Hands-On Geometry Activities for High School Math, Exercises of Mathematics

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Activities for

Class X

Geometry was always considered more as a discipline of the mind than any other part of mathematics, for it could boast closer relations to logic. Genuine deductivity was the privilege of geometry, whereas the business of algebra was substitution into and transforming formulae. On the other hand the pragmatic point of view would require only a few theorems and not the geometry prescribed by Euclidean tradition. Some people are prepared to teach more useless things in mathematics, but object to geometry being a weak system

- H. Freudenthal.

96 Laboratory Manual

3. Stick these strips on the other cardboard sheet as shown in Fig. 6 to Fig. 9.

DEMONSTRATION

As per Euclid Division Lemma, Fig. 6 depicts a = b × 1 + c ( q = 1, r = c ) (1) Fig. 7 depicts b = c × 2 + d ( q = 2, r = d ) (2) Fig.8 depicts c = d × 1 + e ( q = 1, r = e ) (3) and Fig. 9 depicts d = e × 2 + 0 ( q = 2, r = 0) (4)

Fig. 6

Fig. 7

Fig. 8 Fig. 9

Mathematics 97

As per assumptions in Euclid Division Algorithm, HCF of a and b = HCF of b and c = HCF of c and d = HCF of d and e The HCF of d and e is equal to e, from (4) above. So, HCF of a and b = e.

OBSERVATION

On actual measurement (in mm) a =......... , b = ......... , c = ......... , d = ......... , e = ......... So, HCF of __________ and __________ = ......................

APPLICATION

The process depicted can be used for finding the HCF of two or more numbers, which is known as finding HCF of numbers by Division Method.

Mathematics 99

(i) a > 0 (ii) a < 0

4. Find the ordered pairs ( x , f ( x )) for different values of x.
5. Plot these ordered pairs in the cartesian plane.

Fig. 2

Fig. 3

6. Join the plotted points by a free hand curve [Fig. 1, Fig. 2 and Fig. 3].

100 Laboratory Manual

DEMONSTRATION

1. The shape of the curve obtained in each case is a parabola.
2. Parabola opens upward when coefficient of x^2 is positive [see Fig. 2 and Fig. 3].
3. It opens downward when coefficient of x^2 is negative [see Fig. 1].
4. Maximum number of zeroes which a quadratic polynomial can have is 2.

OBSERVATION

1. Parabola in Fig. 1 opens ______
2. Parabola in Fig. 2 opens _______
3. In Fig. 1, parabola intersects x -axis at ______ point(s).
4. Number of zeroes of the given polynomial is ________.
5. Parabola in Fig. 2 intersects x -axis at ______ point(s).
6. Number of zeroes of the given polynomial is ______.
7. Parabola in Fig.3 intersects x -axis at ______ point(s).
8. Number of zeroes of the given polynomial is _______.
9. Maximum number of zeroes which a quadratic polynomial can have is __________.

APPLICATION

This activity helps in

1. understanding the geometrical representation of a quadratic polynomial
2. finding the number of zeroes of a quadratic polynomial.

NOTE

Points on the graph paper should be joined by a free hand curve only.

102 Laboratory Manual

Fig. 1

Fig. 2

Mathematics 103

Fig. 3

DEMONSTRATION

Case I: We obtain the graph as shown in Fig. 1. The two lines are intersecting at one point P. Co-ordinates of the point P ( x,y) give the unique solution for the pair of linear equations (1) and (2).

Therefore, the pair of linear equations with 1 1 2 2

a b a^ ≠^ b is consistent and has the unique solution. Case II: We obtain the graph as shown in Fig. 2. The two lines are coincident. Thus, the pair of linear equations has infinitely many solutions.

Therefore, the pair of linear equations with 1 1 1 2 2 2

a b c a^ =^ b =^ c is also consistent as well as dependent.

Mathematics 105

METHOD OF CONSTRUCTION

1. Take a hardboard of a convenient size and paste a white chart paper on it.
2. Draw a square of side of length x units , on a pink glazed paper and paste it on the hardboard [see Fig. 1]. Divide it into 36 unit squares with a marker.
3. Alongwith each side of the square (outside) paste rectangles of green glazed paper of dimensions x × 1, i.e., 6 × 1 and divide each of them into unit squares with the help of a marker [see Fig. 1].
4. Draw 4 squares each of side 1 unit on a yellow glazed paper, cut them out and paste each unit square on each corner as shown in Fig. 1.

OBJECTIVE MATERIAL REQUIRED

To obtain the solution of a quadratic equation ( x^2 + 4 x = 60) by completing the square geometrically.

