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 27.6 - Segments and Angles of a Circle

In the previous section we saw theorem 27.5 and 27.6. We also saw some solved examples. In this section we will learn about Segments.

• Consider the chord AB in fig.27.36(a) below. It divides the circle into two arcs. Arc AXB and AYB.

Fig.27.36

• Consider the arc AXB and chord AB. We get an area enclosed between them. This area is a segment of the circle. 
• There is another enclosed area also present. It is between the chord AB and arc AYB. It is also a segment. 
• So we can define a segment of a circle as follows:
■ A segment of a circle is the area enclosed between a chord and it’s arc.
• The difference between a segment and a sector can be seen from figs.27.36 b and c. We have earlier seen the details about sectors in chapter 21.8.
■ Just as every arc have an alternate arc, every segment will have an alternate segment. In the fig.27.36(b), segment AYB is the alternate segment of segment AXB and vice versa.

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Now we will learn about the angles inside a segment. 

1. Consider fig.27.37(a) below. The chord AB makes two arcs AXB and AYB.

Fig.27.37

2. Some angles are drawn inside the sector AYB. Let us find the values of those angles:

Using theorem 27.4,

∠APB will be equal to half the central angle of arc AXB
∠AQB will be equal to half the central angle of arc AXB
∠ARB will be equal to half the central angle of arc AXB
3. We find that all three angles are equal to half the central angle of arc AXB
• That means, all the three angles are equal
■ In fact, what ever number of angles we draw in the sector AYB, they will all have the same value
1. Now consider fig(b). 
2. Some angles are drawn inside the sector AXB. Let us find the values of those angles:
Using theorem 27.4,
∠AKB will be equal to half the central angle of arc AYB
∠ALB will be equal to half the central angle of arc AYB
∠AMB will be equal to half the central angle of arc AYB
3. We find that all three angles are equal to half the central angle of arc AYB
• That means, all the three angles are equal
■ In fact, what ever number of angles we draw in the sector AXB, they will all have the same value

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Based on the above results, we can write the following theorem:

Theorem 27.7:
1. Consider a segment and its chord AB
2. Mark any number of points on the arc of the segment
3. Join all those points to A and B
4. All the angles so formed will be equal
5. They are all equal to 'half of the central angle of the alternate arc' 
■ So we can say this:
• Every segment will have a particular value of angle.
• We will call it the unique angle of a segment
• The unique angle of any segment is equal to 'half of the central angle of the alternate arc'

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Now we will see the relationship between two unique angles

1. Consider segment AXB in fig.27.38(a) below:

Fig.27.38

2. An ∠AMB is drawn inside the segment. We know that, where ever be the position of M along the arc AXB, The ∠AMB will be equal to the unique angle of the segment AXB. 

3. Also, by theorem 27.4, ∠AMB will all be equal to half the central angle of the alternate arc AYB. 

4. That is., the unique angle of segment AXB will be equal to ^d⁄₂ as shown in fig(b)

5. In a similar way, the unique angle of segment AYB will be equal to ^c⁄₂ 
6. Let us find the sum of ^c⁄₂ and ^d⁄₂:
We get: ^c⁄₂ + ^d⁄₂ = ^(c+d)⁄₂
7. But (c+d) is the total angle at centre O. The total angle at any point is 360^o. 
That means, (c + d) = 360^o.
8. So we can write: ^c⁄₂ + ^d⁄₂ = ^(c+d)⁄₂ = ³⁶⁰⁄₂ = 180^o
• That means the two angles are supplementary

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We can write the above result in the form of a theorem:

Theorem.27.8:
1. Consider any segment. It will have an unique angle
2. The alternate segment will also have it's own unique angle
3. The sum of the two unique angles will always be 180^o
4. That is., the unique angles of alternate segments are always supplementary.

