SURFACE AREAS AND VOLUMES

SURFACE AREAS AND VOLUMES

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 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)]?

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. 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.

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

So, area of the entire piece of paper

= sum of the areas of all the triangles

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

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

(as \(\left( {{b}_{1}}+{{b}_{2}}+{{b}_{3}}+........ \right)\) 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

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

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 \(\pi {{r}^{2}}\)

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