by himself.
– G. Polya
14/04/18

Mathematics 91
To find experimental probability of each
outcome of a die when it is thrown a large
number of times.
Die, note book, pen.
Activity 34
1. Divide the whole class in ten groups say G1
, G2
, G3
, ..., G10
of a suitable size.
2. Allow each group to throw a die 100 times and ask them to note down the
observations, i.e., the number of times the outcomes 1, 2, 3, 4, 5 or 6 come
up.
3. Count the number of times 1 has appeared in all the groups. Denote it by a.
Similarly, count the number of times each of 2, 3, 4, 5 and 6 has appeared.
Denote them by b, c, d, e and f respectively.
4. Find the probability of each outcome ‘E’ using the formula :
Number of times an outcome occured
P(E)=
Total number of trials
DEMONSTRATION
1. There are 10 groups and each group throws a die 100 times. So, the total
number of trials is 1000.
2. Total number of times 1 has appeared is a
Therefore, experimental probability of 1 is P(1)=
1000
a
Similarly, experimental probability of 2 is P(2)=
1000
b
, of 3 is P(3)=
1000
c
,
14/04/18

of 4 is P(4)=
1000
d
,
of 5 is P(5)=
1000
e
, of 6 is P(6)=
1000
f
OBSERVATION
Fill in the results of your experiment in the following table:
Therefore,
.......
P(1)=
1000
,
.......
P(2)=
1000
,
.......
P(3)=
1000
,
.......
P(4)=
1000
,
.......
P(5)=
1000
,
.......
P(6)=
1000
.
APPLICATION
Concept of probability is used by several statistical institutions to estimate/
predict next action based on available data.
Outcome
Group
Number of times a number comes up on a die
Total
1 2 3 4 5 6
G1
---- ---- ---- ---- ---- ---- 100
G2
---- ---- ---- ---- ---- ---- 100
G3
---- ---- ---- ---- ---- ---- 100
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
G10
---- ---- ---- ---- ---- ---- 100
Total a = ---- b = ---- c = ---- d = ---- e = ---- f = ---- 1000
14/04/18

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