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Mathematics is a subject with not only theory-based knowledge but lots of practical aspects as well. To master mathematics, understanding all mathematical concepts through practical activities is the way. It encourages students to build skills, develop a positive attitude towards math, and learn by doing.
For 9th-grade students, math is a subject that sets the base of their future education and career. The CBSE Board includes practical activities in the curriculum to make math more enjoyable and relatable to everyday life. The manual covers important topics like algebra, geometry, trigonometry, measurement, and probability, helping students to know more about the subject.
NCERT Class 9 Lab Manual PDF
The class 9 lab manual allows students to perform various activities and practicals that hold significant importance in their syllabus and also helps students bridge the gap between theory and practice by helping them apply mathematical concepts through various activities.
The Class 9 math lab manual is a specially designed additional study resource for students that is completely made up of the CBSE Class 9 Math Syllabus. It contains a collection of practical activities aimed at giving students more clarity and a better understanding of mathematical concepts. These activities are given to make students want to experiment, observe, and deduce mathematical principles on their own, making the learning process both easy and a bit fun.
The lab manual activities constitute 12 marks in the final board examinations as part of the internal assessment. NCERT has provided free Class 9 Lab Manual activity PDFs in downloadable format for students as well as teachers.
As per the NCERT Class 9 Lab Manual Activity list, 32 activities are included for the practical examinations.
Objective: To construct a square-root spiral.
Material Required: Coloured threads, adhesive, drawing pins, nails, geometry box, sketch pens, marker, and a piece of plywood.
Method of Construction:
Demonstration
From the figure, AC2 = AB2 + BC2 = 12 + 12 = √2 or AC = 2.
AD2 = AC2 + CD2 = 2 + 1 = 3 or AD = √3 .
Similarly, we get the other lengths AE, AF, AG, ... as √4 or 2, √5, 6...
Observation
On actual measurementAC = ..... , AD = ...... , AE =...... , AF =....... , A
G = ......√2 = AC = ............... (approx.),√3 = AD = ............... (approx.),
√4 = AE = ............... (approx.),√5 = AF = ............... (approx.)
Application
Through this activity, the existence of irrational numbers can be illustrated.
Objective: To represent some irrational numbers on the number line.
Material Required: Two cuboidal wooden strips, thread, nails, a hammer, two photocopies of a scale, a screw with nut, glue, and a cutter.
Method of Construction:
Demonstration
Take 1 unit on the horizontal scale and fix the perpendicular wooden strip at 1 by the screw at the bottom.
Tie the other end of the thread to unit ‘1’ on the perpendicular strip.
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 √a – 1 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 the thread at 2 on the vertical strip.
Objective: To verify the algebraic identity : (a + b)2 = a2 + 2ab + b2
Material Required: Drawing sheet, cardboard, cellotape, coloured papers, cutter and ruler.
Method of Construction:
= a2 + b2 + ab + ba = a2 + b2 + 2ab.
Hence, the algebraic identity (a + b)2 = a2 + 2ab + b2
Observations
On actual measurement: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
Objective: To verify the algebraic identity : (a - b)2 = a2 - 2ab + b2
Material Required: Drawing sheets, cardboard, coloured papers, scissors, ruler and adhesive.
Method of Construction:
Demonstration
According to figure 1, 2, 3, and 4, the Area of square ABCD = a2, the Area of square EBHI = b2
Area of rectangle GDCJ = ab, Area of rectangle IFJH = ab
From Fig. 5, the 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
Here, the area is in square units.
Observation
On actual measurement:
a = .............., b = .............., (a – b) = ..............,
So, a2 = .............., b2 = .............., (a – b)2 = ..............,
ab = .............., 2ab = ..............
Therefore, (a – b)2 = a2 – 2ab + b2
Application
The identity may be used for
Objective: To verify the algebraic identity : a2 – b2 = (a + b)(a – b)
Material Required: Drawing sheets, cardboard, coloured papers, scissors, sketch pen, ruler, transparent sheet and adhesive.
Method of Construction:
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) Here, the area is in square units.
Observation
On actual measurement:
a = .............., b = .............., (a+b) = ..............,
So, a2 = .............., b2 = .............., (a–b) = ..............,
a2–b2 = .............., (a+b) (a–b) = ..............,
Therefore, a2–b2 = (a+b) (a–b)
Application
The identity may be used for
Objective: To verify the algebraic identity : (a+b+c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Material Required: Hardboard, adhesive, coloured papers, white paper.
