12.1 Introduction

From Class IX, you are familiar with some of the solids like cuboid, cone, cylinder, and sphere (see Fig. 12.1). You have also learnt how to find their surface areas and volumes.

Fig. 12.1

In our day-to-day life, we come across a number of solids made up of combinations of two or more of the basic solids as shown above.

You must have seen a truck with a container fitted on its back (see Fig. 12.2), carrying oil or water from one place to another. Is it in the shape of any of the four basic solids mentioned above? You may guess that it is made of a cylinder with two hemispheres as its ends.

Fig. 12.2

Again, you may have seen an object like the one in Fig. 12.3. Can you name it? A test tube, right! You would have used one in your science laboratory. This tube is also a combination of a cylinder and a hemisphere. Similarly, while travelling, you may have seen some big and beautiful buildings or monuments made up of a combination of solids mentioned above.

If for some reason you wanted to find the surface areas, or volumes, or capacities of such objects, how would you do it? We cannot classify these under any of the solids you have already studied.

Fig. 12.3

In this chapter, you will see how to find surface areas and volumes of such objects.

12.2 Surface Area of a Combination of Solids

Let us consider the container seen in Fig. 12.2. How do we find the surface area of such a solid? Now, whenever we come across a new problem, we first try to see, if we can break it down into smaller problems, we have earlier solved. We can see that this solid is made up of a cylinder with two hemispheres stuck at either end. It would look like what we have in Fig. 12.4, after we put the pieces all together.

Fig. 12.4

If we consider the surface of the newly formed object, we would be able to see only the curved surfaces of the two hemispheres and the curved surface of the cylinder.

So, the total surface area of the new solid is the sum of the curved surface areas of each of the individual parts. This gives,

TSA of new solid $=$ CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere

where TSA, CSA stand for ‘Total Surface Area’ and ‘Curved Surface Area’ respectively.

Let us now consider another situation. Suppose we are making a toy by putting together a hemisphere and a cone. Let us see the steps that we would be going through.

First, we would take a cone and a hemisphere and bring their flat faces together. Here, of course, we would take the base radius of the cone equal to the radius of the hemisphere, for the toy is to have a smooth surface. So, the steps would be as shown in Fig. 12.5.

Fig. 12.5

At the end of our trial, we have got ourselves a nice round-bottomed toy. Now if we want to find how much paint we would require to colour the surface of this toy, what would we need to know? We would need to know the surface area of the toy, which consists of the CSA of the hemisphere and the CSA of the cone.

So, we can say:

Total surface area of the toy $=$ CSA of hemisphere + CSA of cone

Now, let us consider some examples.

Example 1 : Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere (see Fig 12.6). The entire top is $5 \mathrm{~cm}$ in height and the diameter of the top is $3.5 \mathrm{~cm}$. Find the area he has to colour. (Take $\pi=\dfrac{22}{7}$ )

Fig. 12.6

Solution : This top is exactly like the object we have discussed in Fig. 12.5. So, we can conveniently use the result we have arrived at there. That is :

$ \text { TSA of the toy }=\text { CSA of hemisphere }+ \text { CSA of cone } $

Now, the curved surface area of the hemisphere $=\dfrac{1}{2}\left(4 \pi r^{2}\right)=2 \pi r^{2}$

$$ =\left(2 \times \dfrac{22}{7} \times \dfrac{3.5}{2} \times \dfrac{3.5}{2}\right) \mathrm{cm}^{2} $$

Also, the height of the cone = height of the top - height (radius) of the hemispherical part

$$ =\left(5-\dfrac{3.5}{2}\right) \mathrm{cm}=3.25 \mathrm{~cm} $$

So, the slant height of the cone $(l)=\sqrt{r^{2}+h^{2}}=\sqrt{\left(\dfrac{3.5}{2}\right)^{2}+(3.25)^{2}} \mathrm{~cm}=3.7 \mathrm{~cm}$ (approx.)

Therefore, CSA of cone $=\pi r l=\left(\dfrac{22}{7} \times \dfrac{3.5}{2} \times 3.7\right) \mathrm{cm}^{2}$

This gives the surface area of the top as

$$ \begin{aligned} & =\left(2 \times \dfrac{22}{7} \times \dfrac{3.5}{2} \times \dfrac{3.5}{2}\right) \mathrm{cm}^{2}+\left(\dfrac{22}{7} \times \dfrac{3.5}{2} \times 3.7\right) \mathrm{cm}^{2} \\ & =\dfrac{22}{7} \times \dfrac{3.5}{2}(3.5+3.7) \mathrm{cm}^{2}=\dfrac{11}{2} \times(3.5+3.7) \mathrm{cm}^{2}=39.6 \mathrm{~cm}^{2} \text { (approx.) } \end{aligned} $$

You may note that ’total surface area of the top’ is not the sum of the total surface areas of the cone and hemisphere.

