CBSE Class 10 Mathematics Activities and Projects

ACTIVITES (TERM-I)

(Any Eight)

• Activity 1: To find the HCF of two Numbers Experimentally Based on Euclid Division Lemma
• Activity 2: To Draw the Graph of a Quadratic Polynomial and observe:
  • i. The shape of the curve when the coefficient of \(x^2\) is positive
  • ii. The shape of the curve when the coefficient of \(x^2\) is negative
  • iii. Its number of zero
• Activity 3: To obtain the zero of a linear Polynomial Geometrically
• Activity 4: To obtain the condition for consistency of system of linear Equations in two variables
• Activity 5: To Draw a System of Similar Squares, Using two intersecting Strips with nails
• Activity 6: To Draw a System of similar Triangles Using Y shaped Strips with nails
• Activity 7: To verify Basic proportionality theorem using parallel line board
• Activity 8: To verify the theorem: Ratio of the Areas of Two Similar Triangles is Equal to the Ratio of the Squares of their corresponding sides through paper cutting.
• Activity 9: To verify Pythagoras Theorem by paper cutting, paper folding and adjusting (Arranging)
• Activity 10: Verify that two figures (objects) having the same shape ( and not Necessarily the same size) are similar figures. Extend the similarity criterion to Triangles.
• Activity 11: To find the Average Height (in cm ) of students studying in a school.
• Activity 12: To Draw a cumulative frequency curve ( or an ogive) of less than type .
• Activity 13: To Draw a cumulative frequency curve (or an ogive ) of more than type.

ACTIVITES (TERM-II)

(Any Eight)

• Activity 1: To find Geometrically the solution of a Quadratic Equation \(ax^2+bx+c=0\), \(a \neq 0\) (where \(a=1\)) by using the method of computing the square.
• Activity 2: To verify that given sequence is an A.P (Arithmetic Progression) by the paper Cutting and Paper Folding.
• Activity 3: To verify that \(\sum n = \frac{n(n+1)}{2}\) by Graphical method
• Activity 4: To verify experimentally that the tangent at any point to a circle is perpendicular to the Radius through that point.
• Activity 5: To find the number of Tangent from a point to the circle
• Activity 6: To verify that lengths of Tangents Drawn from an External Point, to a circle are equal by using method of paper cutting, paper folding and pasting.
• Activity 7: To Draw a Quadrilateral Similar to a given Quadrilateral as per given scale factor (Less than 1)
• Activity 8:
  • (a) To make mathematical instrument clinometer (or sextant) for measuring the angle of elevation/depression of an object
  • (b) To calculate the height of an object making use of clinometers(or sextant)
• Activity 9: To get familiar with the idea of probability of an event through a double color card experiment.
• Activity 10: To verify experimentally that the probability of getting two tails when two coins are tossed simultaneously is \(1/4 = (0.25)\) (By eighty tosses of two coins)
• Activity 11: To find the distance between two objects by physically demonstrating the position of the two objects say two Boys in a Hall, taking a set of reference axes with the corner of the hall as origin.
• Activity 12: Division of line segment by taking suitable points that intersects the axes at some points and then verifying section formula.
• Activity 13: To verify the formula for the area of a triangle by graphical method .
• Activity 14: To obtain formula for Area of a circle experimentally.
• Activity 15: To give a suggestive demonstration of the formula for the surface Area of a circus Tent.
• Activity 16: To obtain the formula for the volume of Frustum of a cone.

PROJECTS

• Project 1 : Efficiency in packing
• Project 2 : Geometry in Daily Life
• Project 3: Experiment on probability
• Project 4: Displacement and Rotation of a Geometrical Figure
• Project 5: Frequency of letters/ words in a language text.
• Project 6: Pythagoras Theorem and its Extension
• Project 7: Volume and surface area of cube and cuboid.
• Project 8: Golden Rectangle and golden Ratio
• Project 9 : Male‐Female ratio
• Project 10 : Body Mass Index(BMI)
• Project 11 : History of Indian Mathematicians and Mathematics
• Project 12 : Career Opportunities
• Project 13 : \(\pi\) (Pie)

Project Work Assignment (Any Eight)

ACTIVITY- 1

TOPIC:- Prime factorization of composite numbers.

