11.1 Surface Area of a Right Circular Cone

We have already studied the surface areas of cube, cuboid and cylinder. We will now study the surface area of cone.

So far, we have been generating solids by stacking up congruent figures. Incidentally, such figures are called prisms. Now let us look at another kind of solid which is not a prism (These kinds of solids are called pyramids.). Let us see how we can generate them.

Activity : Cut out a right-angled triangle $\mathrm{ABC}$ right angled at $\mathrm{B}$. Paste a long thick string along one of the perpendicular sides say AB of the triangle [see Fig. 11.1(a)]. Hold the string with your hands on either sides of the triangle and rotate the triangle about the string a number of times. What happens? Do you recognize the shape that the triangle is forming as it rotates around the string [see Fig. 11.1(b)]? Does it remind you of the time you had eaten an ice-cream heaped into a container of that shape [see Fig. $11.1(\mathrm{c})$ and (d)]?

(a)

(b)

(c)

(d)

Fig. 11.1

This is called a right circular cone. In Fig. 11.1(c) of the right circular cone, the point $\mathrm{A}$ is called the vertex, $\mathrm{AB}$ is called the height, $\mathrm{BC}$ is called the radius and $A C$ is called the slant height of the cone. Here $B$ will be the centre of circular base of the cone. The height, radius and slant height of the cone are usually denoted by $h, r$ and $l$ respectively. Once again, let us see what kind of cone we can not call a right circular cone. Here, you are (see Fig. 11.2)! What you see in these figures are not right circular cones; because in (a), the line joining its vertex to the centre of its base

(a) (b)

Fig. 11.2 is not at right angle to the base, and in (b) the base is not circular.

As in the case of cylinder, since we will be studying only about right circular cones, remember that by ‘cone’ in this chapter, we shall mean a ‘right circular cone.’

Activity : (i) Cut out a neatly made paper cone that does not have any overlapped paper, straight along its side, and opening it out, to see the shape of paper that forms the surface of the cone. (The line along which you cut the cone is the slant height of the cone which is represented by $l$ ). It looks like a part of a round cake.

(ii) If you now bring the sides marked A and B at the tips together, you can see that the curved portion of Fig. 11.3 (c) will form the circular base of the cone.

(a)

(b)

Fig. 11.3

(iii) If the paper like the one in Fig. 11.3 (c) is now cut into hundreds of little pieces, along the lines drawn from the point $\mathrm{O}$, each cut portion is almost a small triangle, whose height is the slant height $l$ of the cone.

(iv) Now the area of each triangle $=\frac{1}{2} \times$ base of each triangle $\times l$.

So, area of the entire piece of paper

$$ \begin{aligned} & =\text { sum of the areas of all the triangles } \\ & =\frac{1}{2} b_{1} l+\frac{1}{2} b_{2} l+\frac{1}{2} b_{3} l+\cdots=\frac{1}{2} l\left(b_{1}+b_{2}+b_{3}+\cdots\right) \\ & =\frac{1}{2} \times l \times \text { length of entire curved boundary of Fig. 11.3(c) } \end{aligned} $$

$\quad$(as $b_{1}+b_{2}+b_{3}+\ldots$ makes up the curved portion of the figure)

But the curved portion of the figure makes up the perimeter of the base of the cone and the circumference of the base of the cone $=2 \pi r$, where $r$ is the base radius of the cone.

So, Curved Surface Area of a Cone $=\frac{1}{2} \times l \times 2 \pi r=\pi r l$

where $r$ is its base radius and $l$ its slant height.

Note that $l^{2}=r^{2}+h^{2}$ (as can be seen from Fig. 11.4), by applying Pythagoras Theorem. Here $h$ is the height of the cone.

Therefore, $l=\sqrt{r^{2}+h^{2}}$

Fig. 11.4

Now if the base of the cone is to be closed, then a circular piece of paper of radius $r$ is also required whose area is $\pi r^{2}$.

So, Total Surface Area of a Cone $=\pi r l+\pi r^{2}=\pi r(l+r)$

11.2 Surface Area of a Sphere

What is a sphere? Is it the same as a circle? Can you draw a circle on a paper? Yes, you can, because a circle is a plane closed figure whose every point lies at a constant distance (called radius) from a fixed point, which is called the centre of the circle. Now if you paste a string along a diameter of a circular disc and rotate it as you had rotated the triangle in the previous section, you see a new solid (see Fig 11.6). What does it resemble? A ball? Yes. It is called a sphere.

Fig. 11.6

Can you guess what happens to the centre of the circle, when it forms a sphere on rotation? Of course, it becomes the centre of the sphere. So, a sphere is a three dimensional figure (solid figure), which is made up of all points in the space, which lie at a constant distance called the radius, from a fixed point called the centre of the sphere.

Note : A sphere is like the surface of a ball. The word solid sphere is used for the solid whose surface is a sphere.

Activity : Have you ever played with a top or have you at least watched someone play with one? You must be aware of how a string is wound around it. Now, let us take a rubber ball and drive a nail into it. Taking support of the nail, let us wind a string around the ball. When you have reached the ‘fullest’ part of the ball, use pins to keep the string in place, and continue to wind the string around the remaining part of the ball, till you have completely covered the ball [see Fig. 11.7(a)]. Mark the starting and finishing points on the string, and slowly unwind the string from the surface of the ball.

Now, ask your teacher to help you in measuring the diameter of the ball, from which you easily get its radius. Then on a sheet of paper, draw four circles with radius equal to the radius of the ball. Start filling the circles one by one, with the string you had wound around the ball [see Fig. 11.7(b)].

