Surface Areas And Volumes

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 ABC right angled at 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 (c) and (d)]?

 Fig. 11.1

This is called a right circular cone. In Fig. 11.1(c) of the right circular cone, the point A is called the vertex, AB is called the height, BC is called the radius and AC 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 is not at right angle to the base, and in (b) the base is not circular.

 Fig. 11.2

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)  (c)

                                      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 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 = \(\frac{1}{2}\times l\times \)base of each triangle × l.

So, area of the entire piece of paper

= sum of the areas of all the triangles

= \(\frac{1}{2}{{b}_{1}}l+\frac{1}{2}{{b}_{2}}l+\frac{1}{2}{{b}_{3}}l+....=\frac{1}{2}({{b}_{1}}+{{b}_{2}}+{{b}_{3}}+....)\)

= \(\frac{1}{2}\times l\times \)length of entire curved boundary of Fig. 11.3(c)

(as \({{b}_{1}}+{{b}_{2}}+{{b}_{3}}+....\)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 = 2πr, where r is the base radius of the cone.

So, Curved Surface Area of a Cone = \(\frac{1}{2}\) × l × 2πr = πrl

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.

 Fig. 11.4

Therefore \(l=\sqrt{{{r}^{2}}+{{h}^{2}}}\)

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 πr ².

So, Total Surface Area of a Cone = πrl + πr ² = πr(l + r)

Example 1 : Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm

Solution : Curved surface area

                      = πrl

                      = \(\frac{22}{7}\)× 7 × 10 cm²

                     = 220 cm²

Example 2 : The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone (Use π = 3.14).

Solution : Here, h = 16 cm and r = 12 cm.

So, from l ² = h ² + r ² ,

we have l = \(\sqrt{{{16}^{2}}+{{12}^{2}}}\) cm = 20 cm

So, curved surface area = πrl                  

                                  = 3.14 × 12 × 20 cm²

                                  = 753.6 cm² Further, total surface area = πrl + πr ² = (753.6 + 3.14 × 12 × 12) cm²

                                                  = (753.6 + 452.16) cm² = 1205.76 cm²

Example 3 : A corn cob (see Fig. 11.5), shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm and length (height) as 20 cm. If each 1 cm² of the surface of the cob carries an average of four grains, find how many grains you would find on the entire cob.

 Fig. 11.5

Solution : Since the grains of corn are found only on the curved surface of the corn cob, we would need to know the curved surface area of the corn cob to find the total number of grains on it. In this question, we are given the height of the cone, so we need to find its slant height.

Here,l =\(\sqrt{{{r}^{2}}+{{h}^{2}}}\)=\(\sqrt{{{2.1}^{2}}+{{20}^{2}}}\)cm

         =\(\sqrt{404.41}\)cm=20.11 cm

Therefore, the curved surface area of the corn cob = πrl

= \(\frac{22}{7}\)× 2.1 × 20.11 cm² = 132.726 cm² = 132.73 cm² (approx.)

Number of grains of corn on 1 cm² of the surface of the corn cob = 4

Therefore, number of grains on the entire curved surface of the cob

       = 132.73 × 4 = 530.92 = 531 (approx.)

So, there would be approximately 531 grains of corn on the cob.

Surface Areas And Volumes

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)].

 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 × (π r ² ) So,

Surface Area of a Sphere = 4 π 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.

 Fig. 11.8

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’)

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 4πr ² .

Therefore, Curved Surface Area of a Hemisphere = 2πr 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 2πr ² + πr ²

So, Total Surface Area of a Hemisphere = 3πr² 

Example 4 : Find the surface area of a sphere of radius 7 cm.

Solution : The surface area of a sphere of radius 7 cm would be

                         4πr ² = 4 × \(\frac{22}{7}\) × 7 × 7 cm² = 616 cm²

Example 5 : Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.

Solution : The curved surface area of a hemisphere of radius 21 cm would be

           = 2πr ² = 2 × \(\frac{22}{7}\) × 21 × 21 cm² = 2772 cm² 

(ii) the total surface area of the hemisphere would be

              3πr ² = 3 × \(\frac{22}{7}\) × 21 × 21 cm² = 4158 cm²

Example 6 : The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.

Solution : Diameter of the sphere = 7 m. Therefore, radius is 3.5 m. So, the riding space available for the motorcyclist is the surface area of the ‘sphere’ which is given by

                4πr ² = 4 × \(\frac{22}{7}\) × 3.5 × 3.5 m² = 154 m²

Example 7 : A hemispherical dome of a building needs to be painted (see Fig. 11.9). If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is ₹ 5 per 100 cm² .

