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:
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| Fig.33.24 |
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.
In the previous section we saw how a sector of a circle is rolled up to form a cone. We also saw some solved examples. In this section, we will learn about curved surface area of cones.
We have seen that the sector of a circle is rolled up to form a cone. So area of that sector will be the curved surface area of the cone. We have seen how to calculate the area of a sector in a previous chapter. Details here.
Let us see an example:
To make a conical hat of base radius 8 cm and slant height 30 cm, how much sq.cm of paper do we need?
Solution:
1. The two main properties of the sector:
(i) Given that slant height of the cone should be 30 cm.
• So radius of the sector rs = 30 cm
(ii) Central angle θ has to be calculated
2. Radius of the base rb is given as 8 cm
• So circumference of the base = 2πrb = 2π×8 = 16π cm
3. This circumference is equal to the length of the arc
For every 1o angle, the length of arc will be πrs⁄180. (Theorem 21.1)
• So for θo, the length of arc will be (θ × πrs⁄180)
• Thus we get:
Length of arc of the sector = [θ × (π×30⁄180)] = [θ × (π×1⁄6)] cm
4. We can equate the results in (2) and (3):
16π = [θ × (π×1⁄6)] ⟹ θ = 96o.
5. When we have the values for the two main properties of the sector, we can easily calculate it's area
• Every 1o central angle in a circle of radius rs will give a sector of area (πrs2⁄360) cm2. (Theorem 21.3)
• So for 96o, the area will be [96 × (πrs2⁄360)] = [96 × 302 × (π⁄360)] = [96 × 30 × (π⁄12)]
= [8 × 30 × π] = 240π cm2.
Another method:
1. We know that, every 1o central angle in a circle of radius rs will give a sector of area (πrs2⁄360) cm2.
2. So if the central angle is θ, the area of the sector will be [θ × (πrs2⁄360)] cm2.
3. We know that length of arc of the sector is (θ × πrs⁄180) cm
• But length of arc of the sector is the circumference of the base of the cone, which is 2πrb.
4. So we can write: 2πrb = (θ × πrs⁄180)
⟹ θ = 360rb⁄rs.
5. Substituting this value of in (2), we get:
• Area of the sector = [(360rb⁄rs) × (πrs2⁄360)] = πrbrs.
♦ But 'area of the sector' is the 'curved surface area of the cone'
♦ And rs is the slant height l
■ So Area of the sector = Curved surface area of the cone = πrbl cm2
Height of a cone
• We have seen that a sector can be completely defined by two properties
♦ It's radius
♦ It's central angle
• In the case of square pyramids, we saw that, it can be completely defined by two properties:
♦ It's height
♦ It's base edge
• In a similar way, a cone can be completely defined by two properties:
♦ It's height
♦ It's base radius
■ That is., if we know the height and base radius of a cone, we will be able to calculate all other properties:
Slant height, circumference of the base, central angle, surface area and volume
Let us see how it is done:
Fig.33.23(a) below shows a cone.
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| Fig.33.23 |
• The apex is marked as O. Center of it's base is marked as O'. So OO' is the height of the cone.
• Now mark any point P on the circumference of the base. Draw O'P and OP
• Obviously, O'P will be rb and OP will be the slant height
• Also OO'P will be a right angled triangle. So we can use Pythagoras theorem to find unknown quantities
An example:
In a cone, the height is 10 cm and base radius is 5 cm. Find the height of the cone
Solution:
1. 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 = 10 cm
♦ O'P = rb = 5 cm
♦ OP = slant height
2. Applying Pythagoras theorem, we get:
OP = √[(O'P)2 + (OO')2] = √[(5)2 + (10)2] = √[25 + 100] = √[125] = 5√5 cm
Now we will see some solved examples
Solved example 33.20
What is the area of the curved surface of a cone of base radius 12 cm and slant height 25 cm
Solution:
1. The two main properties of the sector:
(i) Given that slant height of the cone should be 25 cm.
• So radius of the sector rs = 25 cm
(ii) Central angle θ has to be calculated
2. Radius of the base rb is given as 12 cm
• So circumference of the base = 2πrb = 2π×12 = 24π cm
3. This circumference is equal to the length of the arc
For every 1o angle, the length of arc will be πrs⁄180. (Theorem 21.1)
• So for θo, the length of arc will be (θ × πrs⁄180)
• Thus we get:
Length of arc of the sector = [θ × (π×25⁄180)] = [θ × (π×5⁄36)] cm
4. We can equate the results in (2) and (3):
24π = [θ × (π×5⁄36)] ⟹ θ = (864⁄5)o
5. When we have the values for the two main properties of the sector, we can easily calculate it's area
• Every 1o central angle in a circle of radius rs will give a sector of area (πr2⁄360) cm2. (Theorem 21.3)
• So for (864⁄5)o, the area will be [(864⁄5) × (πrs2⁄360)] = [(864⁄5) × 252 × (π⁄360)]
= [864 × 125 × (π⁄360)] = 300π cm2.
