Chapter 33.4 - Height and Curved surface area of a Cone

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 r_s = 30 cm
(ii) Central angle θ has to be calculated
2. Radius of the base r_b is given as 8 cm
• So circumference of the base = 2πr_b = 2π×8 = 16π cm
3. This circumference is equal to the length of the arc
For every 1^o angle, the length of arc will be ^πr_s⁄₁₈₀. (Theorem 21.1)
• So for θ^o, the length of arc will be (θ × ^πr_s⁄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 = 96^o.
5. When we have the values for the two main properties of the sector, we can easily calculate it's area 
• Every 1^o central angle in a circle of radius r_s will give a sector of area (^πr_s2⁄₃₆₀) cm². (Theorem 21.3)
• So for 96^o, the area will be [96 × (^πr_s2⁄₃₆₀)] = [96 × 30² × (^π⁄₃₆₀)] = [96 × 30 × (^π⁄₁₂)] 
= [8 × 30 × π] = 240π cm².

Another method:
1. We know that,  every 1^o central angle in a circle of radius r_s will give a sector of area (^πr_s2⁄₃₆₀) cm².
2. So if the central angle is θ, the area of the sector will be [θ × (^πr_s2⁄₃₆₀)] cm².
3. We know that length of arc of the sector is (θ × ^πr_s⁄₁₈₀) cm 
• But length of arc of the sector is the circumference of the base of the cone, which is 2πr_b.
4. So we can write:  2πr_b = (θ × ^πr_s⁄₁₈₀)
⟹ θ =  ^360r_b⁄_rs.
5. Substituting this value of in (2), we get:
• Area of the sector = [(^360r_b⁄_rs) × (^πr_s2⁄₃₆₀)] = πr_br_s.
    ♦ But 'area of the sector' is the 'curved surface area of the cone'
    ♦ And r_s is the slant height l
■ So Area of the sector = Curved surface area of the cone =  πr_bl 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

• 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 r_b 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 = r_b = 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

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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 r_s = 25 cm
(ii) Central angle θ has to be calculated
2. Radius of the base r_b is given as 12 cm
• So circumference of the base = 2πr_b = 2π×12 = 24π cm
3. This circumference is equal to the length of the arc
For every 1^o angle, the length of arc will be ^πr_s⁄₁₈₀. (Theorem 21.1)
• So for θ^o, the length of arc will be (θ × ^πr_s⁄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 1^o central angle in a circle of radius r_s will give a sector of area (^πr2⁄₃₆₀) cm². (Theorem 21.3)
• So for (⁸⁶⁴⁄₅)^o, the area will be [(⁸⁶⁴⁄₅) × (^πr_s2⁄₃₆₀)] = [(⁸⁶⁴⁄₅) × 25² × (^π⁄₃₆₀)] 
= [864 × 125 × (^π⁄₃₆₀)] = 300π cm².

Another method using equation:
1. We have: Curved surface area of a cone = πr_bl 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 = r_b = 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 = πr_bl = π×15×√[1825] = 640.8π cm².
3. Surface area of base = πr_b² = 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 = r_b = 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 = πr_bl = π×5×13 = 65π cm².
3. Surface area of base = πr_b² = 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 r_s
(ii) Central angle θ is 180^o (∵ the sector is a semicircle)
2. Let the radius of the base be r_b
• So circumference of the base = 2πr_b 
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 2πr_s. So half of it is πr_s.
4. We can equate the results in (2) and (3):
2πr_b  = πr_s ⟹ 2r_b  = r_s.
5. Now we want the area of the sector
• But the area of the sector is area of the semicircle which is ¹⁄₂ × πr_s²
• Let us substitute for r_s using the result in (4). We get:
• Area of the sector = ¹⁄₂ × πr_s² = ¹⁄₂ × π(2r_b)² = ¹⁄₂ × π × 4 × r_b² =  2πr_b²
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 = 2πr_b².
7. Now we calculate the base area. 
• We have radius of the base of the cone = r_b.
• So area of the base of the cone = πr_b².
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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In the next section, we will see volume of cone.

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Chapter 33.3 - Cone from Sector of a circle

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.

