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:
• 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.
• 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.
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.
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 |
• Fig.33.21 below shows some possible cones.
 |
| Fig.33.21 |
Now let us see how a cone can be made:
Consider the cone in fig.33.22(a) below.
 |
| Fig.33.22 |
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
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