Hardboard, glazed papers, adhesive, scissors, marker, white chart paper.

Activity 4

Fig. 1 Fig. 2

106 Laboratory Manual

5. Draw another square of dimensions 8 × 8 and arrange the above 64 unit squares as shown in Fig. 2.

DEMONSTRATION

1. The first square represents total area x^2 + 4 x + 4.
2. The second square represents a total of 64 (60 + 4) unit squares. Thus, x^2 + 4 x + 4 = 64 or ( x + 2)^2 = (8)^2 or ( x + 2) = + 8 i.e., x = 6 or x = – Since x represents the length of the square, we cannot take x = –10 in this case, though it is also a solution.

OBSERVATION

Take various quadratic equations and make the squares as described above, solve them and obtain the solution(s).

APPLICATION

Quadratic equations are useful in understanding parabolic paths of projectiles projected in the space in any direction.

108 Laboratory Manual

3. Let the lists of numbers be (i) 1, 2, 5, 9, ....... (ii) 1, 4, 7, 10, ......
4. Make strips of lengths 1, 2, 5, 9 units and strips of lengths 1, 4, 7, 10 units and breadth of each strip one unit.
5. Paste the strips of lengths 1, 2, 5, 9 units as shown in Fig. 1 and paste the strips of lengths 1, 4, 7, 10 units as shown in Fig. 2.

DEMONSTRATION

1. In Fig. 1, the difference of heights (lengths) of two consecutive strips is not same (uniform). So, it is not an AP.
2. In Fig. 2, the difference of heights of two consecutive strips is the same (uniform) throughout. So, it is an AP.

OBSERVATION

In Fig. 1, the difference of heights of first two strips = _____________ the difference of heights of second and third strips = _____________ the difference of heights of third and fourth strips = _____________ Difference is _____________ (uniform/not uniform) So, the list of numbers 1, 2, 5, 9 _____________ form an AP. (does/does not) Write the similar observations for strips of Fig.2. Difference is _____________ (uniform/not uniform) So, the list of the numbers 1, 4, 7, 10 _________ form an AP. (does/does not)

APPLICATION

This activity helps in understanding the concept of arithmetic progression.

NOTE

Observe that if the left top corners of the strips are joined, they will be in a straight line in case of an AP.

Mathematics 109

METHOD OF CONSTRUCTION

1. Take a rectangular cardboard of a convenient size and paste a coloured paper on it. Draw a rectangle ABCD of length 11 units and breadth 10 units.
2. Divide this rectangle into unit squares as shown in Fig. 1.
3. Starting from upper left-most corner, colour one square, 2 squares and so on as shown in the figure.

DEMONSTRATION

1. The pink colour region looks like a stair case.
2. Length of 1st stair is 1 unit, length of 2nd stair is 2 units, length of 3rd stair 3 units, and so on, length of 10th stair is 10 units.

OBJECTIVE MATERIAL REQUIRED

To find the sum of first n natural numbers.

Cardboard, coloured papers, white paper, cutter, adhesive.

Activity 6

Fig. 1

Mathematics 111

METHOD OF CONSTRUCTION

1. Take a piece of cardboard of a convenient size and paste a white paper on it.
2. Draw a square of suitable size on it (10 cm × 10 cm).
3. Divide this square into unit squares.
4. Fix a thermocol ball in each square with the help of a pin as shown in Fig. 1.
5. Enclose the balls as shown in the figure.

OBJECTIVE MATERIAL REQUIRED

To find the sum of the first n odd natural numbers.

Cardboard, thermocol balls, pins, pencil, ruler, adhesive, white paper.

Activity 7

Fig. 1

DEMONSTRATION

Starting from the uppermost right corner, the number of balls in first enclosure (blue colour) = 1 (=1^2 ),

112 Laboratory Manual

the number of balls in first 2 enclosures = 1 + 3 = 4 (=2^2 ), the number of balls in first 3 enclosures = 1 + 3 + 5 = 9 (=3^2 ), the number of balls in first 10 enclosures = 1 + 3 + 5 + ... + 19 = 100 (=10^2 ). This gives the sum of first ten odd natural numbers. This result can be generalised for the sum of first n odd numbers as: Sn = 1 + 3 + ......... + (2 n – 1) = n^2 (1)

OBSERVATION

For n = 4 in (1), Sn = ........................ For n = 5 in (1), Sn = ........................ For n = 50 in (1), Sn = ........................ For n = 100 in (1), Sn= ........................

APPLICATION

The activity is useful in determining formula for the sum of the first n odd natural numbers.