Method of Construction:
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
Here, the area is in square units.
Observation
On actual measurement:
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
The identity may be used for
Objective: To verify the algebraic identity : (a+b)3 = a3 + b3 + 3a2b + 3ab2
Material Required: Acrylic sheet, coloured papers, glazed papers, saw, sketch pen, adhesive, Cello tape.
Method of Construction:
Demonstration
Volume of the cube of side a = a×a×a = a3, volume of the cube of side b = b3
The volume of the cuboid of dimensions a×a×b = a2b, the volume of three such cuboids = 3a2b
The volume of the cuboid of dimensions a×b×b = ab2, the volume of three such cuboids= 3ab2
The solid figure obtained in Fig. 5 is a cube of sides (a + b)
Its volume = (a + b)3
Therefore, (a+b)3 = a3 + b3 + 3a2b + 3ab2
Here, the volume is in cubic units.
Observation
On actual measurement:
a = .............., b = ............., a3 = ..............,
So, a3 = .............., b3 = ............., a2b = .............., 3a2b= ..............,
ab2 = .............., 3ab2 = .............., (a+b)3 = ..............,
Therefore, (a+b)3 = a3 + b3 +3a2b + 3ab2
Application
The identity may be used for
Objective: To verify the algebraic identity : (a – b)3 = a3 – b3 – 3(a – b)ab
Material Required: Acrylic sheet, coloured papers, saw, sketch pens, adhesive, Cellotape.
Method of Construction:
Demonstration
The 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
The volume of three cuboids in Fig. 2 = 3 (a–b) ab
The 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, the volume is in cubic units.
Observation
On actual measurement:
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
The identity may be used for
Note
This identity can also be expressed as :
(a – b)3 = a3 – 3a2b + 3ab2 – b3
Objective: To verify the algebraic identity : a3 + b3 = (a + b) (a2 – ab + b2)
Material Required: Acrylic sheet, glazed papers, saw, adhesive, cellotape, coloured papers, sketch pen, etc.
Method of Construction:
Demonstration
The volume of the cube in Fig. 1 = a3
The volume of the cube in Fig. 2 = b3
The volume of the cuboid in Fig. 3 = a2b
Volume of cuboid in Fig. 4 = ab2
Volume of solid in Fig. 5 = a3+b3 + a2b + ab2 = (a+b) (a2 + b2)
Removing cuboids of volumes a2b and ab2, i.e., ab (a + b) from the solid obtained in Fig. 5, we get the solid in Fig. 6.
The volume of solid in Fig. 6 = a3 + b3.
Therefore, a3 + b3 = (a+b) (a2 + b2) – ab (a + b) = (a+b) (a2 + b2 – ab)
Here, volumes are in cubic units.
Observation
On actual measurement:
a = .............., b = ..............,
So, a3 = .............., b3 = .............., (a+b) = .............., (a+b)a2 = ..............,
(a+b) b2 = .............., a2b = .............., ab2 = .............., ab (a+b) = ..............,
Therefore, a3 + b3 = (a + b) (a2 + b2 – ab).
Application
The identity may be used in the simplification and factorisation of algebraic expressions.
Objective: To verify the algebraic identity : a3 – b3 = (a – b)(a2 + ab + b2)
Material Required: Acrylic sheet, sketch pen, glazed papers, scissors, adhesive, cellotape, coloured papers, cutter.
Method of Construction:
Demonstration
The volume of the cuboid in Fig. 1 = (a–b) × a × a cubic units.
The volume of the cuboid in Fig. 2 = (a–b) × a × b cubic units.
The volume of the cuboid in Fig. 3 = (a–b) × b × b cubic units.
The volume of the cube in Fig. 4 = b3 cubic units.
The volume of solid in Fig. 5 = a3 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)
Observation
On actual measurement:
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 the simplification/factorisation of algebraic expressions.
Objective: To find the values of abscissae and ordinates of various points given in a cartesian plane.
Material Required: Cardboard, white paper, graph paper with various given points, geometry box, pen/pencil.
Method of Construction:
Demonstration
To find the abscissa and ordinate of a point, say A, draw perpendiculars AM and AN from A to the 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. Point A is in the first quadrant. The coordinates of A are (2, 9).
Observation
Application
This activity helps locate the position of a particular city/place or country on the map.
