In the previous section we saw how height and base radius of a cone can be used to calculate the curved surface area. We also saw some solved examples. In this section, we will learn about volume of cones.
To find the volume of a cone, we can do an experiment. It is similar to the one that we did for square pyramids. (Details here)
1. Fill a cone with sand.
2. Transfer the sand into a cylinder having the same base area and height as the cone
3. We will find that the sand fills only up to one third height of the cylinder
4. The calculations are also similar. We will get the following result:
• Volume of the cone = 1⁄3 × [πrb2× h]
♦ Where rb is the radius of the base of the cone and h is the height of the cone
• Note that [πrb2× h] is the volume of a cylinder having the same base radius rb and same height h
• We will see the actual derivation of this formula in higher classes
An example:
Volume of a cone of base radius 4 cm and height 6 cm is:
1⁄3 × [πrb2× h] = 1⁄3 × [π × 42× 6] = [π × 16 × 2] = 32π cm3.
Now we will see some solved examples
Solved example 33.24
The base radius and height of a cylindrical block of wood are 15 cm and 40 cm respectively. What is the volume of the largest cone that can be carved out of it?
Solution:
1. The base radius and height of the cone will be same as those of the cylinder. See fig.33.24 below:
So we can write:
Volume of the cone = 1⁄3 × [πrb2× h] = 1⁄3 × [π × 152× 40] = [π × 15 × 5 × 40] = 3000π cm3.
Solved example 33.25
The base radius and height of a solid metal cylinder are 12 cm and 20 cm. By melting it and recasting, how many cones of base radius 4 cm and height 5 cm can be made?
Solution:
1. For the cylinder:
• Radius of base = 12 cm
• Height = 20 cm
• So volume = [πr2× height of cylinder] = [π × 122× 20] = 2880π cm3.
2. For the cone:
• Radius of base = rb = 4 cm
• Height = h = 5 cm
• So volume of one cone = 1⁄3 × [πrb2× h] = 1⁄3 × [π × 42× 5] = 1⁄3 × [80π] cm3.
3. So number of cones that can be obtained = Volume of cylinder⁄Volume of one cone
= {2880π} ÷ {1⁄3 × [80π]}= 108 Nos.
Solved example 33.26
A sector of central angle 216o is cut out from a circle of radius 25 cm and is rolled up into a cone. What are the base radius and height of the cone? What is it's volume?
Solution:
1. The two main properties of the sector:
(i) Given that radius of the circle is 25 cm. This will be same as the radius of the sector. So we can write: rs = 25 cm
• This rs will be the slant height of the cone. So we can write: Slant height = 25 cm
(ii) Central angle θ = 216o
2. From the central angle we can calculate length of arc of the sector:
• For every 1o angle, the length of arc will be πrs⁄180. (Theorem 21.1)
• So for 216o, the length of arc will be 216 × πrs⁄180.
• Thus we get:
Length of arc of the sector = 216 × (π×25⁄180) = 30π cm
3. But this is same as the circumference of the base of the cone
• So if rb is the radius of the base of the cone, we can write:
2πrb = 30π ⟹ rb = 30⁄2 = 15 cm
4. The next step is to find the slant height. Imagine a triangle OO'P.
• It must satisfy the following conditions:
♦ Triangle OO'P must be right angled
♦ O must coincide with the apex
♦ O' must coincide with the base center
♦ P must be a point on the circumference of the base
• Then we get:
♦ OO' = height of the cone
♦ O'P = rb = 15 cm
♦ OP = slant height = 25 cm
• Applying Pythagoras theorem, we get:
OO' = √[(OP)2 - (O'P)2] = √[(25)2 - (15)2] = √[625 + 225] = √[400] = 20 cm
5. So volume of the cone = 1⁄3 × [πrb2× h] = 1⁄3 × [π × 152× 20] = 1500π cm3
Solved example 33.27
The base radii of two cones are in the ratio 3:5 and their heights are in the ratio 2:3. What is the ratio of their volumes?
Solution:
1. Given that rb1⁄rb2 = 3⁄5.
Also given that h1⁄h2 = 2⁄3.
2. We have: V1 ÷ V2 = {1⁄3 × [πrb12× h1]} ÷ {1⁄3 × [πrb22× h2]}
= {[rb12× h1]} ÷ {[rb22× h2]}
= {(rb1⁄rb2)2} × {h1⁄h2}
= {(3⁄5)2} × {2⁄3} = 6⁄25.
⟹ V1⁄V2 = 6⁄25
Solved example 33.28
Two cones have the same volume and their base radii are in the ratio 4:5. What is the ratio of their heights?
Solution:
1. Given that rb1⁄rb2 = 4⁄5.
Also given that V1 = V2
2. We have: V1 = V2 ⟹ {1⁄3 × [πrb12× h1]} = {1⁄3 × [πrb22× h2]}
⟹ {[rb12× h1]} = {[rb22× h2]}
⟹ {(rb1⁄rb2)2} = {h2⁄h1}
⟹ {(4⁄5)2} = {h2⁄h1} = 16⁄25.