Another method using equation:
1. We have: Curved surface area of a cone = πrbl cm2.
2. Substituting the values we get:
Curved surface area of a cone = π×12×25 = 300π cm2.
Solved example 33.21
What is the surface area of a cone of base diameter 30 cm and height 40 cm?
Solution:
1. The first 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 = 40 cm
♦ O'P = rb = 15 cm (∵ diameter = 30 cm)
♦ OP = l = slant height
• Applying Pythagoras theorem, we get:
OP = l = √[(O'P)2 + (OO')2] = √[(15)2 + (40)2] = √[225 + 1600] = √[1825] cm
2. Curved surface area = πrbl = π×15×√[1825] = 640.8π cm2.
3. Surface area of base = πrb2 = π×152= 225π cm2.
4. Total surface area = (640.8+225)π = 865.8π cm2
Solved example 33.22
A conical fire work is of base diameter 10 cm and height 12 cm. 10000 such fire works are to be wrapped in colour paper. The price of colour paper is 2 rupees per sq.m. What is the total cost?
Solution:
1. The first 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 = 12 cm
♦ O'P = rb = 5 cm (∵ diameter = 10 cm)
♦ OP = slant height
• Applying Pythagoras theorem, we get:
OP = l = √[(O'P)2 + (OO')2] = √[(5)2 + (12)2] = √[25 + 144] = √[169] = 13 cm
2. Curved surface area = πrbl = π×5×13 = 65π cm2.
3. Surface area of base = πrb2 = π×52= 25π cm2
4. Total surface area = (65+25)π = 90π cm2 = 90×3.14 = 282.6 cm2
5. Surface area of 10000 fire works = 282.6 × 10000 = 2826000 cm2.
6. 2826000 cm2 = 2826000⁄10000 m2 = 282.6 m2.
7. So cost of colour paper = 282.6 × 2 = Rs 565.20
Solved example 33.23
Prove that for a cone made by rolling up a semicircle, the area of the curved surface is twice the base area
Solution:
1. The two main properties of the sector:
(i) Let the radius of the sector be rs
(ii) Central angle θ is 180o (∵ the sector is a semicircle)
2. Let the radius of the base be rb
• So circumference of the base = 2πrb
3. This circumference is equal to the length of the arc
• We do not need to calculate the length of the arc. The arc length of a semicircle is 'half the circumference of the full circle'
• The 'circumference of the full circle' is 2πrs. So half of it is πrs.
4. We can equate the results in (2) and (3):
2πrb = πrs ⟹ 2rb = rs.
5. Now we want the area of the sector
• But the area of the sector is area of the semicircle which is 1⁄2 × πrs2
• Let us substitute for rs using the result in (4). We get:
• Area of the sector = 1⁄2 × πrs2 = 1⁄2 × π(2rb)2 = 1⁄2 × π × 4 × rb2 = 2πrb2
6. Area of the sector is same as the area of curved surface. So we can write:
Area of the curved surface of the cone = 2πrb2.
7. Now we calculate the base area.
• We have radius of the base of the cone = rb.
• So area of the base of the cone = πrb2.
8. Comparing the results in (6) and (7), we get:
Area of the curved surface of the cone = Twice the base area
In the next section, we will see volume of cone.
In the previous section we saw volume of square pyramids. In this section, we will learn about cones.
In the first section of this chapter we saw how a square prism can be transformed into a square pyramid. See fig.33.2. In the same way, a cylinder can be transformed into a cone. This is shown in fig.33.20 below:
 |
| Fig.33.20 |
• All the points on the circumference of the top circle of a cylinder converge onto a point on the axis. Then we get the cone in fig.b
• Fig.33.21 below shows some possible cones.
 |
| Fig.33.21 |
• For all cones, the base will be a circle and there will be an apex.
Now let us see how a cone can be made:
Consider the cone in fig.33.22(a) below.
 |
| Fig.33.22 |
1. The apex is marked as O. Mark any point P on the circumference of the base of the cone. Draw OP
2. Make a cut through the line OP. The single line OP will become two lines: OP and OP'. This is shown in fig.b
3. The cone can thus be spread out and laid flat on a plane surface. This is shown in fig.c
4. When laid flat, the cone will become a sector OPP' of a circle.
• We have already learned about sectors in an earlier chapter. Details here.