Precaution
The students should be careful while reading the coordinates, otherwise the location of the object will differ.
Objective: To find a hidden picture by plotting and joining the various points with given coordinates in a plane.
Material Required: Cardboard, white paper, cutter, adhesive, graph paper/squared paper, geometry box, pencil.
Method of Construction:
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 ........, ........, ........, ........, ........, ........, ........
The 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.
Objective: To verify experimentally that if two lines intersect, then(i) The vertically opposite angles are equal(ii) the sum of two adjacent angles is 180º(iii) the sum of all the four angles is 360º.
Material Required: Two transparent strips marked as AB and CD, a full protractor, a nail, cardboard, white paper, etc.
Method of Construction:
Demonstration
Observe the adjacent angles and the vertically opposite angles formed in different positions of the strips.
Compare vertically opposite angles formed by the two lines in the strips in different positions.
Check the relationship between the vertically opposite angles.
Check that the vertically opposite angles ∠AOD, ∠COB, ∠COA and ∠BOD are equal.
Compare the pairs of adjacent angles and check that ∠COA + ∠DOA= 180º, etc.
Find the sum of all the four angles formed at point O and see that the sum is equal to 360º.
Observation
On the actual measurement of angles in one position of the strips :
∠COB = ................., ∠BOD = .................
Therefore, ∠AOD = ∠COB and ∠AOC = ............ (vertically opposite angles).
Application
These properties are used in solving many geometrical problems.
Objective: To verify experimentally the different criteria for congruency of triangles using triangle cut-outs.
Material Required: Cardboard, scissors, cutter, white paper, geometry box, pencil/sketch pens, coloured glazed papers.
Method of Construction:
Demonstration
Superpose of DABC on DDEF and see whether one triangle covers the other triangle or not by suitable arrangement. See that △ABC covers DDEF completely only under the correspondence A⟷D, B⟷E, C⟷F. So, △ABC ≅ △DEF, if AB = DE, BC = EF and AC = DF.
This is the SSS criterion for congruency.
Similarly, establish △GHI ≅ △JKL if GH = JK. ∠G = ∠J and GI = JL. This is the SAS criterion for congruency.
Establish △PQR ≅ △STU, if QR = TU, ∠Q = ∠T and ∠R = ∠U. This is the ASA criterion for congruency.
In the same way, △STU ≅ △LMN, if hypotenuse YZ = hypotenuse MN and XZ = LN.
This is the RHS criterion for right triangles.
Observation
On actual measurement :
AB = DE = ..................., BC = EF = ...................,
AC = DF = ..................., ∠A = ...................,
∠D = ..................., ∠B = ..................., ∠E = ...................,
∠C = ..................., ∠F = ....................
Therefore, △ABC ≅ △DEF.
GH = JK = ..................., GI = JL = ...................., HI = ...................,
KL= ..................., ∠G= ...................,
∠J = ..................., ∠H = ..................., ∠K = ...................,
∠L = ...................
Therefore, △GHI ≅ △JKL.
QR = TU = ..................., PQ = ..................., ST = ...................,
PR = ..................., SU = ...................., ∠S= ...................,
∠Q = ∠T = ..................., ∠R = ∠U = ..................., ∠P = ...................,
Therefore, △PQR ≅ △STU.
XZ = LN = ..................., XY = ...................,
LM = ..................., ∠X= ∠L=90°,
∠Y = ..................., ∠M = ..................., ∠Z = ...................,
∠N = ...................,
Therefore, △XYZ ≅ △LMN.
Application
These criteria are useful in solving several problems in geometry.
These criteria are also useful in solving some practical problems such as finding width of a river without crossing it.
Objective: To verify that the sum of the angles of a triangle is 180º
Material Required: Hardboard sheet, glazed papers, sketch pens/pencils, adhesive, cutter, tracing paper, drawing sheet, geometry box.
Method of Construction:
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 the sum of the three angles of a triangle is 180º. Therefore, ∠A + ∠B + ∠C = 180°.
Observation
The measure of ∠A = -------------------.
Measure of ∠B = -------------------.
Measure of ∠C = -------------------.
Sum (∠A + ∠B + ∠C) = -------------------.
Application
This result may be used in many geometrical problems such as to find the sum of the angles of a quadrilateral, pentagon, etc.
Objective: To verify the exterior angle property of a triangle.