Example 2 : The decorative block shown in Fig. 12.7 is made of two solids - a cube and a hemisphere. The base of the block is a cube with edge $5 \mathrm{~cm}$, and the hemisphere fixed on the top has a diameter of $4.2 \mathrm{~cm}$. Find the total surface area of the block. (Take $\pi=\dfrac{22}{7}$ )

Fig. 12.7

Solution : The total surface area of the cube $=6 \times(\text { edge })^{2}=6 \times 5 \times 5 \mathrm{~cm}^{2}=150 \mathrm{~cm}^{2}$.

Note that the part of the cube where the hemisphere is attached is not included in the surface area.

So, the surface area of the block $=$ TSA of cube - base area of hemisphere + CSA of hemisphere

$$ \begin{aligned} & =150-\pi r^{2}+2 \pi r^{2}=\left(150+\pi r^{2}\right) \mathrm{cm}^{2} \\ & =150 \mathrm{~cm}^{2}+\left(\dfrac{22}{7} \times \dfrac{4.2}{2} \times \dfrac{4.2}{2}\right) \mathrm{cm}^{2} \\ & =(150+13.86) \mathrm{cm}^{2}=163.86 \mathrm{~cm}^{2} \end{aligned} $$

Example 3 : A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in Fig. 12.8. The height of the entire rocket is $26 \mathrm{~cm}$, while the height of the conical part is $6 \mathrm{~cm}$. The base of the conical portion has a diameter of $5 \mathrm{~cm}$, while the base diameter of the cylindrical portion is $3 \mathrm{~cm}$. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take $\pi=3.14$ )

Fig. 12.8

Solution : Denote radius of cone by $r$, slant height of cone by $l$, height of cone by $h$, radius of cylinder by $r^{\prime}$ and height of cylinder by $h^{\prime}$. Then $r=2.5 \mathrm{~cm}, h=6 \mathrm{~cm}, r^{\prime}=1.5 \mathrm{~cm}$, $h^{\prime}=26-6=20 \mathrm{~cm}$ and

$$ l=\sqrt{r^{2}+h^{2}}=\sqrt{2.5^{2}+6^{2}} \mathrm{~cm}=6.5 \mathrm{~cm} $$

Here, the conical portion has its circular base resting on the base of the cylinder, but the base of the cone is larger than the base of the cylinder. So, a part of the base of the cone (a ring) is to be painted.

So, the area to be painted orange $=$ CSA of the cone + base area of the cone - base area of the cylinder

$$ \begin{aligned} & =\pi r l+\pi r^{2}-\pi\left(r^{\prime}\right)^{2} \\ & =\pi\left[(2.5 \times 6.5)+(2.5)^{2}-(1.5)^{2}\right] \mathrm{cm}^{2} \\ & =\pi[20.25] \mathrm{cm}^{2}=3.14 \times 20.25 \mathrm{~cm}^{2} \\ & =63.585 \mathrm{~cm}^{2} \end{aligned} $$

Now, the area to be painted yellow $=$ CSA of the cylinder + area of one base of the cylinder

$$ \begin{aligned} & =2 \pi r^{\prime} h^{\prime}+\pi\left(r^{\prime}\right)^{2} \\ & =\pi r^{\prime}\left(2 h^{\prime}+r^{\prime}\right) \\ & =(3.14 \times 1.5)(2 \times 20+1.5) \mathrm{cm}^{2} \\ & =4.71 \times 41.5 \mathrm{~cm}^{2} \\ & =195.465 \mathrm{~cm}^{2} \end{aligned} $$

Example 4 : Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end (see Fig. 12.9). The height of the cylinder is $1.45 \mathrm{~m}$ and its radius is $30 \mathrm{~cm}$. Find the total surface area of the bird-bath. (Take $\pi=\dfrac{22}{7}$ )

Fig. 12.9

Solution : Let $h$ be height of the cylinder, and $r$ the common radius of the cylinder and hemisphere. Then, the total surface area of the bird-bath $=$ CSA of cylinder + CSA of hemisphere

$$ \begin{aligned} & =2 \pi r h+2 \pi r^{2}=2 \pi r(h+r) \\ & =2 \times \dfrac{22}{7} \times 30(145+30) \mathrm{cm}^{2} \\ & =33000 \mathrm{~cm}^{2}=3.3 \mathrm{~m}^{2} \end{aligned} $$

12.3 Volume of a Combination of Solids

In the previous section, we have discussed how to find the surface area of solids made up of a combination of two basic solids. Here, we shall see how to calculate their volumes. It may be noted that in calculating the surface area, we have not added the surface areas of the two constituents, because some part of the surface area disappeared in the process of joining them. However, this will not be the case when we calculate the volume. The volume of the solid formed by joining two basic solids will actually be the sum of the volumes of the constituents, as we see in the examples below.

