In the previous section we completed the discussion on Area of sectors. In this section we will see some solved examples.
Solved example 21.21
Calculate the area of the
green coloured portion in the fig.21.39 below
 |
| Fig.21.39 |
Solution:
1. Area of a sector with radius 3
cm and central angle 120o
= (πr2⁄360) × x = (π×32⁄360) × 120 = (9π⁄3) = 3π cm2
2. Area of a sector with radius 2
cm and central angle 120o = (π×22⁄360) × 120 = (4π⁄3) cm2
3. Area of the green portion =
(1) – (2) = 3π - 4π⁄3 = 5π⁄3 cm2
Solved example 21.22
Centred at each vertex of a
regular hexagon, a part of a circle is drawn and a fig.b is cut out
as shown below. What is the area of the shape in fig.b?
 |
| Fig.21.40 |
Solution:
1. We want the area of the shape in fig.b. Fig.c shows the details of how the shape in fig.b is obtained.
2. First, there is a regular hexagon of side 2 cm. At each corner of it, circles are drawn. So there are 6 circles. All of them have a radius of 1 cm
3. Any two adjacent sides of the hexagon will form a sector inside the circle. So the radius of the sectors will be 1 cm
4. The central angle of the sectors will be equal to the interior angle of a regular hexagon
5. The measure of each interior angle of a regular polygon is ([180(n-2)]⁄n). Where n is the number of sides.
6. We have a regular hexagon. It has 6 sides. So each interior angle is equal to:
([180(6-2)]⁄6) = (180×4⁄6) = 120o.
7. Now we can find the area of each sector:
(πr2⁄360) × x = (π×12⁄360) × 120 = (π⁄3) cm2.
8. Total area of 6 sectors = 6 × (π⁄3) = 2π cm2.
This much area is removed from the total area of the regular hexagon.
9. So we have to find the area of the regular hexagon. We did this in solved example 21.11
Total area of the regular hexagon = 6 × Area of one equilateral triangle of side 's'
Where 's' is the side of the regular hexagon)
10. We can find the area of any equilateral triangle, if we know the side. It is given by:
Area = (√3⁄4)s2. Details here.
11. So total area of our regular hexagon = 6 × (√3⁄4)s2 = 6 × (√3⁄4) × 22 = 6√3 cm2.
12. So the area of the shape in fig.b = (11) - (8) = (6√3 - 2π) cm2
Solved example 21.23
13. In our problem, s = 2 cm. So the area of two triangles
= 2×(√3⁄4)22 = 2√3 cm2.
14. Required area = (11) + (13)
= {[(π⁄6) - (√3⁄4)]×16} + {2√3} = {8π⁄3 - 4√3} + {2√3} = {8π⁄3 - 2√3}
Solved example 21.24
The fig.21.42(a) shows three circles drawn with their centres on each vertex of an equilateral triangle and passing through the other two vertices.
Find the area common to the three circles
Solution:
1. We are given fig.a. Let us add a few more details:
2. The equilateral triangle is named as ΔABC. Circles are drawn centred on each vertex. The three circles have a common area.
3. This common area consists of:
• Three green coloured regions in fig.b
• The equilateral triangle ABC
4. The three green regions are equal. So we need to find the area of only one.
1. We want the area of the shape in fig.b. Fig.c shows the details of how the shape in fig.b is obtained.
2. First, there is a regular hexagon of side 2 cm. At each corner of it, circles are drawn. So there are 6 circles. All of them have a radius of 1 cm
3. Any two adjacent sides of the hexagon will form a sector inside the circle. So the radius of the sectors will be 1 cm
4. The central angle of the sectors will be equal to the interior angle of a regular hexagon
5. The measure of each interior angle of a regular polygon is ([180(n-2)]⁄n). Where n is the number of sides.
6. We have a regular hexagon. It has 6 sides. So each interior angle is equal to:
([180(6-2)]⁄6) = (180×4⁄6) = 120o.
7. Now we can find the area of each sector:
(πr2⁄360) × x = (π×12⁄360) × 120 = (π⁄3) cm2.
8. Total area of 6 sectors = 6 × (π⁄3) = 2π cm2.
This much area is removed from the total area of the regular hexagon.
9. So we have to find the area of the regular hexagon. We did this in solved example 21.11
Total area of the regular hexagon = 6 × Area of one equilateral triangle of side 's'
Where 's' is the side of the regular hexagon)
10. We can find the area of any equilateral triangle, if we know the side. It is given by:
Area = (√3⁄4)s2. Details here.