Material Required: Hardboard sheet, adhesive, glazed papers, sketch pens/pencils, drawing sheet, geometry box, tracing paper, cutter, etc.
Method of Construction:
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.
Objective: To verify experimentally that the sum of the angles of a quadrilateral is 360º.
Material Required: Cardboard, white paper, coloured drawing sheet, cutter, adhesive, geometry box, sketch pens, tracing paper.
Method of Construction:
Demonstration
The vertex of each cut-out angle coincides at point O.
Such an 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
The 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
Objective: To verify experimentally that in a triangle, the longer side has the greater the angle opposite to it.
Material Required: Coloured paper, scissors, tracing paper, geometry box, cardboard sheet, sketch pens.
Method of Construction:
Demonstration
Take the cut-out angle and compare it with the 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.
Application
Length of side AB = .......................
Length of side BC = .......................
Length of side CA = .......................
The measure of the angle opposite to the longest side = .......................
The 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.
Objective: To verify experimentally that the parallelograms on the same base and between the same parallels are equal in area.
Material Required: A piece of plywood, two wooden strips, nails, elastic strings, and graph paper.
Method of Construction:
Demonstration
Put a string along A1, A2, B8, B2 which forms a parallelogram A1A2B8B2. By counting a number of squares, find the area of this parallelogram.
Keeping the same base A1A2, make another parallelogram A1A2B9B3 and find the area of this parallelogram by counting the squares.
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, and more than half squares. Less than half squares may be ignored.
Objective: To verify that the triangles on the same base and between the same parallels are equal in area.
Material Required: A piece of plywood, graph paper, a pair of wooden strips, a colour box, scissors, a cutter, an adhesive, geometry box.
Method of Construction:
Demonstration
Observation
The number of squares in triangle ABC =.........., Area of DABC = ........ units
The number of squares in triangle ABC’ =......., Area of DABC’ = ........ units
The number of squares in triangle ABC’’ =....... , Area of DABC’’ = ........ units
Therefore, area (DABC) = ar(ABC’) = ar(ABC’’).
Application
This result helps in solving various geometric problems. It also helps in finding the formula for the area of a triangle.
Objective: To verify that the ratio of the areas of a parallelogram and a triangle on the same base and between the same parallels is 2:1.
Material Required: Plywood sheet of convenient size, graph paper, colour box, a pair of wooden strips, scissors, cutter, adhesive, geometry box.
Method of Construction:
Demonstration
AB is parallel to CD and P is any point on CD.
Triangle PAB and parallelogram ABCD are on the same base AB and between the same parallels.
Count the number of squares contained in each of the above triangles and parallelograms, keeping half a square as ½ and more than half as 1 square, leaving those squares which contain less than half a square.
See that the area of the triangle PAB is half of the area of parallelograms ABCD.
Observation
So, the area of parallelogram ABCD = 2 [Area of triangle PAB]
Thus, area of parallelogram ABCD : area of DPAB = ........ : ...........
Application
This activity is useful in deriving formulas for the area of a triangle and also in solving problems on mensuration.
Note
You may take different triangles PAB by taking different positions of point P and the two parallel strips as shown in Fig. 2.
Objective: 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.
Material Required: Cardboard, coloured drawing sheets, scissors, sketch pens, adhesive, geometry box, transparent sheet.
Method of Construction:
Demonstration
Place the two cut-outs of ∠BAC on the cut-out of angle BOC, adjacent to each as shown in 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.
Objective: To verify that the angles in the same segment of a circle are equal.
Material Required: Geometry box, coloured glazed papers, scissors, cardboard, white paper and adhesive.
Method of Construction:
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
On actual measurement:
∠ACB = ---------------, ∠ADB = ---------------
So, ∠ACB = ∠ADB. Thus, angles in the same segment are ---------.
Objective: To verify that the opposite angles of a cyclic quadrilateral are supplementary.
Material Required: Chart paper, geometry box, scissors, sketch pens, adhesive, transparent sheet.
Method of Construction:
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
On actual measurement:
∠1 = ................; ∠2 = ................; ∠3 = ................; ∠4 = .................
So, ∠1 + ∠3 = ..........; ∠2 + ∠4 = ..........;
Therefore, the sum of each pair of opposite angles of a cyclic quadrilateral is .........................
Application
The concept may be used to solve various problems in geometry.
Objective: To find the formula for the area of a trapezium experimentally.