Example 5 : Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (see Fig. 12.12). If the base of the shed is of dimension $7 \mathrm{~m} \times 15 \mathrm{~m}$, and the height of the cuboidal portion is $8 \mathrm{~m}$, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of $300 \mathrm{~m}^{3}$, and there are 20 workers, each of whom occupy about $0.08 \mathrm{~m}^{3}$ space on an average. Then, how much air is in the shed? (Take $\pi=\dfrac{22}{7}$ )

Fig. 12.12

Solution : The volume of air inside the shed (when there are no people or machinery) is given by the volume of air inside the cuboid and inside the half cylinder, taken together.

Now, the length, breadth and height of the cuboid are $15 \mathrm{~m}, 7 \mathrm{~m}$ and $8 \mathrm{~m}$, respectively. Also, the diameter of the half cylinder is $7 \mathrm{~m}$ and its height is $15 \mathrm{~m}$.

So, the required volume $=$ volume of the cuboid $+\dfrac{1}{2}$ volume of the cylinder

$$ =\left[15 \times 7 \times 8+\dfrac{1}{2} \times \dfrac{22}{7} \times \dfrac{7}{2} \times \dfrac{7}{2} \times 15\right] \mathrm{m}^{3}=1128.75 \mathrm{~m}^{3} $$

Next, the total space occupied by the machinery $=300 \mathrm{~m}^{3}$

And the total space occupied by the workers $=20 \times 0.08 \mathrm{~m}^{3}=1.6 \mathrm{~m}^{3}$

Therefore, the volume of the air, when there are machinery and workers

$$ =1128.75-(300.00+1.60)=827.15 \mathrm{~m}^{3} $$

Example 6 : A juice seller was serving his customers using glasses as shown in Fig. 12.13. The inner diameter of the cylindrical glass was $5 \mathrm{~cm}$, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was $10 \mathrm{~cm}$, find the apparent capacity of the glass and its actual capacity. (Use $\pi=3.14$.)

Fig. 12.13

Solution : Since the inner diameter of the glass $=5 \mathrm{~cm}$ and height $=10 \mathrm{~cm}$, the apparent capacity of the glass $=\pi r^{2} h$

$$ =3.14 \times 2.5 \times 2.5 \times 10 \mathrm{~cm}^{3}=196.25 \mathrm{~cm}^{3} $$

But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass.

i.e., $\quad$ it is less by $\dfrac{2}{3} \pi r^{3}=\dfrac{2}{3} \times 3.14 \times 2.5 \times 2.5 \times 2.5 \mathrm{~cm}^{3}=32.71 \mathrm{~cm}^{3}$

So, the actual capacity of the glass $=$ apparent capacity of glass - volume of the hemisphere

$$ \begin{aligned} & =(196.25-32.71) \mathrm{cm}^{3} \\ & =163.54 \mathrm{~cm}^{3} \end{aligned} $$

Example 7 : A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is $2 \mathrm{~cm}$ and the diameter of the base is $4 \mathrm{~cm}$. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take $\pi=3.14$ )

Fig. 12.14

Solution : Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere (see Fig. 12.14). The radius BO of the hemisphere (as well as of the cone) $=\dfrac{1}{2} \times 4 \mathrm{~cm}=2 \mathrm{~cm}$.

So, volume of the toy $=\dfrac{2}{3} \pi r^{3}+\dfrac{1}{3} \pi r^{2} h$

$$ =\left[\dfrac{2}{3} \times 3.14 \times(2)^{3}+\dfrac{1}{3} \times 3.14 \times(2)^{2} \times 2\right] \mathrm{cm}^{3}=25.12 \mathrm{~cm}^{3} $$

Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of the base of the right circular cylinder $=\mathrm{HP}=\mathrm{BO}=2 \mathrm{~cm}$, and its height is

$$ \mathrm{EH}=\mathrm{AO}+\mathrm{OP}=(2+2) \mathrm{cm}=4 \mathrm{~cm} $$

So, the volume required $=$ volume of the right circular cylinder - volume of the toy

$$ \begin{aligned} & =\left(3.14 \times 2^{2} \times 4-25.12\right) \mathrm{cm}^{3} \\ & =25.12 \mathrm{~cm}^{3} \end{aligned} $$

Hence, the required difference of the two volumes $=25.12 \mathrm{~cm}^{3}$.

12.4 Summary

In this chapter, you have studied the following points:

1. To determine the surface area of an object formed by combining any two of the basic solids, namely, cuboid, cone, cylinder, sphere and hemisphere.

2. To find the volume of objects formed by combining any two of a cuboid, cone, cylinder, sphere and hemisphere.