Solution : Since only the rounded surface of the dome is to be painted, we would need to find the curved surface area of the hemisphere to know the extent of painting that needs to be done. Now, circumference of the dome = 17.6 m. Therefore, 17.6 = 2πr.

So, the radius of the dome = 17.6 × \(\frac{7}{2\times 22}\)× m = 2.8 m

The curved surface area of the dome = 2πr ²

                                               = 2 × \(\frac{22}{7}\) × 2.8 × 2.8 m²

                                               = 49.28 m²

Now, cost of painting 100 cm² is ₹5.

So, cost of painting 1 m² = ₹500

Therefore, cost of painting the whole dome

               = ₹500 × 49.28

               = ₹24640   

Surface Areas And volumes

Volume of a Right Circular Cone

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

 Fig 11.11

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?

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!

          So, let us start like this.

 Fig. 11.12(a)  Fig. 11.12(b)  Fig. 11.12(c)

      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, Volume of a Cone =  πr ²h

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

Example 8 : The height and the slant height of a cone are 21 cm and 28 cm respectively. Find the volume of the cone.

Solution : From l ² = r ² + h ² ,

we have r = \(\sqrt{{{l}^{2}}-{{h}^{2}}}\)= \(\sqrt{{{28}^{2}}-{{21}^{2}}}cm= 7 \sqrt{7}\) cm

So, volume of the cone =πr²h = \(\frac{1}{3}\times \frac{22}{7}\times 7\sqrt{7}\times 7\sqrt{7}\times 21\)cm³ = 7546 cm³  

Example 9 : Monica has a piece of canvas whose area is 551 m2 . She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m2 , find the volume of the tent that can be made with it

Solution : Since the area of the canvas = 551 m² and area of the canvas lost in wastage is 1 m² , therefore the area of canvas available for making the tent is (551 – 1) m² = 550 m² . Now, the surface area of the tent = 550 m² and the required base radius of the conical tent = 7 m Note that a tent has only a curved surface (the floor of a tent is not covered by canvas!!).

Therefore, curved surface area of tent = 550 m² .

That is, πrl = 550

or, \(\frac{22}{7}\) × 7 × l = 550

or, l = 3 \(\frac{550}{22}\)m = 25 m

Now, l ² = r ² + h ²

Therefore, h = \(\sqrt{{{l}^{2}}-{{r}^{2}}}\)= \(\sqrt{{{25}^{2}}-{{7}^{2}}}\)m =\(\sqrt{625-49}\)m =\(\sqrt{576}\)

                                                                               = 24 m

So, the volume of the conical tent = \(\frac{1}{3}\times \pi {{r}^{2}}h=\frac{1}{3}\times \frac{22}{7}\times 7\times 7\times 24\) m³ = 1232 m³

Surface Areas And Volumes

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 \(\frac{4}{3}\) πr 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 \(\frac{4}{3}\pi {{R}^{3}}\) 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 \(\frac{4}{3}\) π times the cube of its radius. This gives us the idea that

                                        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, Volume of a Hemisphere = \(\frac{2}{3}\pi {{r}^{3}}\)

      where r is the radius of the hemisphere.

Let us take some examples to illustrate the use of these formulae.

Example 10 : Find the volume of a sphere of radius 11.2 cm.

Solution : Required volume = \(\frac{4}{3}\pi {{r}^{3}}\)

= \(\frac{4}{3}\times \frac{22}{7}\times 11.2\times 11.2\times 11.2\) cm³

= 5887.32 cm³

Example 11 : A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm³ , find the mass of the shot-putt.

Solution : Since the shot-putt is a solid sphere made of metal and its mass is equal to the product of its volume and density, we need to find the volume of the sphere.

Now, volume of the sphere = \(\frac{4}{3}\times \pi {{r}^{3}}\)

=\(\frac{4}{3}\times \frac{22}{7}\times 4.9\times 4.9\times 4.9\)cm³

= 493 cm³ (nearly)

Further, mass of 1 cm³ of metal is 7.8 g.

Therefore, mass of the shot-putt = 7.8 × 493 g = 3845.44 g

= 3.85 kg (nearly)

Example 12 : A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?

Solution : The volume of water the bowl can contain

= \(\frac{2}{3}\times \pi {{r}^{3}}\)

= \(\frac{2}{3}\times \frac{22}{7}\times 3.5\times 3.5\times 3.5\)cm³ = 89.8 cm³