Material Required: Hardboard, thermocol, coloured glazed papers, adhesive, scissors.
Method of Construction:
Demonstration
The figure formed by the two trapeziums [see Fig. 2] is a parallelogram ABCD.
Side AB of the parallelogram = (a + b) units and its corresponding altitude= h units.
Area of each trapezium= ½ x (area of the parallelogram)= ½ x a + b × h
Therefore, area of trapezium= ½ a + b × h
= ½ (sum of parallel sides) × perpendicular distance.
Here, the area is in square units.
Observation
Lengths of parallel sides of the trapezium = -------,-------.
Length of altitude of the parallelogram = --------.
Area of parallelogram = ---------------.
Area of the trapezium = ½ (Sum of ---------- sides) × ---------.
Application
This concept is used for finding the formula for the 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.
Objective: To form a cube and find the formula for its surface area experimentally.
Material Required: Cardboard, ruler, cutter, cellotape, sketch pen/pencil.
Method of Construction:
Demonstration
Each face of the cube so obtained is a square of side a units. Therefore, the area of one face of the cube is a2 square units.
Thus, the surface area of the cube with side a units = 6a2 square units.
Observation
On actual measurement:
Length of side a = ..................
Area of one square / one face = a2 = ................
So, the sum of the areas of all the squares = ..........+............+..........+ ..........+ .......... + ............
Therefore, the surface area of the cube = 6a2
Application
This result is useful in estimating the materials required for making cubical boxes needed for packing.
Note
Instead of making six squares separately as done in the activity, a net of a cube be directly prepared on the cardboard itself.
Objective: To form a cuboid find the formula for its surface area experimentally.
Material Required: Cardboard, cellotape, cutter, ruler, sketch pen/pencil.
Method of Construction:
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.
The surface area of the cuboid so formed = (2 × ab + 2 × bc + 2 × ca) square units
= 2 (ab + bc + ca) square units.
Observation
On actual measurement:
a = ....................., b = ....................., c = .....................,
So, ab = ....................., bc = ....................., ca = .....................,
2ab = ....................., 2bc = ....................., 2ca = .....................
The sum of areas of all the six rectangles = ..............
Therefore, the 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.
Objective: To form a cone from a sector of a circle and to find the formula for its curved surface area.
Material Required: Wooden hardboard, acrylic sheets, cellotape, glazed papers, sketch pens, white paper, nails, and markers.
Method of Construction:
Demonstration
The slant height of the cone = radius of the circle = l.
Radius of the base of the cone = r.
Circumference of the base of the cone = Arc length of the sector = 2πr.
Curved surface area of the cone = Area of the sector
= (Arc length / Circumference of the circle) x Area of the circle
= (2πr / 2πl) x πl2
= πrl
Observation
On actual measurement :
The slant height l of the cone = -----------------------, r = ------------------------
So, arc length l = ----------------------,
Area of the sector = ----------------, Curved surface area of the cone = ---------
------------------
Therefore, the curved surface area of the cone = the Area of the sector.
Here, the area is in square units.
Application
The result is useful in
Objective: 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.
Material Required: Cardboard, acrylic sheet, cutter, a hollow ball, adhesive, marker, sand, or salt.
Method of Construction:
Demonstration
Therefore, the volume of the cone = ½ volume of hemisphere.
Therefore, the volume of the cone = ⅓ the volume of a cylinder.
Observation
Radius of cone = Height of the cone = ----------.
The volume of cone= ½ Volume of ---------------.
The volume of cone = ⅓ Volume of ---------------.
Volume of cone: Volume of a hemisphere = --------: ----------
Volume of cone: Volume of a cylinder = --------: ----------
The volume of cone: Volume of hemisphere: Volume of cylinder = -------- : ----------: ----------
Application
Objective: To find a formula for the curved surface area of a right circular cylinder, experimentally.
Material Required: Coloured chart paper, cellotape, ruler.
Method of Construction:
Demonstration
Length of the rectangular paper = l = circumference of the base of the cylinder = 2πr, where r is the radius of the cylinder.
The breadth of the rectangular paper = b = height (h) of the cylinder.
The curved surface area of the cylinder is equal to the area of the rectangle = l × b = 2πr × h = 2πrh square units.
Observation
On actual measurement :
l = ...................., b = .....................,
2πr = l = ...................., h = b = ....................,
Area of the rectangular paper = l × b = .................
