In the previous section we saw some solved examples related to the Perimeter of circle. In this section we will see Area of circle.
Consider fig.21.19 below. Some polygons are drawn inside circles. All the circles in the fig.21.19 are given yellow colour, and they all have the same diameter.
The polygons are given red colour. We can see that, when the number of sides 'n' increases, the yellow colour decreases. That means, the polygon covers more and more area of the circle. In other words, the area of the polygon becomes closer and closer to the area of the circle.
Just as we saw in the case of perimeter, if 'n' is very large, the area of the polygon will be approximately equal to the area of the circle. So we want to find the area of the regular polygon which has a very large 'n'. For finding it, we will first see the area of a regular polygon with a smaller 'n'. Say n = 5. That is., a pentagon. Fig.21.20(a) below shows a regular pentagon drawn inside a circle.
The centre of the circle is joined to all the five corners. Thus we get 5 equal triangles. How are they equal?
They are equal because, they all have the same base and same sides. So we need only find the area of any one of those triangles.
1. Let the base of the triangle be 's'
2. Let the height be 'h'
[Note that, 's' forms a chord. The perpendicular bisector of this chord will pass through the centre of the circle. Thus we can easily draw the height 'h']
3. Now we have the base and height of the triangle. So area = 1⁄2 × s × h
4. Area of the pentagon = Total area of the five triangles = 5 × 1⁄2 × s × h
5. Let us group the '5' and 's' together. We will get:
6. Area of the pentagon = 1⁄2 × (5 × s) × h
7. But (5 × s) is the perimeter of the pentagon. Let us denote this perimeter as p5 (The subscript '5' denotes the five sides of the pentagon). So we get:
8. Area of the pentagon = 1⁄2 × p5 × h
Next consider the octagon in fig.21.20(b). We will get a similar result. Let us write the steps:
1. Let the base of the triangle be 's'
2. Let the height be 'h'
[Note that, 's' forms a chord. The perpendicular bisector of this chord will pass through the centre of the circle. Thus we can easily draw the height 'h']
3. Now we have the base and height of the triangle. So area = 1⁄2 × s × h
4. Area of the octagon = Total area of the eight triangles = 8 × 1⁄2 × s × h
5. Let us group the '8' and 's' together. We will get:
6. Area of the pentagon = 1⁄2 × (8 × s) × h
7. But (8 × s) is the perimeter of the pentagon. Let us denote this perimeter as p8 (The subscript '8' denotes the eight sides of the pentagon). So we get:
8. Area of the octagon = 1⁄2 × p8 × h
In a similar way, we will get the area of a regular polygon of 'n' sides as: 1⁄2 × pn × h
1. If 'n' is very large, we have seen that, the perimeter of the regular polygon will be equal to the perimeter of the circle. If 'r' is the radius of the circle, it's perimeter = 2πr
2. So we can write:
When 'n' is very large,
Area of the regular polygon = Area of the circle = 1⁄2 × 2πr × h = πrh
3. So now we have to find 'h'
Look again at the pentagon and octagon in fig.21.20 above. When the pentagon (n=5) became an octagon (n=8), the sides came closer to the periphery of the circle. If 'n' is increased further,
• the sides will become closer and closer to the periphery
• the length of the chords 's' becomes smaller and smaller
• the height of the triangles 'h' will become larger and larger.
4. So, when 'n' is very large, the chord will lie almost on the periphery, and 'h' will be almost equal to 'r'. So we can put 'r' in the place of 'h'
5. Thus, from (2) we get: Area of the circle = πrh = πr2
A sample calculation:
Area of a circle with radius 4 cm = π × 42 = π × 16 = 16π cm2
In the fig.b, the inside hollow area is filled up by a green circle. Now it is clear that the required area is the difference between the 'area of the outer circle' and 'area of the inner circle'
1. Area of the outer circle = πr12
2. Area of the inner circle = πr22
3. Area of the ring = πr12 - πr22 = π(r12 - r22)
A sample calculation:
Area of a ring with outer diameter 2.5 cm and inner diameter 2.0 cm = π(r12 - r22) = π(2.52 - 2.02)
= π(6.25 - 4.0) = π(2.25) = 2.25π cm2
Now we will see some solved examples:
Solved example 21.11
⇒ 16 = 2(s)2 ⇒ s2= 8
3. But s2 is the area of the square. So we get:
4. Area of the square = s2 = 8 cm2
⇒ h2 = (2)2 - (1)2 = 4 - 1 = 3
⇒ h = √3
Consider fig.21.19 below. Some polygons are drawn inside circles. All the circles in the fig.21.19 are given yellow colour, and they all have the same diameter.
 |
| Fig.21.19 |
Just as we saw in the case of perimeter, if 'n' is very large, the area of the polygon will be approximately equal to the area of the circle. So we want to find the area of the regular polygon which has a very large 'n'. For finding it, we will first see the area of a regular polygon with a smaller 'n'. Say n = 5. That is., a pentagon. Fig.21.20(a) below shows a regular pentagon drawn inside a circle.
 |
| Fig.21.20 |
They are equal because, they all have the same base and same sides. So we need only find the area of any one of those triangles.