11. So total area of our regular hexagon = 6 × (√3⁄4)s2 = 6 × (√3⁄4) × 22 = 6√3 cm2.
12. So the area of the shape in fig.b = (11) - (8) = (6√3 - 2π) cm2
Solved example 21.23
In fig.21.41(a), two circles are
drawn, each passing through the centre of the other.
 |
| Fig.21.41 |
Calculate the
area of the region which is common to both the circles.
Solution:
1. We are given fig.a. Let us add some more details
to it. In the fig.b, names are given to important points:
• The centre of the left circle
is A. The centre of the right circle is B
• The top point of intersection
is C. The bottom point of intersection is D
2. Consider the circle with
centre A on the left. The point B lies on this circle. But it is
given that AB = 2 cm. Thus we get the radius of the left circle as 2
cm
3. Consider the circle with
centre B on the right. The point A lies on this circle. But it is
given that AB = 2 cm. Thus we get the radius of the right circle also
as 2 cm
4. Points C and D lies on the
circle with centre A. So we get AC = AD = 2 cm
5. Points C and D lies on the
circle with centre B also. So we get BC = BD = 2 cm
6. So ΔABC and ΔABD are two
equilateral triangles. These details are shown in fig.c
7. Consider the sector ABC in the
left circle. It's boundaries are:
The minor arc BC, and the two
radial lines AC and AB
8. The equilateral triangle ABC can be separated out from this sector. The remaining portion will be a curved region.
9. We know how to calculate the area of this curved region. We can use the equation:
■ Area after the removal of equilateral triangle from a sector of radius 'r' and central angle 60o
10. In our problem, r = 2 cm. So we get:
Area = [(π⁄6) - (√3⁄4)]22 = [(π⁄6) - (√3⁄4)]×4
11. Four such areas are present. So the total curved area = [(π⁄6) - (√3⁄4)]×4×4 = [(π⁄6) - (√3⁄4)]×16
12. In addition, we have two equilateral triangles of side 2 cm. We can find the area of any equilateral triangle, if we know the side. It is given by: Area = (√3⁄4)s2. Details here.13. In our problem, s = 2 cm. So the area of two triangles
= 2×(√3⁄4)22 = 2√3 cm2.
14. Required area = (11) + (13)
= {[(π⁄6) - (√3⁄4)]×16} + {2√3} = {8π⁄3 - 4√3} + {2√3} = {8π⁄3 - 2√3}
Solved example 21.24
The fig.21.42(a) shows three circles drawn with their centres on each vertex of an equilateral triangle and passing through the other two vertices.
 |
| Fig.21.42 |
Solution:
1. We are given fig.a. Let us add a few more details:
2. The equilateral triangle is named as ΔABC. Circles are drawn centred on each vertex. The three circles have a common area.
3. This common area consists of:
• Three green coloured regions in fig.b
• The equilateral triangle ABC
4. The three green regions are equal. So we need to find the area of only one.
5. Consider the sector ABC in the circle centred at A. It's boundaries are:
The minor arc BC, and the two radial lines AC and AB
6. The equilateral triangle ABC can be separated out from this sector. The remaining portion will be a curved region.
7. We know how to calculate the area of this curved region. We can use the equation:
■ Area after the removal of equilateral triangle from a sector of radius 'r' and central angle 60o
8. In our problem, r = 2 cm. So we get:
Area = [(π⁄6) - (√3⁄4)]22 = [(π⁄6) - (√3⁄4)]×4
9. Three such areas are present. So the total curved area = [(π⁄6) - (√3⁄4)]×4×3 = [(π⁄6) - (√3⁄4)]×12
10. We can find the area of any equilateral triangle, if we know the side. It is given by: Area = (√3⁄4)s2. Details here.
11. In our problem, s = 2 cm. So the area of ΔABC
= (√3⁄4)22 = √3 cm2.
12. Required area = (9) + (11)
= {[(π⁄6) - (√3⁄4)]×12} + {√3} = {2π - 3√3} + {√3} = {2π - 2√3} = 2(π-√3) cm2
10. We can find the area of any equilateral triangle, if we know the side. It is given by: Area = (√3⁄4)s2. Details here.
11. In our problem, s = 2 cm. So the area of ΔABC
= (√3⁄4)22 = √3 cm2.
12. Required area = (9) + (11)
= {[(π⁄6) - (√3⁄4)]×12} + {√3} = {2π - 3√3} + {√3} = {2π - 2√3} = 2(π-√3) cm2
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