Therefore, the curved surface area of the cylinder = 2πrh.
Application
This result can be used to find the material used in making cylindrical containers, i.e., powder tins, drums, oil tanks used in industrial units, overhead water tanks, etc.
Objective: To obtain the formula for the surface area of a sphere.
Material Required: A ball, cardboard/wooden strips, thick sheet of paper, ruler, cutter, string, measuring tape, adhesive.
Method of Construction:
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 = 4pr2
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 _____ = 4pr2.
Application
This result is useful in finding the cost of painting, repairing, constructingspherical and hemispherical objects.
Precautions
Objective: To draw histograms for classes of equal widths and varying widths.
Material Required: Graph paper, geometry box, sketch pens, scissors, adhesive, cardboard.
Method of Construction:
Case I: For classes of equal widths
Case II: For classes of varying widths
Here : d – f = 2 (a – b)
Demonstration
Observation
Case I
a-b = ................., b-c = ................., c-d = .................,
d-e = ................., e– f = .................
f4 = ................., f5 = .................
Case II
d-e = ................., e– f = .................
f4 = ................., F’= (f4/2).................
Application
Histograms are used in presenting large data in a concise form pictorially
Objective: To find the experimental probability of unit’s digits of telephone numbers listed on a page selected at random of a telephone directory.
Material Required: Telephone directory, note book, pen, ruler.
Method of Construction:
Demonstration
Prepare a frequency distribution table (using tally marks) for digits 0, 1, ..., 8, 9 as shown below:
P(E)= (Number of trials in which the event occured)/Total number of trials)
P(0)= n0/N , P(1)= n/N , P(2)= n2/N, ..., P(9)= n9/N
Observation
Total number of telephone numbers on a page (N) = .......................... .
Number of times 0 occurring at unit’s place (n0) = ........................ .
Number of times 1 occurring at unit’s place (n1) = ........................ .
Number of times 2 occurring at unit’s place (n2) = ........................ .
---------------------3 -------------------------- (n3) = ......................... .
--------------------- 4 ------------------------ (n4) = .........................
Number of times 9 occurring at unit’s place (n9)= ........................
Therefore, the experimental probability of occurrence of 0= P(0)= n0/N
Experimental probability of occurrence of 1= P(1)= n/N
P(2)= n2/N, ...,
...
P(9)= n9/N
Application
The concept of experimental probability is used for deciding premium tables by insurance companies, by meteorological departments to forecast weather, and for forecasting the performance of a company in the stock market.
Objective: To find the experimental probability of each outcome of a die when it is thrown a large number of times.
Material Required: Die, notebook, pen.
Method of Construction:
P(E)= (Number of times an outcome occurred/Total number of trials)
Demonstration
There are 10 groups and each group throws a die 100 times. So, the total number of trials is 1000.Total number of times 1 has appeared is a Therefore, experimental probability of 1 is P(1)= (a/1000)
Similarly, experimental probability of 2 is P(2)= (b/1000), of 3 is P(3)= (c/1000)
of 4 is P(4)= (d/1000,
of 5 is P(5)= (e/1000,
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
The concept of probability is used by several statistical institutions to estimate/ predict next action based on available data.
The Class 9 Math Lab Manual is so organised and neat, all systematically to cover important topics appearing in the syllabus. While the content inside may differ a little bit as it depends on the publisher, the general structure is as follows:
Introduction to the Lab Manual: The first part of the lab manual usually starts with an introduction, which tells us about practical mathematics, its importance, and how to use the manual correctly.
Explained Concepts: Before telling us more about the activities part, the manual provides a detailed explanation of the mathematical concepts that might be used in the activities. These explanations work as revision capsules for students and provide theory-based background for activities.
Details about Activities: Take note that all activities are explained in detail in the manual. The details include the objective of the activity, the materials required, and the step-by-step procedure to be followed. The activities are designed to be self-explanatory and easy to follow.
Viva Voce: To make sure that students are not just performing the activities but also understanding the concepts and logic lying behind the activities, the manual includes a section on Viva Voce. This section consists of questions that the teacher may ask during practical exams. It allows students to think critically about the activities they perform and to enhance their understanding.
Mental Maths: There is a section on mental maths included in the class 9 CBE math lab manual. This section has questions that challenge students to apply the concepts they learnt and practiced in other different scenarios. This helps in improving their ability to think quickly and accurately.