OBJECTIVE:- To verify the prime factorization 150 in the form \(5^2 \times 3 \times 2\) i.e \(150 = 5^2 \times 3 \times 2\).

PRE-REQUISITE KNOWLEDGE:- For a prime number P, \(P^2\) can be represented by the area of a square whose each side of length P units.

MATERIALS REQUIRED:-

• i. A sheet of graph paper ( Pink / Green)
• ii. Colored (black) ball point pen.
• iii. A scale

TO PERFORM THE ACTIVITY:-

Steps:-

1. Draw a square on the graph paper whose each side is of length \(5\text{ cm}\) and then make partition of this square into 25 small squares as shown in fig 1.1 each small square has its side of length \(1\text{ cm}\).

Here, we observe that the area of the square having side of length \(5\text{ cm} = 5^2\text{ cm}^2 = 25\text{ cm}^2\)

2. As shown in Fig 1.2 draw there equal squares where each square is of same size as in figure 1.1 then the total area in the fig1.2
\(= 5^2 + 5^2 + 5^2\text{ cm}^2\)
\(= 5^2 \times 3\text{ cm}^2\text{ ie}, 75\text{ cm}^2\)

Fig=1.1

Fig=1.2

3. As shown in fig 1.3 draw six equal square where each square is as same size as in Fig 1.1 Here , three squares are in one row and three squares in the second row.

We observe that the total area of six squares
\(= 5^2 \times (3+3)\text{ cm}^2\)
\(= 5^2 \times 3 \times 2\text{ cm}^2\)

Also observe that the total area
\(= 75\text{ cm}^2 + 75\text{ cm}^2 = 150\text{ cm}^2\)

Hence, we have verified that
\(150 = 5^2 \times 3 \times 2\)

Fig-1.3

ACTIVITY-2

TOPIC:- Ratio of the areas of two similar triangles

STATEMENT:- The ratio of the area of two similar triangle is equal to the ratio of the squares of their corresponding sides.

OBJECTIVE:- To verify the above statement through activity.

PRE-REQUISITE KNOWLEDGE:-

• 1. The concept of similar triangles.
• 2. Division of a line segment into equal parts.
• 3. Construction of lines parallel to given line.

MATERIAL REQUIRED:-

• 1. White paper sheet
• 2. Scale /Rubber
• 3. Paint box
• 4. Black ball point pen or pencil

TO PERFORM THE ACTIVITY:-

STEPS:-

1. On the poster paper sheet, draw two similar triangle ABC and DEF. We have the ratio of their corresponding sides same and let as have
\(AB: DE = BC: EF = CA: FD = 5:3\)
ie , \(AB/DE = 5/3\) , \(BC/EF = 5/3\) , \(CA/FD = 5/3\),
ie \(DE = 3/5 AB\), \(EF = 3/5 BC\), \(FD = 3/5 CA\)

2. Divide each side of \(\Delta ABC\) into 5 equal parts and those of \(\Delta DEF\) into 3 equal parts as shown in Fig (i) and (ii).

3. By drawing parallel lines as shown in Fig (i) and (ii)., we have partition \(\Delta ABC\) into 25 smaller triangle of same size and also each smaller triangle in fig (i) has same size and as that of the smaller triangle fig (ii).

4. Paint the smaller triangle as shown in Fig (i) and (ii)..

OBSERVATION:-

1. Area of \(\Delta ABC = \text{Area of 25 smaller triangle in fig (i)} = 25\text{ square unit}\)
Where area of one smaller triangle in fig (i) \(= P\) (square unit )

2. Area of \(\Delta DEF = \text{Area of a smaller triangle in Fig (ii)} = 9P\) where area of one smaller triangle in fig (ii) \(= P\) square units.

