In the previous section we completed the discussion on Length of arcs. In this section we will see Area of sectors.
In fig.21.35 below, AB is an arc. We can see two radii: OA and OB. They are drawn from the two ends of the arc towards the centre of the circle.
■ The area enclosed by an arc, and two radii is called a sector. It's symbol is ⌔. So, in fig.21.35, OAB is a sector. Let us find the method to calculate the area of a sector.
In fig.21.35 below, AB is an arc. We can see two radii: OA and OB. They are drawn from the two ends of the arc towards the centre of the circle.
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| Fig.21.35 |
• Look at fig.21.36 below. The circle with radius ‘r’ is divided into four equal parts.
So P, Q, R and S are the quadrant points. Let us see four cases:
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| Fig.21.36 |
Case 1:
1. In fig.a, the area of sector OPQ is one fourth of the total area of the 'circle with radius r cm'.
2. We know that the total area = πr2. So area of the sector OPQ = (1⁄4 × πr2)
3. The central angle of this sector is 90o
4. So we can write: 90o → (1⁄4 × πr2)
5. So 1o → [(1⁄4 × πr2) ÷ 90]
6. [(1⁄4 × πr2) ÷ 90] = πr2⁄(4×90) = πr2⁄360
7. So 1o → πr2⁄360. That means, If the central angle of a sector with radius 'r' is 1o, then it's area will be equal to πr2⁄360 cm2
8. Another form of writing this is: In a circle of radius 'r' cm, every 1o will give a sector of area πr2⁄360 cm2.
Case 2:
1. In fig.b, the area of sector OPQR is one half of the total area of the 'circle with radius r cm'.
2. We know that the total area = πr2. So area of the sector OPQR = (1⁄2 × πr2)
3. The central angle of this sector is 180o
4. So we can write: 180o → (1⁄2 × πr2)
5. So 1o → [(1⁄2 × πr2) ÷ 180]
6. [(1⁄2 × πr2) ÷ 180] = πr2⁄(2×180) = πr2⁄360
7. So 1o → πr2⁄360. That means, If the central angle of a sector with radius 'r' is 1o, then it's area will be equal to πr2⁄360 cm2
8. Another form of writing this is: In a circle of radius 'r' cm, every 1o will give a sector of area πr2⁄360 cm2.
Case 3:
1. In fig.c, the area of sector OPQRS is three fourth of the total area of the 'circle with radius r cm'.
2. We know that the total area = πr2. So area of the sector OPQRS = (3⁄4 × πr2)
3. The central angle of this sector is 270o
4. So we can write: 270o → (1⁄2 × πr2)
5. So 1o → [(3⁄4 × πr2) ÷ 270]
6. [(3⁄4 × πr2) ÷ 270] = 3πr2⁄(4×270) = πr2⁄360
7. So 1o → πr2⁄360. That means, If the central angle of a sector with radius 'r' is 1o, then it's area will be equal to πr2⁄360 cm2
8. Another form of writing this is: In a circle of radius 'r' cm, every 1o will give a sector of area πr2⁄360 cm2.
Case 4:
1. In fig.d, the area of sector OPQRSP is one full of the total area of the 'circle with radius r cm'.
2. We know that the total area = πr2. So area of the sector OPQRSP = πr2
3. The central angle of this sector is 360o
4. So we can write: 360o → πr2
5. So 1o → [(πr2) ÷ 360]
6. [(πr2) ÷ 360] = πr2⁄360
7. So 1o → πr2⁄360. That means, If the central angle of a sector with radius 'r' is 1o, then it's area will be equal to πr2⁄360 cm2
8. Another form of writing this is: In a circle of radius 'r' cm, every 1o will give a sector of area πr2⁄360 cm2.
■ In all the four cases, we get the same result. We can write it in the form of a theorem:
Theorem 21.3
• The radius of a circle is 'r' cm
• Every 1o central angle in this circle will give a sector of area (πr2⁄360) cm2.
This is shown in the fig.21.37 below:
• So the area of a sector with radius 'r' cm, and central angle xo will be equal to
(πr2⁄360) × x = (πr2x⁄360)cm2
A sample calculation:
■ Area of a sector of central angle 120o in a circle of radius 3 cm:
• Given: r = 3 cm, and central angle x = 120o
• Area of the sector = (π×32⁄360) × 120 = (9π⁄3) = 3π cm2
Now we will see a special case: Sectors with central angle 60o.
Consider the minor arc AB in fig.21.38(a) below:
• The points A and B are joined by a straight line. Then we get a separate area:
• The area enclosed between the minor arc AB and the straight line AB. This area is shown in blue colour. We want to calculate this area. We can proceed as follows:
1. Consider the whole sector OAB (fig.b). It's area can be easily calculated:
(πr2⁄360) × 60 = (πr2⁄6) cm2
2. Now join A and B by a straight line (fig.c). Now we have a triangle OAB. What is the peculiarity of this triangle?
• OA = OB = r. So it is an isosceles triangle
3. In isosceles triangles, base angles are equal. So ∠A = ∠B
• Since they are equal, we will put ∠A = ∠B = xo
4. Sum of the three interior angles in any triangle is 180o. So we can write:
60 + x + x = 180 ⇒ 60+2x = 180 ⇒ 2x = 120 ⇒ x = 60o
5. So we get ∠A = ∠B = 60o. So all the three angles are 60o. It is an equilateral triangle
6. We can find the area of any equilateral triangle, if we know the side. Details here.
So area of ΔOAB = (√3⁄4)r2.
7. Area of the blue region = Area of ⌔ OAB - Area of ΔOAB = [(πr2⁄6) - (√3⁄4)r2]
= [(π⁄6) - (√3⁄4)]r2 cm2.
■ In all the four cases, we get the same result. We can write it in the form of a theorem:
• The radius of a circle is 'r' cm
• Every 1o central angle in this circle will give a sector of area (πr2⁄360) cm2.
This is shown in the fig.21.37 below:
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| Fig.21.37 |
(πr2⁄360) × x = (πr2x⁄360)cm2
■ Area of a sector of central angle 120o in a circle of radius 3 cm:
• Given: r = 3 cm, and central angle x = 120o
• Area of the sector = (π×32⁄360) × 120 = (9π⁄3) = 3π cm2
Consider the minor arc AB in fig.21.38(a) below:
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| Fig.21.38 |
• The area enclosed between the minor arc AB and the straight line AB. This area is shown in blue colour. We want to calculate this area. We can proceed as follows:
1. Consider the whole sector OAB (fig.b). It's area can be easily calculated:
(πr2⁄360) × 60 = (πr2⁄6) cm2
2. Now join A and B by a straight line (fig.c). Now we have a triangle OAB. What is the peculiarity of this triangle?
• OA = OB = r. So it is an isosceles triangle
3. In isosceles triangles, base angles are equal. So ∠A = ∠B
• Since they are equal, we will put ∠A = ∠B = xo
4. Sum of the three interior angles in any triangle is 180o. So we can write:
60 + x + x = 180 ⇒ 60
6. We can find the area of any equilateral triangle, if we know the side. Details here.
So area of ΔOAB = (√3⁄4)r2.
7. Area of the blue region = Area of ⌔ OAB - Area of ΔOAB = [(πr2⁄6) - (√3⁄4)r2]
= [(π⁄6) - (√3⁄4)]r2 cm2.
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