Test yourself Sheets: The class 9 math lab manual includes assessment sheets at the end of each activity or section given in the manual. These are given to let the students test their understanding of the concept and have feedback on their performance.
The use of the Math Lab Manual provides lots of benefits, especially in CBSE Class 9, where all the students are preparing their bases for their higher studies. A few benefits of the following are given below:
Making Conceptual Understanding Better: Practice over learning, always! By doing and getting involved in practical activities, students get a better and deeper understanding of the math concepts used. This hands-on approach makes it easier for them to get hold of complex ideas and try applying them to real-world problems and situations.
Improved Problem-Solving Skills: The activities in the class 9 lab manual math are made so that the students are motivated to think critically and solve problems on their own. This not only helps in making their problem-solving skills better but also gives a boost to their confidence level in handling mathematical challenges.
Better retention of concepts: Research shows that students are more likely to remember information when they learn through practical experience in comparison to theory. The Maths Lab Manual helps students remember the theory through practical experience, which will eventually lead to better memory and understanding.
Engagement and Motivation: The interactive nature of the activities keeps students engaged and motivated to learn. This is particularly important in subjects like mathematics, where students often struggle to stay interested.
Preparation for Practical Exams: The manual prepares students for the practical aspects of their exams. The inclusion of Viva Voce and assessment sheets ensures that students are well-prepared to demonstrate their knowledge in a practical setting.
When you climb from middle school to high school, many things change. Be it regarding syllabus, books, exam pattern, or… life. Well, focussing on math, there’s a lot of change in the complex level of chapters and mathematical concepts. The math lab manual for class 9 plays an important role in this transition to the next level, like:
Solid Understanding of Concepts: High school-level mathematics can often include concepts that are hard to follow and understand through just the textbooks and all that theory given. That’s why the lab manual helps in putting these concepts out there through practical activities, making them easier to understand.
Develop your skills: The manual is not just about solving numerical problems and practicing; it’s more than that. Through it, students can develop certain skills, such as logical reasoning, analytical thinking, or problem-solving. By doing the lab activities, students learn to solve problems systematically and know the principles of mathematics better.
Learning Made Interactive: Lab Manual Class 9 is basically a good approach to make learning easy and interactive. This method keeps the students busy and engaged and makes things and concepts easy to learn and retain in the memory better.
Prepare for Higher Studies: CBSE Class 9 is the base class and the most important class for students as it lays the foundation for higher classes. The lab manual helps in knowing the basics, making sure that students are preparing well for more advanced topics in the coming years.
Self-assess and Evaluate: The manual has various exercises that allow students to self-assess their understanding. This continuous assessment helps in knowing areas of strength and weakness, allowing for targeted improvement.
To get all the benefits from the class 9 math lab manual, students should know the right strategy and have the right mindset. Here are some tips on how to make the most of this important resource:
Pay attention to theory: Before starting to do any activity, you first have to be sure that you have a good understanding of the concepts that lie behind the activity and its procedure. The manual provides a detailed explanation of the concepts, so take the time to read and understand them.
Follow each step given: There are several steps given along with each activity that tells you how to perform a certain activity to get your desired results. Skipping steps or not following instructions can lead to incorrect conclusions.
Ask questions and solve your doubts: If you don’t understand something, don’t hesitate to ask questions. Whether it’s during the activity or in the Viva Voce section, asking questions is a very, very important part of the learning process, guys!
Practiced mental math regularly is the key. The mental math section is important as it is especially there to improve your calculation skills. Regular practice will not only help you in exams but will also improve your overall mathematics.
Self-analyse your work: Always, I repeat, Always review your work after completing an activity. Check and compare your results against the solutions that are already given in the manual, and then understand where you made mistakes and why you made them.
Prepare for Viva Voce: The Viva Voce section is a great way when it comes to test your understanding. Prepare for it by reviewing the questions provided and thinking about how you would answer them. Go for it!
The 9th-grade math lab manual is not just a supplementary resource, but it's also a very important tool in case you want to master Mathematics Class 9 CBSE. It’s a perfect mixture of all the theoretical knowledge and practical application, which eventually helps the students develop a good and detailed understanding of the math basic concepts used in the activities. The Maths Lab Manual Class 9 is a must-have resource that can help you achieve your academic goals. Make the most of it by doing the activities fully, practicing regularly, and doing each task with determination. Happy learning!
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