3. \(\frac{\text{Area of }\Delta ABC}{\text{Area of }\Delta DEF} = \frac{25P}{9P} = \frac{25}{9}\)

4. \(\frac{(AB)^2}{(DE)^2} = \frac{(AB)^2}{\left(\frac{3}{5}AB\right)^2} = \frac{(AB)^2}{\frac{9}{25}(AB)^2} = \frac{25}{9}\)
Similarly
\(\frac{(BC)^2}{(EF)^2} = \frac{25}{9}\quad\text{and}\quad\frac{(CA)^2}{(FD)^2} = \frac{25}{9}\)

5. From steps (3) and (4) , we conclude that
\(\frac{\text{Area of }\Delta ABC}{\text{Area of }\Delta DEF} = \frac{(AB)^2}{(DE)^2} = \frac{(BC)^2}{(EF)^2} = \frac{(CA)^2}{(FD)^2}\)

Hence the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Fig(i)  Fig (ii)

ACTIVITY-3

TOPIC:- Trigonometric identities.

STATEMENT:- \(\sin^2\theta + \cos^2\theta = 1, 0^\circ < \theta < 90^\circ\)

OBJECTIVE:- To verify the above identity

PRE-REQUISITE KNOWLEDGE:- In a right angled triangle.
\(\sin\theta = \frac{\text{Side opposite to angle }\theta}{\text{Hypotenuse of the triangle}}\)
\(\cos\theta = \frac{\text{Side adjacent to angle }\theta}{\text{Hypotenuse of the triangle}}\)

MATERIAL REQUIRED:-

• 1. Drawing sheet
• 2. Black ball point pen
• 3. Geometry box
• 4. Scale

TO PERFORM THE ACTIVITY

Step:-

1. On the drawing sheet, draw horizontal ray AX .

2. Construct any arbitrary \(\angle CAX = \theta\) (say)

3. Construct \(AC = 10\text{ cm}\).

4. From C draw \(CB \perp AX\).

5. Measure the length sides of sides AB and BC of the right angled \(\Delta ABC\) (see fig)

6. We measure that \(AB = 8.4\text{ cm (approx)}\) and \(BC = 5.4\text{ cm (approx)}\)

OBSERVATION

1. \(\sin\theta = BC/AC = 5.4/10 = .54\text{ (Approx)}\)

2. \(\cos\theta = AB/AC = 8.4/10 = .84\text{ (approx)}\)

3. \(\sin^2\theta + \cos^2\theta = (.54)^2 + (.84)^2\)
\(= .2916 + .7056\)
\(= .9972\text{ (Approx)}\)

Ie. \(\sin^2\theta + \cos^2\theta\) is nearly equal to 1. Hence the identity is verified.

ACTIVITY-4

Topics:- Measure of the central tendencies of a data.

STATEMENT:- We have an empirical relationship for statistical data as \(3 \times \text{median} = \text{Mode} + 2 \times \text{mean}\).

OBJECTIVE :- To verify the above statement for a data.

PRE-REQUISITE KNOWLEDGE:-
Method to find central tendencies for grouped data.

MATERIAL REQUIRED:-

• 1. A data about the heights of students of a class and arranged in grouped form.
• 2. A ball point pen.
• 3. A scale.

TO PERFORM THE ACTIVITY:-

Step:-

1. Count the number of girl students in the class. The number is 51

2. Record the data about their height in centimeter.

3. Write the data in grouped form as below:-

4. On three different sheets of paper find mean height on the sheet of paper , median height on the second and the modal height on the third sheet of paper.

5. Let us find mean by step deviation method:-

\(\text{Mean} = a + h \times \frac{\sum f_i u_i}{\sum f_i} = 147.5 + 5 \times 23/51 = 147.5 + 115/51\)
\(= (147.5 + 2.255)\text{ cm} = 149.755\text{ cm}\)

6. Let us find median of the data:-

\(n/2 = 25.5\)

we have median class (145-150) it gives \(l=145, h=5, f=18, cf=11\)

\(\text{median} = l + \left\{ \frac{\frac{n}{2} - cf}{f} \right\} \times h = 145 + \left\{ \frac{25.5 - 11}{18} \right\} \times 5\)
\(= 145 + \frac{14.5 \times 5}{18}\)
\(= 145 + 4.028\)
\(= 149.028\text{ cm}\)

