Class 09 To form a cuboid and find the formula for its surface area experimentally.

 

Activity 27 

OBJECTIVE	MATERIAL REQUIRED
To form a cuboid and find the formula	Cardboard, cellotape, cutter, ruler,
for its surface area experimentally.	sketch pen/pencil.

 METHOD OF CONSTRUCTION

 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].

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

 On actual measurement:

 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

Class 09 To form a cube and find the formula for its surface area experimentally.

 Activity 26 

OBJECTIVE	MATERIAL REQUIRED
To form a cube and find the formula for	Cardboard, ruler, cutter, cellotape,
its surface area experimentally.	sketch pen/pencil.

 METHOD OF CONSTRUCTION

 1.   Make six identical squares each of side a units, using cardboard and join them as shown in Fig. 1 using a cellotape.

 Fold the squares along the dotted markings to form a cube [see Fig. 2].

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

 On actual measurement:

 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.

Note: Instead of making six squares separately as done in the activity, a net of a cube be directly prepared on the cardboard itself.

Class 09 To find the formula for the area of atrapezium experimentally.

 Activity 25 

OBJECTIVE	MATERIAL REQUIRED
To find the formula for the area of a	Hardboard, thermocol, coloured
trapezium experimentally.	glazed papers, adhesive, scissors.

 METHOD OF CONSTRUCTION

 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].

 Place them on the hardboard as shown in Fig. 2

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	(area of parallelogram)	=	1	( a + b ) × h
	2			2

1Therefore, area of trapezium =     ( a + b)× h 

1= ₂ (sum of parallel sides) × perpendicular distance.

 Here, 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 =	1	(Sum of	sides) ×	.
2

 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.

Class 09 To verify that the opposite angles of a cyclic quadrilateral are supplementary.

 

Activity 24

OBJECTIVE	MATERIAL REQUIRED
To verify that the opposite angles of a	Chart  paper,  geometry  box,scissors, sketch pens, adhesive, transparent sheet.

 cyclic quadrilateral are supplementary.

 METHOD OF CONSTRUCTION

 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.

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, 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.

Class 09 To verify that the angles in the same segment of a circle are equal.

 Activity 23 

OBJECTIVE                                                                     

To verify that the angles in the same segment of a circle are equal.

 METHOD OF CONSTRUCTION

MATERIAL REQUIRED 

Geometry box, coloured glazed papers, scissors, cardboard, white paper and adhesive.

 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].

 Take replicas of the angles ∠ACB and ∠ADB.

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 ---------.

 APPLICATION

 This result may be used in proving other theorems/riders of geometry related to circles.

Class 09 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.

Activity 22

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

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.                                                                                       Fig. 1

 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].

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.

Class 09 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.

 Activity 21

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.

METHOD OF CONSTRUCTION

 MATERIAL REQUIRED

Plywood sheet of convenient size, graph paper, colour box, a pair of wooden strips, scissors, cutter, adhesive, geometry box.

 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.

 Take any point P on the second strip and join it to A and B [see Fig. 1].

DEMONSTRATION

 1.   AB is parallel to CD and P is any point on CD.

 2.   Triangle PAB and parallelogram ABCD are on the same base AB and between the same parallels.

 3.   Count the number of squares contained in each of the above triangle and

 1parallelograms, keeping half square as 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 parallelograms ABCD.

 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 = ........ : ...........

APPLICATION

 This activity is useful in deriving formula 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.