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So far, we know the following two points:
• Every segment has it's own unique angle
• This unique angle is equal to 'half of the central angle of the alternate arc'      
■ That means, to find the unique angle, we must first know the 'central angle of the alternate arc'.
Would it not be better, if we do not need to know the 'central angle of the alternate arc'?
Let us try:
1. Consider the segment AYB in fig.27.39(a) below. Mark a point P any where on the arc AYB

Fig.27.39

2. We have seen that, wherever we mark P, the angle APB will be the same, and it is the unique angle of the segment AYB. Let us denote this unique angle as u^o. 
3. In fig(b), the central angle of 'it's own arc', the arc AYB is marked as d^o.
• We want the relation between u and d
4. Let the central angle of the alternate arc AXB be c^o. Then, by converse of theorem 27.4, we get:
c = 2u. This is shown in fig(c).
5. Now, c and d are angles around a point. So there sum will be 360^o
• Thus we get: c+d = 360 ⇒ d = 360 – c ⇒ d = 360 – 2u
• This is the required relation.
■ That is., if we know the unique angle (u^o) of a segment, the 'central angle of the arc of that segment' will be equal to (360 -2u)^o 

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Let us see if this is true for the other arc also:

1. Consider the segment AXB in fig.27.40(a) below. Mark a point M any where on the arc AXB

Fig.27.40

2. We have seen that, wherever we mark M, the angle AMB will be the same, and it is the unique angle of the segment AMB. Let us denote this unique angle as v^o. 
3. In fig(b), the central angle of 'it's own arc', the arc AXB is marked as c^o.
• We want the relation between v and c
4. Let the central angle of the alternate arc AYB be c^o. Then, by converse of theorem 27.4, we get:
d = 2v. This is shown in fig(c).
5. Now, c and d are angles around a point. So there sum will be 360^o
• Thus we get: c+d = 360 ⇒ c = 360 – d ⇒ c = 360 – 2v
• This is the required relation.
■ That is., if we know the unique angle (v^o) of a segment, the 'central angle of the arc of that segment' will be equal to (360 -2v)^o 

• So we proved the relation in both cases

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Now we will see a solved example

Solved example 27.14

In the fig.27.41(a) below, ΔABC is equilateral and O is it's circumcentre. 

Fig.27.41

Prove that, the length of AD is equal to the radius of the circle.

Solution:
• In the given problem, ABC is an equilateral triangle. And O is the circumcentre of that triangle
• O is joined to vertex C. So OC is a radius
• This OC is extended until it meets the circle at D. So CD is a diameter and OD is a radius
• Point D is then joined to A. 
• We have to prove that, AD is equal to the radius of the circle
1. Consider the three segments: ADB, BYC and CXA. [fig.27.41(b)]
• Since ΔABC is equilateral, AB = BC = CA
• That means, the three segments are equal
• So they have the same central angle   
2. Join A and B to O. (Note that C is already joined to O)
• OA, OB and OC shows the central angles of the three segments
• It is clear that each of the three segments have a ¹⁄₃ share at the centre
• So ∠AOB = ∠BOC = ∠COA = ¹⁄₃ × 360 = 120^o.
3. Now consider fig(c)
• B is joined to D by a red dotted line. This gives us ∠ABD
• This ∠ABD is the unique angle of segment ABD. The segment ABD has a central angle of 120^o.
• So 120 = [360 - (2 ×∠ABD)] ⇒ 2×∠ABD = 360-120  ⇒ 2×∠ABD = 240 ⇒ ∠ABD = 120^o.
4. AB is a chord and OD is a radius. So by theorem 17.1, OD is the perpendicular bisector of chord AB.
So we get: ∠BDO = ∠ADO = ¹⁄₂×120 = 60^o.
5. Now consider ΔOAD
• We have OA = OD (∵ radii of same circle)
• So ΔOAD is isosceles. The base angles are equal. 
• From (4) we already have one base angle ∠ADO = 60^o
• So the other base angle ∠DAO = 60^o
• If these two angles are 60^o each, the third angle ∠DOA is also 60^o. (∵ 180 -60 -60 = 60)    
• So we find that all the three angles in OAD is 60^o
• It is an equilateral triangle. All three sides are equal
• Thus we get AD = AO

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Some additional solved examples can be seen here.