7. Let us find mode of the data:-

Modal class is 145-150
Thus \(l=145\), \(h=5\), \(f_m=18\), \(f_1=7\), \(f_2=11\)

\(\text{Mode} = l + \left\{ \frac{f_m - f_1}{2f_m - f_1 - f_2} \right\} \times h = 145 + \left\{ \frac{18 - 7}{36 - 7 - 11} \right\} \times 5\)
\(= 145 + 55/18 = 145 + 3.055\)
\(= 148.055\text{ cm}\)

8. CONCLUSION:-
Mean=149.755, median=149.028 and mode=148.055
\(3 \times \text{median} = 3 \times 149.028 = 447.084\)
\(\text{Mode} + 2 \times \text{mean} = 148.055 + 2 \times 149.755\)
\(= 148.055 + 299.510 = 447.565\)

Thus we have verified that \(3 \times \text{median} = \text{mode} + 2 \times \text{mean}\) (Approx)

ACTIVITY – 5

TOPIC : Angle of Elevation

OBJECTIVE : To find the angle of elevation of the sun at a particular time on a sunny day.

PRE-REQUISITE KNOWLEDGE: knowledge of trigonometric ratios.

MATERIAL REQUIRED :

• 1. A metre rod
• 2. Measuring tape
• 3. Table for tangent of angles .

TO PERFORM THE ACTIVITY:

STEPS :

1. On the particular sunny day at the given time, put the metre rod on the level ground with one end on the ground and the other vertically upward.

2. Measure the length of the shadow of the metre rod from the beginning to the end. Let the length of the shadow be 58cm = 0.58m.

3. The length of the metre rod = 1m or 100cm.

OBSERVATION:

1. If \(\theta\) be the angle of elevation of the sun at the given moment, then we have the following figure on a sheet of paper by taking a suitable scale.

2. From the right angle \(\Delta OMP\) drawn in figure, we have
\(\tan \theta = \frac{MP}{OM} = \frac{100}{58} = 1.724\text{ (approx)}\)
\(\tan \theta = \sqrt{3}\text{ (approx.)}\)
i.e. \(\tan \theta = \tan 60^\circ\)
\(\theta = 60^\circ\)

Hence, the required angle of elevation of the sun is \(60^\circ\). For better result, we can take the help of the table of tangent of angles.

ACTIVITY – 6

TOPIC - Probability of events of a random experiment.

STATEMENT: For an event E of a random experiment, \(P(\text{not } E) = 1 - P(E)\).

OBJECTIVE: To verify the above statement by tossing three coins of different denominations simultaneously for head and tail. Event E happens if we get at least two heads and the event not-E happens if we do not get two or more than two heads.

PRE-REQUISITE KNOWLEDGE:

• 1. Probability of an event : \(\frac{\text{Number of outcome which favour the happening of the event } E}{\text{Total number of outcome}}\)
• 2. Event not-E happens when the outcome is not favourable for the event E to happen.

TO PERFORM THE ACTIVITY:

STEPS:

1. Take three fair coins of different denominations and toss these coins simultaneously.

2. We imaging about the possible outcomes as below.
HHH, HHT, HTH, THH, HTT, TTH, TTT
i.e. there can be 8 possible outcomes
favourable outcomes to the event E are
HHH, HHT, HTH, THH
Then \(P(E) = 4/8 = 1/2\)
Now, favourable outcomes to the event not-E are HTT, THT, TTH, TTT
Then \(P(\text{not-}E) = 1 - 1/2 = 1 - P(E)\)

3. Repeating above random experiment, we record the observation of 20 trials as below:

4. From table in step 3, we observe that for 2 heads or for 3 heads, the event E happens i.e. there are \(5+4=9\) chances out of 20 which favour E
Thus, we have \(P(E) = \frac{9}{20}\)
Also we observe that for 0 head or for 1 head the event not-E happens. There are \(4+7=11\) chances out of 20 which favour not-E.
So, \(P(\text{not-}E) = 11/20 = 1 - 9/20 = 1 - P(E)\).