The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm. Find its volume, the curved surface area and

Solution

The frustum can be viewed as a difference of two right circular cones OAB and OCD. Let the height (in cm) of the cone OAB be

h1 and its slant height l1 i.e., OP= h1, and OA = OB = l1. Let h2 be the height of cone OCD and l2 its slant height.

We have: r1 = 28 cm, r2 = 7 cm and the height of frustum (h) = 45 cm. Also, h1 = 45 + h2 ………. (i)

We first need to determine the respective heights h1 and h2 of the cone OAB and OCD. Since the triangles OPB and OQD are similar (Why?), we have

From (i) and (ii), we get h2 = 15 and h1 = 60.

Now, the volume of the frustum = volume of the cone OAB – volume of the cone OCD

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The respective slant height l2 and l1 of the cones OCD and OAB are given by

.

Thus, the curved surface area of the frustum = πr1l1 – πr2l2

  .

Now, the total surface area of the frustum = the curved surface area + πr12+πr22

= 5461.5 cm2 + 2464 cm2 + 154 cm2 = 8079.5 cm2.

Let h be the height, l the slant height and r1 and r2 the radii of the ends (r1 > r2) of the frustum of a cone. Then we can directly find the volume, the curved surface area and the total surface area of frustum by using the formulae given below:

(i) Volume of the frustum of the cone = .

(ii) The curved surface area of the frustum of the cone = π(r1 + r2) l, where .

(iii) Total surface area of the frustum of the cone = πl (r1 + r2) + πr12 + πr22, where .

These formulae can be derived using the idea of similarity of triangles but we shall not be doing derivations here.

Let us solve this example, by using these formulae:

(i) Volume of the frustum

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(ii) We have

.

So, the curved surface area of the frustum .

(iii) Total curved surface area of the frustum

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