# Surface Areas And Volumes Chapter 13 Class 10 Maths NCERT PDF

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### Chapter 13: Surface Areas and Volumes

#### 13.1 Introduction

From Class IX, you are familiar with some of the solids like cuboids, cones, cylinders, and spheres (see Fig. 13.1).

You have also learned 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.

#### 13.2 Surface Area of a Combination of Solids

Let us consider the container seen in Fig. 13.2. 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. 13.4 after we put the pieces all together.

#### 13.3 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.

### NCERT Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes

1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

The diagram is given as:

Given,

The Volume (V) of each cube is = 64 cm3

This implies that a3 = 64 cm3

∴ a = 4 cm

Now, the side of the cube = a = 4 cm

Also, the length and breadth of the resulting cuboid will be 4 cm each. While its height will be 8 cm.

So, the surface area of the cuboid = 2(lb+bh+lh)

= 2(8×4+4×4+4×8) cm2

= 2(32+16+32) cm2

= (2×80) cm2 = 160 cm2

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

The diagram is as follows:

Now, the given parameters are:

The diameter of the hemisphere = D = 14 cm

The radius of the hemisphere = r = 7 cm

Also, the height of the cylinder = h = (13-7) = 6 cm

And, the radius of the hollow hemisphere = 7 cm

Now, the inner surface area of the vessel = CSA of the cylindrical part + CSA of hemispherical part

(2πrh+2πr2) cm2 = 2πr(h+r) cm2

2×(22/7)×7(6+7) cm2 = 572 cm2

3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

The diagram is as follows:

Given that the radius of the cone and the hemisphere (r) = 3.5 cm or 7/2 cm

The total height of the toy is given as 15.5 cm.

So, the height of the cone (h) = 15.5-3.5 = 12 cm

∴ The curved surface area of cone = πrl

(22/7)×(7/2)×(25/2) = 275/2 cm2

Also, the curved surface area of the hemisphere = 2πr2

2×(22/7)×(7/2)2

= 77 cm2

Now, the Total surface area of the toy = CSA of cone + CSA of hemisphere

= (275/2)+77 cm2

= (275+154)/2 cm2

= 429/2 cm2 = 214.5cm2

So, the total surface area (TSA) of the toy is 214.5cm2