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In the next section, we will see Cyclic quadrilaterals.

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Chapter 21.10 - Solved examples on Circles, Arcs and Sectors

In the previous section we completed the discussion on Area of sectors. In this section we will see some solved examples related to this chapter in general.

Solved example 21.25
Perimeter of two circles are in the ratio 2:3
(i) What is the ratio of their radii?
(ii) What is the ratio of their areas?
Solution:
1. Let the radii of the two circles be r₁ and r₂. Then the perimeters are 2πr₁ and 2πr₂.
2. Given that 2πr₁ : 2πr₂ = 2:3
3. Dividing both sides by 2π we get r₁ : r₂ = 2:3 [This is the answer for part(i)]
4. Ratio of areas = πr₁² : πr₂² 
5. Dividing both sides of the ratio by π, we get: 
Ratio of areas = r₁² : r₂²  
6. From (3) we have: ^r₁⁄_r2 = ²⁄₃ ⇒ r₁ = ^2r₂⁄₃ . Substituting this value of r₁ in (5), we get:
7. Ratio of areas = (^2r₂⁄₃)² : r₂²
8. Dividing both sides of the ratio by r₂² we get:
Ratio of areas = ⁴⁄₉ :1 ⇒ Ratio of areas = 4:9 [This is the answer for part(ii)]

Solved example 21.26
A wheel of radius 20 cm rolls along. How much would it move ahead after 10 rotations?
Solution:
Consider the circle in fig.21.43 below. The circle represents a wheel

Fig.21.43

1. Position (1) shows the position when the wheel is just about to roll to the right. At that moment, the point of contact of the wheel with the ground is marked as 'A'. 
2. When the wheel rolls, point A moves along the circumference of the wheel. When A touches the ground again, one rotation is completed. This is shown as position (2)
3. So the distance travelled after one rotation is equal to the perimeter of the wheel
4. So the distance travelled after 10 rotations = 10 × perimeter = 10 × 2πr = 20πr
5. In this problem, r is given as 20 cm. So the distance travelled = 20 × 3.14 × 20 = 1256 cm

Solved example 21.27
Two semicircular pieces are cut off from a rectangular sheet as shown in the fig.21.44 below.

Fig.21.44

 Find the area of the remaining part
Solution:
1. Total area = area of the rectangle = 50 × 20 = 1000 cm².
2. Consider any one semicircular part. It reaches from the top edge to bottom edge of the rectangle.
3. So the diameter of the semicircle = height of the rectangle = 20 cm
4. So radius = 10 cm
5. Area of the semicircular part = half the area of a circle with 10 cm radius
= ¹⁄₂ × πr² = ¹⁄₂ × π × 10² = 50π cm²
6. There are two semicircular parts. Their total area = 2 × 50π = 100π cm².
7. So area of the remaining part = 1000 - 100π = 1000 - (100×3.14) = 1000 - 314 = 686 cm²

Solved example 21.28
The length of an arc of a circle is 4 cm. What is the length of an arc with the same central angle, in a circle of double the radius?
Solution:
1. We are given the length of an arc. No other details are given. So let us assume the required details:
Let the radius be r, and let the central angle be x
2. So length of arc will be equal to (^πr⁄₁₈₀ × x) cm
3. But the length of arc is given as 4 cm. So we can write:
(^πr⁄₁₈₀ × x) = 4 
4. Now we are given another arc. It has the same central angle. But radius = 2r. We want the length of this new arc
5. We can write: length of new arc = (^π(2r)⁄₁₈₀ × x)
⇒ length of new arc = 2 × (^πr⁄₁₈₀ × x)
6. But from (3), the quantity inside brackets is 4 cm. So we get:
7. length of new arc = 2 × 4 = 8 cm

Solved example 21.29
In the fig.21.45 below, the sector OAB has a radius of 6 cm, and central angle 60^o.

