This document discusses surface areas and volumes of combinations of 3D shapes. It begins by explaining that combinations of shapes can be broken down into their individual components to calculate surface areas and volumes. Specific examples are given of calculating the total surface area and volume of objects made of multiple solids by finding the surface area or volume of each individual solid and adding them together. New formulas for calculating the surface area and volume of frustums are also introduced.
2. INTRODUCTION
In earlier classes, we have studied about the surface
areas and volumes of different 3-D figures. But, the
figures about which we studied were single figures.
In this chapter, we will study about the surface areas
and volumes of some joined figures such as the surface
area and volume of a cuboid joined with a hemisphere.
We can notice these type of figures in our daily life also.
For ex.- an oil tanker, which is made up of a cylinder
joined with two hemispheres at it’s both ends.
3. Surface Area of a Combination of
Solids To solve a complex problem, we first try to break it
down into simpler and more simpler parts, so that we
can solve the problem easily.
Similarly, to find the surface areas and volumes of the
joint figures, we first break it down into different
figures.
4. For example, to find the Total Surface Area of
these two joined figures, we first find the T.S.A of
the first figure and then of the another.
5. Volume of a Combination of Solids
In finding the surface areas of some combinations, we found
the individual surface areas of the figures and then, added
them.
Similarly, in order to find the volume of the combinations, we
first find the volume of the first figure and then of the other.
Then, we add the these two volumes to get the volume of the
combination of solids.
For Ex-
6. Suppose we have to find the volume of that figure, we
would first find the volume of the cylinder and then of the
cuboids.
Then we would add these up, in order to find the total
volume of the figure.
7. Conversion of Solid from One Shape to
Another We all have seen candles of different shapes, like cylindrical,
in the shape of some animals, statues etc.
You might be thinking that how they are made? First a normal
candle of a cylindrical shape is taken, then it is heated and
molten in a container, and then it is allowed to cool in a
container which of the shape of a bird or an animal.
But do you think that if there’s a change in the dimensions of
the candle, the SA and the Volume of the candle will also
change.
No, the surface and the volume of the candle will remain
same, irrespective of its dimensions.
8. Frustum Of A Cone
We have seen objects which are made up of the combination
of two solids.
Now, let us take an example of the right circular cone.
A
BC
9. A Cone is also made up of two parts, a smaller right circular
cone and an another solid.
First let’s do the practical:
A Right Circular Cone The Two parts Separated
Small Right Circular Cone Frustum of the cone
10. So, given a cone, when we slice through with a plane
parallel to its base and remove the cone that is formed on
one side of that plane, the part that is now left over is
called the Frustum Of The Cone.
11. Some Questions Related To The Chapter
The decorative block here, is made up of two solids – a cube and a hemisphere. The
base of the block is a cube with edge 5cm, and the hemisphere fixed on the top has a
diameter of 4.2cm. Find the total surface area of the block.
The total surface area of the cube = 6 x (edge)² = 6 x 5 x 5 cm² = 150 cm²
Curved surface area of the hemisphere = 2Πr² = 2 x 22/7 x 2.1 x 2.1 cm²
= 13.86 cm²
.˙. Total Surface Area of the Block = (150 + 13.86) cm² = 163.86cm²
5 cm
4.2 cm
12. A juice seller was serving his customers using glasses as shown in the fig. The inner
diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a
hemispherical raised which reduced the capacity of the glass. If height the glass was
10 cm, find apparent capacity of the glass and it’s capacity. ( Taking Π=3.14).
The inner diameter of the glass = 5 cm and height of the glass = 10 cm
.˙. The Apparent Capacity of the glass = Πr²h
= 3.14 x 2.5 x 2.5 x 10 cm³ = 196.25 cm³
But, because of the hemisphere, the actual capacity of the glass is less than it’s volume.
.˙. The volume of the hemisphere = 2/3Πr²h = 2/3 x 196.25 cm³ = 32.71 cm³
Now, the actual capacity of the glass = apparent capacity – volume of the hemisphere
= (196.25 – 32.71) cm³
= 163.54 cm³
13. A Cone of height 24 cm and of base diameter 12 cm is made up of modelling
clay. A Child reshapes it in the form of a sphere. Find the radius of the
sphere.
The Diameter of the cone = 12 cm so, radius of the cone = 12/2 = 6 cm
.˙.Volume of the cone = 1/3Πr²h = 1/3 x Π x 6 x 6 x 24 cm³
Now, taking ‘r’ as the radius of the sphere, the volume will be = 4/3 Πr³
Since, the volume of clay in the form of the cone and the sphere remains the
same, we have :
4/3 x Π x r³ = 1/3 x Π x 6 x 6 x 24 cm
r³ = 3 x 3 x 24 = 3² x 2³
.˙. r = 3 x 2 = 6
24 cm
12 cm
_ cm
14. Hanumappa and his wife Gangamma are busy making jaggery out of sugarcane
juice. They have processed the sugarcane juice to the molasses, which is poured
into molds in the shape of a frustum of a cone having the diameters of it’s two
circular faces as 30 cm and 35 cm and the vertical height of the mold is 14cm. If
each cm³ of molasses has about 1.2 g, find the mass of molasses that can be
poured into each mold. (Taking Π = 22/7)
Since the mold is in the shape of a frustum of a cone, the volume of molasses that
can be poured into it = Π/3 x h( r₁² + r₂² + r₁ r₂ )
where, ‘r’ is the radius of the target base and ‘r₂’ is the radius of the smaller base
= 1/3 x 22/7 x 14[(35/2)² + (30/2)² + (35/2 x 30/2)] cm³ = 11641.7 cm³
Now, we know that : 1cm³ of molasses = 1.2 g
.˙. The mass of molasses that can be poured into each mold =(11641 x 1.2) g
= 13970.04 g = 13.97 kg = 14 kg (approx.)
15. New Formulas Learnt In This Chapter
Volume of the Frustum = 1/3Πh(r₁² + r₂² + r₁ r₂ )
Curved Surface Area of the Frustum = Πl(r₁+ r₂)
Total Surface Area of the Frustum =Πl(r₁+ r₂) + Π(r₁+ r₂)
where, h = vertical height of the frustum
l = slant height of the frustum
r₁+ r₂ = radii of the two bases(ends) of the frustum
