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. 13 (a)]. Hold the string with your hands on either sides of the triangle and rotate the triangle
This is called a right circular cone. In Fig. 13 (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. 14)! 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.
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. 15 (c) will form the circular base of the cone.
(iii) If the paper like the one in Fig. 15 (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 = ½ × base of each triangle × l.
So, area of the entire piece of paper
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 l2 = r2 + h2 (as can be seen from Fig. 16), by applying Pythagoras Theorem. Here h is the height of the cone.
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 πr2.
Example 4 : 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 = 22/7 × 7 × 10 cm2
= 220 cm2
Example 5 : 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 l2 = h2 + r2, we have
cm = 20 cm
So, curved surface area = πrl
= 3.14 × 12 × 20 cm2
= 753.6 cm2
Further, total surface area = πrl + πr2
= (753.6 + 3.14 × 12 × 12) cm2
= (753.6 + 452.16) cm2
= 1205.76 cm2
Example 6 : A corn cob (see Fig. 17), 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.
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.
Therefore, the curved surface area of the corn cob = πrl
= 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.