Fig.21.45

Find the area of the blue region
Solution:
1. If a sector has central angle 60^o, we can separate out an equilateral triangle from it.

2. Area after the removal of equilateral triangle from a sector of radius 'r' and central angle 60^o

= [(^π⁄₆) -  (^√3⁄₄)]r²  cm². Details here.

3. In our problem, r = 6 cm. So we get:

Area = [(^π⁄₆) -  (^√3⁄₄)]6² =  [6π - 9√3] cm²

Solved example 21.30
The area of a circular table is 31400 cm². What is it's radius? And it's perimeter?
Solution:
1. Area of a circle = πr² = 3.14 × r² = 31400
⇒ r² = ³¹⁴⁰⁰⁄_3.14 = 10000 ⇒ r = √10000 = 100 cm
2. Perimeter = 2πr = 2 × 3.14 × 100 = 628 cm 

Solved example 21.31
In fig.21.46 below, OPQR is a sector of a circle with centre O and radius 8 cm.

Fig.21.46

Find the area of the yellow region.
Solution:
1. The central angle of the sector OPQR is given as 90^o. So this sector will cover one fourth of the area of the full circle. We can write:
Area of sector OPQR = ¹⁄₄ × πr² = ¹⁄₄ × π × 8² = 16π cm².
2. Now, OPR is a triangle. It is a right triangle. We have it's base and height. So area of ΔOPR
= ¹⁄₂ × b × h = ¹⁄₂ × 8 × 8 = 32 cm².
3. So area of the yellow region = (16π - 32) = 16(π - 2) cm².


Solved example 21.32
The length of the pendulum of a clock is 12 cm. It makes an angle of 45^o in one swing. What is the distance travelled by the tip of the pendulum in on swing?
Solution:
1. Fig.21.47 below shows one swing of the pendulum

Fig.21.47

2. In one swing, the pendulum travels from P to Q. It is an arc of radius 12 cm, and central angle 45^o.
3. So we want the length of arc PQ. We can use theorem 21.1
According to the theorem, For every 1^o central angle, the length of arc will be (^πr⁄₁₈₀) cm 
4. So for 45^o, the length of arc will be (^π×12⁄₁₈₀) × 45 =  (^π×12⁄₄) = 3π cm

Solved example 21.33
In the fig.21.48 below, 'O' is the centre of the circle, and OPQR is a rectangle.

Fig.21.48

OP = 8 cm and OR = 15 cm. Find the area of the yellow region
Solution:
1. Consider a diagonal OQ of the rectangle. It's length can be obtained as follows:
OQ² = OP² + PQ²
⇒ OQ² = (8)² + (15)² = 6•4 + 225 = 289 (∵ PQ = OR = 15 cm)

⇒ OQ = √289 = 17
2. But OQ is the radius of the circle. So we have all the details required to find the area of the yellow region:
• It's radius = 17 cm
• It's central angle = 90^o
3. The yellow region is one fourth of the total area of the circle. So it's area
=  ¹⁄₄ × πr² = ¹⁄₄ × π × 17² = 72.25π cm².

Solved example 21.34
In the fig.21.49 below, both sectors OAB and OPQ have the same centre O, and the same central angle 45^o.

