Class 09 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.

 

Activity 29

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

 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].

DEMONSTRATION

 1.   Fill the cone with sand (or salt) and pour it twice into the hemisphere. The hemisphere is completely filled with sand.

 1Therefore, volume of cone = 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.

 1Therefore, volume of cone = 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		Volume of	.
	2
Volume of cone	=	1			Volume of ---------------	.
	3
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.

Class 09 To form a cone from a sector of a circle and to find the formula for its curved surface area

 

Activity 28

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, marker.

METHOD OF CONSTRUCTION

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].

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r	× l 2 = rl.
	2l
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, curved surface area of the cone = Area of the sector.

 Here, area is in square units.

 APPLICATION

 The result is useful in

 1.   estimating canvas required to make a conical tent

 estimating material required to make Joker’s cap, ice cream cone, etc.

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.