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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 r_b is the 'radius of the base of the cone', the circumference of the base will be 2πr_b
    ♦ And we can write: PP' = 2πr_b

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Let us see an example:
From a circle of radius 12 cm, a sector of central angle 45^o 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 r_s
(ii) Central angle is already given as 45^o
• 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 1^o angle, the length of arc will be ^πr_s⁄₁₈₀. (Theorem 21.1)
• So for 45^o, the length of arc will be 45 × ^πr_s⁄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 r_b is the radius of the base of the cone, we can write:
2πr_b = 3π ⟹ r_b = 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 r_s.
(ii) The central angle is not given. We have to find it. The steps are given below:
1. Base radius of the cone = r_b = 5 cm
So circumference of the base = 2πr_b = 2π×5 = 10π cm
2. But this circumference is the arc length of the sector
Let θ be the central angle.
For every 1^o angle, the length of arc will be ^πr_s⁄₁₈₀. (Theorem 21.1)
• So for θ^o, the length of arc will be (θ × ^πr_s⁄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 = 120^o
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 120^o

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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 60^o 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: r_s = 10 cm
• This r_s will be the slant height l of the cone. So we can write: Slant height l = 10 cm
(ii) Central angle θ = 60^o
2. From the central angle we can calculate length of arc of the sector:
• For every 1^o angle, the length of arc will be ^πr_s⁄₁₈₀. (Theorem 21.1)
• So for 60^o, the length of arc will be 60 × ^πr_s⁄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 r_b is the radius of the base of the cone, we can write:
2πr_b = ^π×10⁄₃ ⟹ r_b = 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 r_s = 25 cm
(ii) Central angle θ has to be calculated
2. Radius of the base r_b is given as 10 cm
• So circumference of the base = 2πr_b = 2π×10 = 20π cm
3. This circumference is equal to the length of the arc
For every 1^o angle, the length of arc will be ^πr_s⁄₁₈₀. (Theorem 21.1)
• So for θ^o, the length of arc will be (θ × ^πr_s⁄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 = 144^o

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 r_s be the radius of the sector.
• Then slant height of the cone will be r_s.
(ii) Central angle θ of a semi circle = 180^o
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 2πr_s. So half of it is πr_s.
3. This is same as the circumference of the base of the cone 
• So if r_b is the radius of the base of the cone, we can write:
2πr_b = πr_s ⟹ ^r_b⁄_rs = ¹⁄₂
• But rs is the slant height. So we can write:
^{base radius}⁄_{slant height} = ¹⁄₂

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In the next section, we will see surface area of cone.

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Chapter 32.3 - Tangent and Chord

In the previous section we saw tangents giving cyclic quadrilaterals. In this section we will learn the relations between tangents and chords.

1. A circle is drawn with center at O. See fig.32.28(a)

Fig.32.28

2. Two green radial lines OP and OQ are drawn in such a way that, ∠POQ = 100
3. A red tangent is drawn at P. Another red tangent is drawn at Q
4. The red tangents meet at T. So the tangents are named as AT and BT
5. A magenta line is drawn joining P and Q. So the magenta line is a chord
• Clearly, the central angle of the chord is 100^o  
6. We know that ∠PTQ will be equal to 80^o [Theorem 32.5]
• This is shown in fig(b)
7. We also know that PT = QT. [Theorem 32.2]
8. So ΔPTQ is an isosceles triangle. It's base angles are equal. Let them be x
9. So in ΔPTQ, we get: (2x + 80) = 180^o ⟹ x = 50^o
10. But this 50^o is half of the central angle (made by the chord PQ) 100^o 

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Let us see if this is true for any chord:
1. A circle is drawn with center at O. See fig.32.29(a) 

Fig.32.29

2. Two green radial lines OP and OQ are drawn in such a way that, ∠POQ = c^o
3. A red tangent is drawn at P. Another red tangent is drawn at Q
4. The red tangents meet at T. So the tangents are named as AT and BT
5. A magenta line is drawn joining P and Q. So the magenta line is a chord
• Clearly, the central angle of the chord is c^o
6. We know that ∠PTQ will be equal to (180-c)^o  [Theorem 32.5]
7. We also know that PT = QT. [Theorem 32.2]
8. So ΔPTQ is an isosceles triangle. It's base angles are equal. Let them be x^o
9. So in ΔPTQ, we get: [2x + (180-c)] = 180^o ⟹ (2x - c)  = 0 ⟹ x = (^c⁄₂)^o

---

We can write the above result as a theorem:
Theorem 32.6
• P and Q are two points on the circle
• A tangent is drawn at P
    ♦ The angle between this tangent and the chord PQ is half of the central angle of the chord
• A tangent is drawn at Q
    ♦ The angle between this tangent and the chord PQ is also half of the central angle of the chord

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Now we will see a more interesting case:
1. The chord PQ in fig.32.29 above, divides the circle into two arcs: 
• A minor arc PQ and a major arc PQ
2. Consider any point R on the major arc PQ. This is shown in fig.32.30(a)

