Volume of a Combination of Solids
We saw how when calculating surface areas sometimes we add and sometimes we remove some areas as some surface area disappeared in the process of joining the seperate shapes. Lets see how we can calculate volumes of combination of solids.
Let's say you have the below treasure chest with you. You need to find our how much volume is holds so that you can fill up your treasure accordingly.
In the previous section, we have discussed how 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.
Following the same approach as earlier, how can we break up this image so that we get some basic shapes.
By separating the different sections, we have the two basic shapes,
But observe closely that we actually dont have a full cylinder. We have just
So the volume of our treasure chest is volume of the cuboid +
Now, if the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively and if the diameter of the half cylinder is 7 m and its height is 15 m the volume of the treasure chest is
Say, we are pouring a liquid in a beaker first into a cuboid container and next we are pouring the same liquid into a cylindrical container. Will the volume of the liquid and in the cuboid and the cylinder be the same?
That's right. The volume is retained though the matter is in different shapes and in different containers. Let us take this insight and solve some problems.
A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.
We know that the volume of the solid remains the same even if changes it's shape. So the volume of the cone is equal to the volume of the sphere.
Volume of cone=
Volume of sphere=
Volume of cone= Volume of sphere
Solving for this r=
Example 5 : Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (just like treasure chest shown earlier). If the base of the shed is of dimension 7 m × 15 m, and the height of the cuboidal portion is 8 m, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of 300
- The volume of air inside the shed (when there are no people or machinery) is given by the volume of air inside the
and (taken together). - Total volume of shed =
+ 1 2 where l,b,h is length,breadth and height of cuboidal space while r and t are the radius and height of the cylindrical space. - Substituting we get the evaluated volume =
m 3 - Volume occupied by machinery and workers:
m 3 - Remaining volume :
m 3 - Thus, we have found the answer.
Example 6 : A juice seller was serving his customers using glasses as shown in Fig. 12.13. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Use π = 3.14)
- Apparent capacity of glass:
where r and h are radius and height of glass - Substituting we get apparent volume =
cm 3 - Given actual capacity of the glass is less by the volume of the hemisphere at the base of the glass i.e.
where t is radius of hemispherical shape. - Substituting we get the evaluated hemispherical volume =
cm 3 - Actual volume of glass :
cm 3 - Thus, we have found the answer.
Example 7 : A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take π = 3.14)
- Radius of hemisphere (and cone) =
cm - Thus, volume of toy:
+ where r is radius of hemisphere while d and l are radius and height of cone. - Substituting we get volume of toy =
cm 3 - To find the volume of the right circular cylinder circumscribing the toy: radius =
cm while height = cm. - Finding the volume, we get:
cm 2 - Thus, the difference in volume =
cm 2 - Thus, we have found the answer.