1. Let the base of the triangle be 's'
2. Let the height be 'h'
[Note that, 's' forms a chord. The perpendicular bisector of this chord will pass through the centre of the circle. Thus we can easily draw the height 'h']
3. Now we have the base and height of the triangle. So area = 1⁄2 × s × h
4. Area of the pentagon = Total area of the five triangles = 5 × 1⁄2 × s × h
5. Let us group the '5' and 's' together. We will get:
6. Area of the pentagon = 1⁄2 × (5 × s) × h
7. But (5 × s) is the perimeter of the pentagon. Let us denote this perimeter as p5 (The subscript '5' denotes the five sides of the pentagon). So we get:
8. Area of the pentagon = 1⁄2 × p5 × h
Next consider the octagon in fig.21.20(b). We will get a similar result. Let us write the steps:
1. Let the base of the triangle be 's'
2. Let the height be 'h'
[Note that, 's' forms a chord. The perpendicular bisector of this chord will pass through the centre of the circle. Thus we can easily draw the height 'h']
3. Now we have the base and height of the triangle. So area = 1⁄2 × s × h
4. Area of the octagon = Total area of the eight triangles = 8 × 1⁄2 × s × h
5. Let us group the '8' and 's' together. We will get:
6. Area of the pentagon = 1⁄2 × (8 × s) × h
7. But (8 × s) is the perimeter of the pentagon. Let us denote this perimeter as p8 (The subscript '8' denotes the eight sides of the pentagon). So we get:
8. Area of the octagon = 1⁄2 × p8 × h
In a similar way, we will get the area of a regular polygon of 'n' sides as: 1⁄2 × pn × h
1. If 'n' is very large, we have seen that, the perimeter of the regular polygon will be equal to the perimeter of the circle. If 'r' is the radius of the circle, it's perimeter = 2πr
2. So we can write:
When 'n' is very large,
Area of the regular polygon = Area of the circle = 1⁄2 × 2πr × h = πrh
3. So now we have to find 'h'
Look again at the pentagon and octagon in fig.21.20 above. When the pentagon (n=5) became an octagon (n=8), the sides came closer to the periphery of the circle. If 'n' is increased further,
• the sides will become closer and closer to the periphery
• the length of the chords 's' becomes smaller and smaller
• the height of the triangles 'h' will become larger and larger.
4. So, when 'n' is very large, the chord will lie almost on the periphery, and 'h' will be almost equal to 'r'. So we can put 'r' in the place of 'h'
5. Thus, from (2) we get: Area of the circle = πrh = πr2
A sample calculation:
Area of a circle with radius 4 cm = π × 42 = π × 16 = 16π cm2
Area of a ring
Consider a ring shown in the fig.21.21(a) below. It has an outer radius of r1 and inner radius of r2. How do we find the area of the ring? |
| Fig.21.21 |
1. Area of the outer circle = πr12
2. Area of the inner circle = πr22
3. Area of the ring = πr12 - πr22 = π(r12 - r22)
A sample calculation:
Area of a ring with outer diameter 2.5 cm and inner diameter 2.0 cm = π(r12 - r22) = π(2.52 - 2.02)
= π(6.25 - 4.0) = π(2.25) = 2.25π cm2
Now we will see some solved examples:
Solved example 21.11
(i) In fig.21.22(a) below, find the
difference between the area of the circle and area of the square up to two decimal places
(ii) In fig.21.22(b) below, find the
difference between the area of the circle and area of the hexagon up to two decimal places
 |
| Fig.21.22 |
Solution:
Part (i):
1. In fig.21.23(a) below, we can see
that, the diagonal splits the square into two right triangles.
 |
| Fig.21.23 |
2. Let the sides of the square be
‘s’. Then, applying Pythagoras theorem, we get:
42 = s2 + s2⇒ 16 = 2(s)2 ⇒ s2= 8
3. But s2 is the area of the square. So we get:
4. Area of the square = s2 = 8 cm2
5. Now we want the area of the
circle. The diameter is given as 4 cm. So radius = 2 cm
6. Thus area = πr2 = π22 = 4π = 4 × 3.14 = 12.56 cm2
7. Thus difference in area = 12.56 - 8 = 4.56 cm2
Part (ii):
1. In fig.21.23(b), all the three
diagonals of the hexagon are drawn. This gives 6 equilateral
triangles
2. The sides of the hexagon are
chords of the circle. Perpendicular bisector of any chord will pass
through the centre of the circle
3. One such perpendicular
bisector is shown as dashed green line. It is perpendicular to the base, and so will
become the altitude ‘h’ of the equilateral triangle. We have to
find the value of this ‘h’
4. The altitude splits the
equilateral triangle into two right triangles. Consider any one of
the two.
Applying Pythagoras theorem,
we get:
22 = 12 + h2⇒ h2 = (2)2 - (1)2 = 4 - 1 = 3
⇒ h = √3
6. So total area of the hexagon = 6 × √3 = 6 × 1.7321 = 10.39 cm2
7. In (6) of part (i), we have already calculated the
area of a circle of 4 cm diameter as 12.56 cm2
8. So the difference = 12.56 - 10.39 = 2.17 cm2.
• The answer for part (i) is 4.56
• The answer for part (i) is 2.17
• The diameters of the two circles are the same.
• So we find that, when the number of sides of the regular polygon increases, it's area becomes closer to the area of the circle.
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