(a)

(b)

Fig. 11.7

What have you achieved in all this?

The string, which had completely covered the surface area of the sphere, has been used to completely fill the regions of four circles, all of the same radius as of the sphere.

So, what does that mean? This suggests that the surface area of a sphere of radius $r$ $=4$ times the area of a circle of radius $r=4 \times\left(\pi r^{2}\right)$

So,$\quad$Surface Area of a Sphere $=4 \pi r^{2}$

where $r$ is the radius of the sphere.

How many faces do you see in the surface of a sphere? There is only one, which is curved.

Now, let us take a solid sphere, and slice it exactly ’through the middle’ with a plane that passes through its centre. What happens to the sphere?

Yes, it gets divided into two equal parts (see Fig. 11.8)! What will each half be called? It is called a hemisphere. (Because ‘hemi’ also means ‘half’)

Fig. 11.8

And what about the surface of a hemisphere? How many faces does it have?

Two! There is a curved face and a flat face (base).

The curved surface area of a hemisphere is half the surface area of the sphere, which is $\frac{1}{2}$ of $4 \pi r^{2}$.

Therefore, Curved Surface Area of a Hemisphere $=\mathbf{2} \boldsymbol{\pi} r^{2}$

where $r$ is the radius of the sphere of which the hemisphere is a part.

Now taking the two faces of a hemisphere, its surface area $2 \pi r^{2}+\pi r^{2}$

So, Total Surface Area of a Hemisphere $=3 \pi r^{2}$

11.3 Volume of a Right Circular Cone

In earlier classes we have studied the volumes of cube, cuboid and cylinder

In Fig 11.11, can you see that there is a right circular cylinder and a right circular cone of the same base radius and the same height?

Fig. 11.11

Activity : Try to make a hollow cylinder and a hollow cone like this with the same base radius and the same height (see Fig. 11.11). Then, we can try out an experiment that will help us, to see practically what the volume of a right circular cone would be!

(a)

(b)

(c)

Fig. 11.12

So, let us start like this.

Fill the cone up to the brim with sand once, and empty it into the cylinder. We find that it fills up only a part of the cylinder [see Fig. 11.12(a)].

When we fill up the cone again to the brim, and empty it into the cylinder, we see that the cylinder is still not full [see Fig. 11.12(b)].

When the cone is filled up for the third time, and emptied into the cylinder, it can be seen that the cylinder is also full to the brim [see Fig. 11.12(c)].

With this, we can safely come to the conclusion that three times the volume of a cone, makes up the volume of a cylinder, which has the same base radius and the same height as the cone, which means that the volume of the cone is one-third the volume of the cylinder.

So,$\quad \text { Volume of a Cone }=\frac{1}{3} \pi r^{2} h$

where $r$ is the base radius and $h$ is the height of the cone.

11.4 Volume of a Sphere

Now, let us see how to go about measuring the volume of a sphere. First, take two or three spheres of different radii, and a container big enough to be able to put each of the spheres into it, one at a time. Also, take a large trough in which you can place the container. Then, fill the container up to the brim with water [see Fig. 11.13(a)].

Now, carefully place one of the spheres in the container. Some of the water from the container will over flow into the trough in which it is kept [see Fig. 11.13(b)]. Carefully pour out the water from the trough into a measuring cylinder (i.e., a graduated cylindrical jar) and measure the water over flowed [see Fig. 11.13(c)]. Suppose the radius of the immersed sphere is $r$ (you can find the radius by measuring the diameter of the sphere). Then evaluate $\frac{4}{3} \pi r^{3}$. Do you find this value almost equal to the measure of the volume over flowed?

(a)

(b)

(c)

Fig. 11.13

Once again repeat the procedure done just now, with a different size of sphere. Find the radius $R$ of this sphere and then calculate the value of $\frac{4}{3} \pi R^{3}$. Once again this value is nearly equal to the measure of the volume of the water displaced (over flowed) by the sphere. What does this tell us? We know that the volume of the sphere is the same as the measure of the volume of the water displaced by it. By doing this experiment repeatedly with spheres of varying radii, we are getting the same result, namely, the volume of a sphere is equal to $\frac{4}{3} \pi$ times the cube of its radius. This gives us the idea that

$\quad$Volume of a Sphere $=\frac{4}{3} \pi r^{3}$

where $r$ is the radius of the sphere.

Later, in higher classes it can be proved also. But at this stage, we will just take it as true.

Since a hemisphere is half of a sphere, can you guess what the volume of a hemisphere will be? Yes, it is $\frac{1}{2}$ of $\frac{4}{3} \pi r^{3}=\frac{2}{3} \pi r^{3}$.

So, $\quad$ Volume of a Hemisphere $=\frac{\mathbf{2}}{\mathbf{3}} \pi r^{3}$

where $r$ is the radius of the hemisphere.

11.5 Summary

In this chapter, you have studied the following points:

1. Curved surface area of a cone $=\pi r l$

2. Total surface area of a right circular cone $=\pi r l +\pi r^{2}$, i.e., $\pi r(l+r)$

3. Surface area of a sphere of radius $r=4 \pi r^{2}$

4. Curved surface area of a hemisphere $=2 \pi r^{2}$

5. Total surface area of a hemisphere $=3 \pi r^{2}$

6. Volume of a cone $=\frac{1}{3} \pi r^{2} h$

7. Volume of a sphere of radius $r=\frac{4}{3} \pi r^{3}$

8. Volume of a hemisphere $=\frac{2}{3} \pi r^{3}$

[Here, letters $l, b, h, a, r$, etc. have been used in their usual meaning, depending on the context.]