⟹ h1⁄h2 = 25⁄16
In the next section, we will see spheres.
To find the volume of a cone, we can do an experiment. It is similar to the one that we did for square pyramids. (Details here)
1. Fill a cone with sand.
2. Transfer the sand into a cylinder having the same base area and height as the cone
3. We will find that the sand fills only up to one third height of the cylinder
4. The calculations are also similar. We will get the following result:
• Volume of the cone = 1⁄3 × [πrb2× h]
♦ Where rb is the radius of the base of the cone and h is the height of the cone
• Note that [πrb2× h] is the volume of a cylinder having the same base radius rb and same height h
• We will see the actual derivation of this formula in higher classes
An example:
Volume of a cone of base radius 4 cm and height 6 cm is:
1⁄3 × [πrb2× h] = 1⁄3 × [π × 42× 6] = [π × 16 × 2] = 32π cm3.
Now we will see some solved examples
Solved example 33.24
The base radius and height of a cylindrical block of wood are 15 cm and 40 cm respectively. What is the volume of the largest cone that can be carved out of it?
Solution:
1. The base radius and height of the cone will be same as those of the cylinder. See fig.33.24 below:
 |
| Fig.33.24 |
Volume of the cone = 1⁄3 × [πrb2× h] = 1⁄3 × [π × 152× 40] = [π × 15 × 5 × 40] = 3000π cm3.
Solved example 33.25
The base radius and height of a solid metal cylinder are 12 cm and 20 cm. By melting it and recasting, how many cones of base radius 4 cm and height 5 cm can be made?
Solution:
1. For the cylinder:
• Radius of base = 12 cm
• Height = 20 cm
• So volume = [πr2× height of cylinder] = [π × 122× 20] = 2880π cm3.
2. For the cone:
• Radius of base = rb = 4 cm
• Height = h = 5 cm
• So volume of one cone = 1⁄3 × [πrb2× h] = 1⁄3 × [π × 42× 5] = 1⁄3 × [80π] cm3.
3. So number of cones that can be obtained = Volume of cylinder⁄Volume of one cone
= {2880π} ÷ {1⁄3 × [80π]}= 108 Nos.
Solved example 33.26
A sector of central angle 216o is cut out from a circle of radius 25 cm and is rolled up into a cone. What are the base radius and height of the cone? What is it's volume?
Solution:
1. The two main properties of the sector:
(i) Given that radius of the circle is 25 cm. This will be same as the radius of the sector. So we can write: rs = 25 cm
• This rs will be the slant height of the cone. So we can write: Slant height = 25 cm
(ii) Central angle θ = 216o
2. From the central angle we can calculate length of arc of the sector:
• For every 1o angle, the length of arc will be πrs⁄180. (Theorem 21.1)
• So for 216o, the length of arc will be 216 × πrs⁄180.
• Thus we get:
Length of arc of the sector = 216 × (π×25⁄180) = 30π cm
3. But this is same as the circumference of the base of the cone
• So if rb is the radius of the base of the cone, we can write:
2πrb = 30π ⟹ rb = 30⁄2 = 15 cm
4. The next step is to find the slant height. Imagine a triangle OO'P.
• It must satisfy the following conditions:
♦ Triangle OO'P must be right angled
♦ O must coincide with the apex
♦ O' must coincide with the base center
♦ P must be a point on the circumference of the base
• Then we get:
♦ OO' = height of the cone
♦ O'P = rb = 15 cm
♦ OP = slant height = 25 cm
• Applying Pythagoras theorem, we get:
OO' = √[(OP)2 - (O'P)2] = √[(25)2 - (15)2] = √[625 + 225] = √[400] = 20 cm
5. So volume of the cone = 1⁄3 × [πrb2× h] = 1⁄3 × [π × 152× 20] = 1500π cm3
Solved example 33.27
The base radii of two cones are in the ratio 3:5 and their heights are in the ratio 2:3. What is the ratio of their volumes?
Solution:
1. Given that rb1⁄rb2 = 3⁄5.
Also given that h1⁄h2 = 2⁄3.
2. We have: V1 ÷ V2 = {1⁄3 × [πrb12× h1]} ÷ {1⁄3 × [πrb22× h2]}
= {[rb12× h1]} ÷ {[rb22× h2]}
= {(rb1⁄rb2)2} × {h1⁄h2}
= {(3⁄5)2} × {2⁄3} = 6⁄25.
⟹ V1⁄V2 = 6⁄25
Solved example 33.28
Two cones have the same volume and their base radii are in the ratio 4:5. What is the ratio of their heights?
Solution:
1. Given that rb1⁄rb2 = 4⁄5.
Also given that V1 = V2
2. We have: V1 = V2 ⟹ {1⁄3 × [πrb12× h1]} = {1⁄3 × [πrb22× h2]}
⟹ {[rb12× h1]} = {[rb22× h2]}
⟹ {(rb1⁄rb2)2} = {h2⁄h1}
⟹ {(4⁄5)2} = {h2⁄h1} = 16⁄25.
⟹ h1⁄h2 = 25⁄16
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