5. If we know the central angle θ, and radius of a sector, we can completely define a sector.
• The radius of the sector will be the slant height of the cone. It is usually represented by the letter 'l'
• The arc length P'P of the sector will be the circumference of the base of the cone
♦ So if rb is the 'radius of the base of the cone', the circumference of the base will be 2πrb
♦ And we can write: PP' = 2πrb
Let us see an example:
From a circle of radius 12 cm, a sector of central angle 45o is cut out and made into a cone. What is the slant height and base radius of the cone?
Solution:
• Let us write the two important properties of the sector:
(i) Radius is already given as 12 cm.
♦ To avoid confusion with the radius of the base of the cone, let us denote it as rs
(ii) Central angle is already given as 45o
• The third property is the 'arc length'. That we can find using the given radius and central angle. The steps are given below:
1. For every 1o angle, the length of arc will be πrs⁄180. (Theorem 21.1)
• So for 45o, the length of arc will be 45 × πrs⁄180.
• Thus we get:
Length of arc of the sector = 45 × (π×12⁄180) = 3π cm
2. 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 = 3π ⟹ rb = 3⁄2 = 1.5 cm
Another example:
How do we make a cone of base radius 5 cm and slant height 15 cm?
Solution:
• This is a sort of 'reverse' of the previous example
• To make a cone, we need a sector of a circle. Let us try to write the two important properties of the required sector:
(i) The radius of the sector can be straight away written as 15 cm. Because, that radius will become the slant height of the cone
♦ To avoid confusion with the radius of the base of the cone, let us denote it as rs.
(ii) The central angle is not given. We have to find it. The steps are given below:
1. Base radius of the cone = rb = 5 cm
So circumference of the base = 2πrb = 2π×5 = 10π cm
2. But this circumference is the arc length of the sector
Let θ be the central angle.
For every 1o angle, the length of arc will be πrs⁄180. (Theorem 21.1)
• So for θo, the length of arc will be (θ × πrs⁄180)
• Thus we get:
Length of arc of the sector = [θ × (π×15⁄180)] = [θ × (π⁄12)] cm
3. We can equate the results in (1) and (2):
10π = [θ × (π⁄12)] ⟹ θ = 120o
4. Now we have all the details.
• To make a cone of base radius 5 cm and slant height 15 cm:
From a circular sheet of 12 cm radius, cut out a sector with central angle 120o
Now we will see some solved examples
Solved example 33.17
What are the radius of the base and slant height of a cone made by rolling up a sector of central angle 60o cut out from a circle of radius 10 cm?
Solution:
1. The two main properties of the sector:
(i) Given that radius of the circle is 10 cm. This will be same as the radius of the sector. So we can write: rs = 10 cm
• This rs will be the slant height l of the cone. So we can write: Slant height l = 10 cm
(ii) Central angle θ = 60o
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 60o, the length of arc will be 60 × πrs⁄180.
• Thus we get:
Length of arc of the sector = 60 × (π×10⁄180) = π×10⁄3 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 = π×10⁄3 ⟹ rb = 10⁄6 = 1.67 cm
Solved example 33.18
What is the central angle of the sector to be used to make a cone of base radius 10 cm and slant height 25 cm?
Solution:
1. The two main properties of the sector:
(i) Given that slant height of the cone should be 25 cm.
• So radius of the sector rs = 25 cm
(ii) Central angle θ has to be calculated
2. Radius of the base rb is given as 10 cm
• So circumference of the base = 2πrb = 2π×10 = 20π cm
3. This circumference is equal to the length of the arc
For every 1o angle, the length of arc will be πrs⁄180. (Theorem 21.1)
• So for θo, the length of arc will be (θ × πrs⁄180)
• Thus we get:
Length of arc of the sector = [θ × (π×25⁄180)] = [θ × (π×5⁄36)] cm
4. We can equate the results in (2) and (3):
20π = [θ × (π×5⁄36)] ⟹ θ = 144o
Solved example 33.19
What is the ratio of the base radius and slant height of a cone made by rolling up a semicircle?
Solution:
1. The two main properties of the sector:
In this problem, the sector is a semicircle
(i) Let rs be the radius of the sector.
• Then slant height of the cone will be rs.
(ii) Central angle θ of a semi circle = 180o
2. From the central angle we can calculate length of arc of the sector.
• But we do not need to calculate it. The arc length of a semicircle is 'half the circumference of the full circle'
• The 'circumference of the full circle' is 2πrs. So half of it is πrs.
3. 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 = πrs ⟹ rb⁄rs = 1⁄2
• But rs is the slant height. So we can write:
base radius⁄slant height = 1⁄2
In the next section, we will see surface area of cone.