Fig.21.49

The sum of the radii is 18 cm. Area of the shaded portion ABQP is 18π. Find the radius of both the sectors.
Solution:
1. If we subtract the area of the inner sector OPQ from the area of the outer sector, we will get the area of the shaded portion ABQP
2. Area of the inner sector OPQ = (^πr2⁄₃₆₀) × x = (^π×OP2⁄₃₆₀) × 45 = (^π⁄₈)×OP² cm²
3. Area of the outer sector OAB = (^πr2⁄₃₆₀) × x = (^π×OA2⁄₃₆₀) × 45 = (^π⁄₈)×OA² cm². 
4. So area of ABPQ = [(^π⁄₈)×OA² - (^π⁄₈)×OP² ] = (^π⁄₈)×[OA² - OP² ] cm².
5. But area of ABPQ is given as 18π. So we can equate the two:
 (^π⁄₈)×[OA² - OP² ] = 18π.
⇒[OA² - OP² ] = 18 × 8 = 144  
⇒(OA-OP)(OA+OP) = 144 [∵ (a² - b²) = (a+b)(a-b)]
6. But (OA+OP) is given as 18 cm. So we can write:
(OA-OP) × 18 = 144 ⇒ (OA-OP) = 8 cm. Now we have two equations:
7. OA + OP = 18
8. OA - OP = 8 
9. From (8) we get: OA = 8 + OP
10. Substituting this value of OA in (7), we get:
8 + OP + OP = 18 ⇒ 8 + 2OP = 18 ⇒ 2OP = 10 ⇒ OP = 5 cm
11. Substituting this value of OP on (8) we get: OA = 8 + 5 = 13 cm

Solved example 21.35
Find the area of the yellow portion in fig.21.50 below

Fig.21.50

Solution:
1. Area of the green sector = (^πr2⁄₃₆₀) × x = (^π×62⁄₃₆₀) × 60 = 6π cm² 
2. Total area = πr² = π × 6²  = 36π cm².
3. So area of the yellow portion = 36π - 6π = 30π cm².
Another method:
1. Central angle of the yellow portion = 360 -60 = 300^o.
2. So area = (^πr2⁄₃₆₀) × x = (^π×62⁄₃₆₀) × 300 = 30π cm² 

Solved example 21.36
Given a circular metal disc of perimeter 30 cm. A regular hexagon of maximum possible size is to be cut out from the disc. What will be the perimeter of the hexagon?
Solution:
Consider fig.21.51 below

Fig.21.51

1. A regular hexagon can be divided into 6 equilateral triangles by drawing it's three diagonals.
2. Each diagonal will be a diameter of the circle. So our first step is to find the diameter of the circle
3. Given that perimeter = 30 cm. That means 2πr = 30. So diameter = 2r = ³⁰⁄_π .
4. Each side of the equilateral triangle = half of diameter = radius = ¹⁄₂ × ³⁰⁄_π = ¹⁵⁄_π .
5. So side of the regular hexagon = ¹⁵⁄_π cm.
6. Thus perimeter of the regular hexagon = 6 × ¹⁵⁄_π = ⁹⁰⁄_π cm

Solved example 21.37
In the fig.21.52 below, O is the centre of the circle, and AB is the diameter 

Fig.21.52

The length of an arc with central angle 72^o is 4π cm.
(i) Find the diameter of the circle.
(ii) Find the area of the green portion
Solution:
1. Given that: length of an arc with central angle 72^o is 4π cm.
2. From this, we can find the radius: We can use theorem 21.1. According to the theorem, For every 1^o central angle, the length of arc will be (^πr⁄₁₈₀) cm 
3. So for 72^o, the length of arc will be (^π×r⁄₁₈₀) × 72 =  (^π×r⁄₅) × 2 cm
4. But this length is given as 4π cm. So we can equate the two:
(^π×r⁄₅) × 2 cm = 4π. ⇒ r = 10 cm 
5. So diameter of the circle = 20 cm. This is the answer of part (i)
6. AB is the diameter. So central angle of the yellow portion = 180^o.
7. So central angle of the green portion = [360 -(180 + 72)] = 108^o.
8. So area of the green portion = (^πr2⁄₃₆₀) × x = (^π×102⁄₃₆₀) × 108 = 30π cm². This is the answer of part (ii)

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In the next section we will see Real numbers.

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