Fig.32.30

• The chord PQ will subtend  ∠PRQ on that major arc. 
3. We know that ∠PRQ will be equal to half the central angle of the chord PQ. [Theorem 27.4] 
• So we get: ∠PRQ = (^c⁄₂)^o
4. But based on the theorem 32.6 that we wrote above, the angle between the chord and the tangent is also (^c⁄₂)^o
5. Thus we get an interesting result: ∠PRQ = ∠TPQ = ∠TQP

Let us see some examples:
1. In fig.32.30(b), PQ is a chord. 
• Tangent is drawn at P
• Tangent is drawn at Q
• The two tangents meet at T
2. The meeting point T is on the right side of the chord
• We will call the right side as 'Tangent side of the chord'
• So the left side of the chord can be called: 'Non-tangent side of the chord'
3. A point R is marked on the circle on the 'Non-tangent side of the chord'
■ Wherever we mark R on the major arc, the ∠PRQ will be the same. [Details here]
• In our present case, it is 65^o. 
4. Then, from what we have seen based on fig.32.30(a), we can directly write:
■ The angle between the chord and tangent will be 65^o at both ends of the chord
• That is., ∠TPQ and ∠TQP will be equal to 65
• We do not need to know the position of the center 'O' for writing the above result
5. Note that, ∠TPQ and ∠TQP are the angles on the Tangent side of the chord PQ
• Which are the angles on the Non-tangent side?
They are: ∠APQ and ∠BQP
6. What are their values?
• We know that ∠APQ and ∠TPQ form a linear pair
• So ∠APQ = (180 - ∠TPQ) = (180 - 65) = 115^o. This is shown in fig (c) 
• Similarly, ∠BQP = 1(80 - ∠TQP) = (180 - 65) = 115^o
7. Now consider any point S on the circle on the tangent side of the chord
• PRQS is a cyclic quadrilateral. So ∠PSQ = (180 - ∠PRQ) = (180 - 65) = 115^o. This is shown in fig(c)
• Note that, in figs.(b) and (c), we do not know where the center of the circle is. Even then we are able to find various angles

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We will write the above findings as a theorem:
Theorem 32.7
1. A chord PQ subtends ∠PSQ  at a point S on the circle, on the Tangent side
• At the end P, the chord makes ∠APQ (with the tangent AT) on the Non-tangent side 
• At the end Q, the chord makes ∠BQP (with the tangent BT) on the Non-tangent side 
■ All the above three angles are equal
2. The chord PQ subtends ∠PRQ  at a point R on the Non-tangent side
• At the end P, the chord makes ∠QPT (with the tangent AT) on the Tangent side 
• At the end Q, the chord makes ∠TQP (with the tangent BT) on the Tangent side 
■ All the above three angles are equal
• In (1), point S is on the Tangent side. The chord angles are on the Non-tangent side
• In (2), point R is on the Non-tangent side. The chord angles are on the Tangent side

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Now we will see the practical application of the above theorem 32.7
We know how to draw the tangent at a given point on a circle. 
1. Draw a radial line through the given point
2. Draw a line perpendicular to the radial line through the given point. This line is the required tangent.
• So the procedure is simple. But it will not be so simple if the centre of the circle is not known. Because, we will not be able to draw the radial line.
• In such a situation, we can use the above theorem 32.7. Let us see the method:
• In fig.32.31(a) below, a circle is shown and a point P is marked on it. Center of the circle is not given. We are required to draw the tangent at P. 

Fig.32.31

We can use the following procedure:
1. From the point P, draw a convenient chord PQ. This is shown in fig(b)
2. Mark a convenient point R on the circle on the Non-tangent side. Complete the triangle PQR
3. Measure ∠PRQ. Let it be x^o
4. Draw a red line through P in such a way that the angle (at P) between the red line and the chord is x^o. This is shown in fig(c)
5. Then the red line is the required tangent
An easier method:
• In any case, we will need to draw PQR. This is to obtain the value of x
1. We will draw PQR in such a way that, it is isosceles. For that, with P as centre, draw two arcs cutting the circle at Q and R. This is shown in fig.32.32(a) below:

Fig.30.32

2. Now complete ΔPQR. It is an isosceles triangle because PR = PQ. 
• The base angles at R and Q will be the same.
3. Draw a red line through P in such a way that it is parallel to RQ. Then we have:
• RQ and the red line are two parallel lines. They are cut by a transversal PQ
• So the following two angles are co-interior angles and hence will be equal:
    ♦ ∠RQP
    ♦ The angle (at P) between the red line and PQ
4. So we get:
• ∠PRQ = x = ∠RQP = The angle (at P) between the red line and PQ
• So the angle at P is also x. This is shown in fig c. 
■ Thus the red line is the tangent at P

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An actual construction can be seen in the form of a video presentation here.

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In the next section, we will see some solved examples.

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