Please Wait... While Loading Full Video#### class-10 chapter-13

### Surface Area of a Combination of Solids

♦ Introduction

♦ Surface Area of a Combination of Solids

♦ Surface Area of a Combination of Solids

We're familiar with some of the solids like cuboid, cone, cylinder, and sphere (see Fig. 13.1). You have also learnt how to find their surface areas and volumes.

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. 13.2), 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.

Again, you may have seen an object like the one in Fig. 13.3. 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.

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

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. 13.2), 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.

Again, you may have seen an object like the one in Fig. 13.3. 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.

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

Let us consider the container seen in Fig. 13.2. To find the surface area of such a solid , 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. 13.4, after we put the pieces all together.

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,

` text (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 step 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. 13.5.

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:

`text ( Total surface area of the toy = CSA of hemisphere + CSA of cone )`

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. 13.4, after we put the pieces all together.

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,

` text (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 step 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. 13.5.

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:

`text ( Total surface area of the toy = CSA of hemisphere + CSA of cone )`

Q 3210601519

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 13.6). The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he

has to colour. (Take `π = 22/7` )

Class 10 Chapter 13 Example 1

it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere

(see Fig 13.6). The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he

has to colour. (Take `π = 22/7` )

Class 10 Chapter 13 Example 1

This top is exactly like the object we have discussed in Fig. 13.5. So, we

can conveniently use the result we have arrived at there. That is :

TSA of the toy = CSA of hemisphere + CSA of cone

Now, the curved surface area of the hemisphere ` = 1/2 (4 pi r^2 ) = 2 pi r^2`

`= (2 xx 22/7 xx 3.5/2 xx 3.5/2 ) cm^2`

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

`= (5- 3.5/2) cm = 3.25 cm`

So, the slant height of the cone` ( l ) = sqrt (r^2 + h^2) = sqrt ( (3.5/2)^2 + (3.25)^2 ) = 3.7 cm` (approx )

Therefore, CSA of cone =` πrl = (22/7 xx 3.5/2 xx 3.7) cm^2`

This gives the surface area of the top as

`= ( 2 xx 22/7 xx 3.5/2 xx 3.5/2 ) cm^2 + (22/7 xx 3.5/2 xx 3.7) cm^2`

` = 22/7 xx 3.5 /2 (3.5 +3.7) cm^2 = 11/2 xx (3.5 +3.7) cm^2 = 39.6 cm^2` (aaprox)

You may note that ‘total surface area of the top’ is not the sum of the total

surface areas of the cone and hemisphere.

Q 3220701611

The decorative block shown in Fig. 13.7 is made of two solids — a cube

and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere

fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block.

(Take ` pi = 22/7` )

Class 10 Chapter 13 Example 2

and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere

fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block.

(Take ` pi = 22/7` )

Class 10 Chapter 13 Example 2

The total surface area of the cube `= 6 × text ( (edge) )^2 = 6 × 5 × 5 cm^2 = 150 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

`= 150 – πr^2 + 2 πr^2 = (150 + πr^2) cm^2`

`= 150 cm^2 + (22/7 xx 4.2/2 xx 4.2/2) cm^2 `

`== (150 + 13.86) cm^2 = 163.86 cm^2`

Q 3230701612

A wooden toy rocket is in the shape of a cone mounted on a cylinder, as

shown in Fig. 13.8. The height of the entire rocket is 26 cm, while the height of the conical

part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base

diameter of the cylindrical portion is 3 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 `π = 3.14` )

Class 10 Chapter 13 Example 3

shown in Fig. 13.8. The height of the entire rocket is 26 cm, while the height of the conical

part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base

diameter of the cylindrical portion is 3 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 `π = 3.14` )

Class 10 Chapter 13 Example 3

Denote radius of cone by `r`, slant

height of cone by `l`, height of cone by `h`, radius

of cylinder by `r′` and height of cylinder by `h′`.

Then `r = 2.5 cm, h = 6 cm, r′ = 1.5 cm,

h′ = 26 – 6 = 20 cm `and

` l = sqrt (r^2 + h^2) = sqrt(2.5^2 + 6^2) cm = 6.5 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

`= πrl + πr^2 – π(r′)^2`

`= π[(2.5 × 6.5) + (2.5)^2 – (1.5)^2] cm^2`

`= π[20.25] cm^2 = 3.14 × 20.25 cm^2`

`= 63.585 cm^2`

Now, the area to be painted yellow = CSA of the cylinder

+ area of one base of the cylinder

`= 2πr′h′ + π(r′)^2`

`= πr′ (2h′ + r′)`

`= (3.14 × 1.5) (2 × 20 + 1.5) cm^2`

`= 4.71 × 41.5 cm^2`

`= 195.465 cm^2`

Q 3260701615

Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical

depression at one end (see Fig. 13.9). The height of the cylinder is 1.45 m and its radius is 30 cm. Find the

toal surface area of the bird-bath. (Take `π = 22/7` )

Class 10 Chapter 13 Example 4

depression at one end (see Fig. 13.9). The height of the cylinder is 1.45 m and its radius is 30 cm. Find the

toal surface area of the bird-bath. (Take `π = 22/7` )

Class 10 Chapter 13 Example 4

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

`= 2πrh + 2πr^2 = 2πr (h + r)`

` = 2 xx 22/7 xx 30 (145 + 30 ) cm^2`

` = 33000 cm^2 =3.3 m^2`