📖 generic · CBSE Class 10 ENGLISH MEDIUM · MATHEMATICS · Page 1example

V OLUMES · Part 6

Chapter 12: SURFACE AREAS AND VOLUMES · MATHEMATICS

to find the surface area of solids made up of a combination of two basic solids. Here, we shall see how to calculate their volumes. It may be noted that in calculating the surface area, we have not added the surface areas of the two constituents, because some part of the surface area disappeared in the process of joining them. However, this will not be the case when we calculate the volume.

The volume of the solid formed by joining two basic solids will actually be the sum of the volumes of the constituents, as we see in the examples below. Example : Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (see Fig. . ).

If the base of the shed is of dimension m × m, and the height of the cuboidal portion is m, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of m , and there are workers, each of whom occupy about . m space on an average. Then, how much air is in the shed?

Solution : The volume of air inside the shed (when there are no people or machinery) is given by the volume of air inside the cuboid and inside the half cylinder, taken together. Now, the length, breadth and height of the cuboid are m, m and m, respectively. Also, the diameter of the half cylinder is m and its height is m. So, the required volume = volume of the cuboid + volume of the cylinder m       = .

m Next, the total space occupied by the machinery = m And the total space occupied by the workers = × . m = . m Therefore, the volume of the air, when there are machinery and

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