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
♦ 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.
• 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.
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.
 |
| Fig.33.23 |
• 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
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