Class 9 Volume of a Right Circular Cone

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Volume of a Right Circular Cone

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`color{red} ♦` Volume of a Right Circular Cone

`color{red} ♦` Volume of a Right Circular Cone

Volume of a Right Circular Cone

In Fig 13.28, can you see that there is a right circular cylinder and a right circular cone of the same base radius and the same height?

Activity : Try to make a hollow cylinder and a hollow cone like this with the same base radius and the same height (see Fig. 13.28).

Then, we can try out an experiment that will help us, to see practically what the volume of a right circular cone would be!

So, let us start like this.

Fill the cone up to the brim with sand once, and empty it into the cylinder. We find that it fills up only a part of the cylinder [see Fig. 13.29(a)].

When we fill up the cone again to the brim, and empty it into the cylinder, we see that the cylinder is still not full [see Fig. 13.29(b)].

When the cone is filled up for the third time, and emptied into the cylinder, it can be seen that the cylinder is also full to the brim [see Fig. 13.29(c)].

With this, we can safely come to the conclusion that three times the volume of a cone, makes up the volume of a cylinder, which has the same base radius and the same height as the cone, which means that the volume of the cone is one-third the volume of the cylinder.

So,

Volume of a Cone ` = 1/3 pi r^2 h`

where `r` is the base radius and `h` is the height of the cone.

In Fig 13.28, can you see that there is a right circular cylinder and a right circular cone of the same base radius and the same height?

Activity : Try to make a hollow cylinder and a hollow cone like this with the same base radius and the same height (see Fig. 13.28).

Then, we can try out an experiment that will help us, to see practically what the volume of a right circular cone would be!

So, let us start like this.

Fill the cone up to the brim with sand once, and empty it into the cylinder. We find that it fills up only a part of the cylinder [see Fig. 13.29(a)].

When we fill up the cone again to the brim, and empty it into the cylinder, we see that the cylinder is still not full [see Fig. 13.29(b)].

When the cone is filled up for the third time, and emptied into the cylinder, it can be seen that the cylinder is also full to the brim [see Fig. 13.29(c)].

With this, we can safely come to the conclusion that three times the volume of a cone, makes up the volume of a cylinder, which has the same base radius and the same height as the cone, which means that the volume of the cone is one-third the volume of the cylinder.

So,

Volume of a Cone ` = 1/3 pi r^2 h`

where `r` is the base radius and `h` is the height of the cone.

Q 3225734661

The height and the slant height of a cone are `21` cm and `28` cm respectively. Find the volume of the cone.
Class 9 Chapter 13 Example 15

Solution:

From `l^2 = r^2 + h^2`, we have

`r = sqrt (l^2 − h^2) = sqrt (28^2 − 21^2) cm = 7 sqrt (7) cm`

So, volume of the cone `= 1/3 π r ^2 h = 1/3 xx (22)/7 xx 7 sqrt (7) xx 7 sqrt (7) xx 21 cm^3`

`= 7546 cm^3`

Q 3235734662

Monica has a piece of canvas whose area is `551 m^2`. She uses it to have a conical tent made, with a base radius of `7 m`. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately `1 m^2`, find the volume of the tent that can be made with it.
Class 9 Chapter 13 Example 16

Solution:

Since the area of the canvas `= 551 m^2` and area of the canvas lost in wastage is `1 m^2`, therefore the area of canvas available for making the tent is `(551 – 1) m^2 = 550 m^2`.
Now, the surface area of the tent `= 550 m^2` and the required base radius of the conical tent `= 7 m`
Note that a tent has only a curved surface (the floor of a tent is not covered by canvas!!).

Therefore, curved surface area of tent `= 550 m^2`.

That is, `πrl = 550`

or, `(22)/7 xx 7 xx l = 550`

or, `l = 3 (550)/(22) m = 25 m`

Now, `l^2 = r^2 + h^2`

Therefore, `h = sqrt (l^2 − r^2) = sqrt (25^2 − 7^2) m = sqrt (625 − 49) m = sqrt (576) m`

`= 24 m`

So, the volume of the conical tent ` = 1/3 π r^2 h = 1/3 xx (22)/7 xx 7 xx 7 xx 24 m^3 = 1232 m^3`

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