Chapter 21.3 - Derivation of the formula for Area of a circle

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.

Fig.21.19

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. 

Fig.21.20

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 × p_n × 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

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

(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

5. Now, area of the equilateral triangle = ¹⁄₂ × b × h = ¹⁄₂ × 2 × √3 = √3 cm² 

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.

---

In the next section we will see a few more solved examples.

PREVIOUS      CONTENTS       NEXT



                        Copyright©2016 High school Maths lessons. blogspot.in - All Rights Reserved

Chapter 21 - Circle Measures - Perimeter

In the previous section we completed the discussion on polynomials. In this chapter we will see Circle measures.

In this section, we will try to find a method for calculating the perimeter of any given circle. We know that perimeter of any polygon can be calculated if it’s sides are known. All we have to do is add the sides. For example, all the sides of a 5 sided polygon is given in fig.21.1(a). 

Fig.21.1

It’s perimeter is equal to 3 + 5 + 4 + 5 + 6 = 23 cm. In fig.b, another polygon with 6 sides are given. It is a ‘regular hexagon’. So all it’s 6 sides will be equal. Thus the calculation of perimeter becomes easier. We will get perimeter as 6 × 3 cm = 18 cm

Now we will consider perimeter of circles. Suppose there is a circular ground. The owner of the ground wants to make a fence around it. He will want to know how much barb wire will be needed to make the fence. In such a case there should be a method to calculate the perimeter of circles. One method is to use a simple procedure as follows:

Fig.21.2

1. Fix a peg on the periphery of the circle (fig.21.2)

2. Tie one end of a rope to the peg.

3. Place the rope along the periphery to make one complete circle upto the peg.

4. Put a mark on the rope where it meets the peg

5. Measure the length of the rope from the mark upto the peg. This will give the perimeter.

But in maths we want actual calculations. Mathematicians from very early days have tried to derive a method for calculating perimeter of circles. We will now try to understand how they derived the method that we now use commonly in Science and Engineering.

Consider fig.21.3. Some polygons are drawn inside circles. All the circles in the fig.21.3 have the same diameter. 

Fig.21.3

1. The first fig.a shows the polygon with the smallest possible number of sides. The smallest possible number of sides to form a polygon is of course three. Because, with two2 sides, we cannot form a closed figure. So, in fig.a, we have a triangle inside a circle. Note that, it is a regular polygon. So all sides have to be equal. Thus it is an equilateral triangle.

2. Next, in fig.b, we have a regular polygon with 4 sides. That is., we have a square. Note that, a rectangle, though have 4 sides, is not a regular polygon.

3. Continuing like this, in fig c, d, e and f, we have regular pentagon, hexagon, septagon and octagon. The number of sides 'n' increase by 1 in each successive fig.

4. We can see that, as the number of sides n increases, the polygon inside gets closer and closer to the circle.

5. If n is very large, the polygon will become so close to the circle that, It will be difficult to distinguish between the two. Such a polygon is shown in the fig.21.4(a) below. It has n = 14.

Fig.21.4(a)

A small portion of this fig.a is enlarged and is shown in fig.21.4(b) below:

Fig.21.4(b)

In this enlarged fig.b, we can distinguish between the two. But if we increase n further, even an enlarged fig. will not show much difference.

■ So we can say this:

1. We want to calculate the perimeter of a circle.

2. For that, we draw a regular polygon inside the circle.

3. This regular polygon has a very large ‘n’ that, it is hard to distinguish between the circle and the polygon

4. In such a situation, if we calculate the perimeter of that regular polygon, that perimeter will be approximately equal to the perimeter of the circle.

This seems to be a very good method to calculate the perimeter of circles. But before proceeding further, we must be sure of one thing: We must make sure that, we are able to calculate the perimeter of any regular polygon drawn inside a circle of known diameter. We will start with the polygon with the lowest possible n. That is., the triangle. We will do it as a solved example.

Solved example 21.1

Fig.21.5(a) below shows an equilateral triangle drawn inside a circle of diameter 1m.

Fig.21.5

Calculate the perimeter of the equilateral triangle.

Solution:

We have only two information:

• The triangle is equilateral

• The diameter of the circle is 1 m

With these, we have to calculate the perimeter of the triangle.

1. Let us draw two medians as in fig.b

• Median AE (E is the midpoint of BC)

• Median CD (D is the midpoint of AB)

Now look at the two medians carefully.

• The median CD is drawn from the vertex C to the midpoint D of the opposite side AB. This median will be perpendicular to the side AB. This is because ABC is an equilateral triangle. Also, CD passes through the midpoint D. So CD is the perpendicular bisector of AB

• Similarly, AE is the perpendicular bisector of BC.

So we can write:

The 'medians' in any equilateral triangle will serve another purpose also:

They will bisect the side perpendicularly. In other words, they are perpendicular bisectors also. 

• We know that the point of intersection of any two perpendicular bisectors of a triangle is it’s circumcentre. (details here)

2. So O is the circumcentre of the given equilateral triangle. Let us mark the radius. OB is the radius marked with a dashed green arrow. OB = half of diameter = ¹⁄₂ of 1 m = ¹⁄₂ m = 0.5 m

3. Now we use one important property of medians: The point of intersection splits the medians in the ratio 2:1, measured from the vertex. (Theorem 18.6)

4. So we can write: OC:OD = 2:1

5. That means: If we divide CD into 3 equal parts, OC will constitute 2 such parts, and OD will constitute 1 such part

6. When we divide CD into 3 equal parts, each part will be ^CD⁄₃

7. Two such parts = ^2CD⁄₃

8. So OC = ^2CD⁄₃

9. But OC = radius = ¹⁄₂ m

10. Substituting this in (8) we get: ¹⁄₂ m = ^2CD⁄₃. So we get CD = ³⁄₄ m

11. Now, AD = ^AB⁄₂. But AB = AC (∵ ABC is an equilateral triangle)

12. So AD = ^AC⁄₂

13. Now consider the right triangle ADC

14. Applying Pythagoras theorem, we get: AC² = AD² + CD².

⇒ AC² = (^AC⁄₂)² + (³⁄₄)². ⇒ AC² = (^AC2⁄₄) + (⁹⁄₁₆).
⇒ AC² - (^AC2⁄₄) = (⁹⁄₁₆) ⇒ (^3AC2⁄₄) = (⁹⁄₁₆)
⇒ AC² = ³⁄₄ ⇒ AC = ^√3⁄₂.
15. Thus we got one side of the equilateral triangle.
So perimeter = 3 × (^√3⁄₂) = ^(3√3)⁄₂

Solved example 21.2

Fig.21.6(a) below shows a square drawn inside a circle of diameter 1 m. 

Fig.21.6

Calculate the perimeter of the square.

Solution:

We have only two information:

• The polygon is a square

• The diameter of the circle is 1 m

With these, we have to calculate the perimeter of the square.

1. Let us draw the two diagonals AC and BD of the square.

2. Length of each of these diadonals are same as the diameter of the circle = 1 m

3. The diagonals bisect each other at right angles. So OA = OB = OC = OD = 0.5 m = ¹⁄₂ m

4. As the diagonals are perpendicular to each other, angle at O = 90^o

5. Consider any of the right triangles. Say ⊿OCD. Applying Pythagorus theorem we get:
CD² = OC² + OD²..
⇒ CD² = (¹⁄₂)² + (¹⁄₂)² = ¹⁄₄ + ¹⁄₄ = ¹⁄₂
⇒ CD = ¹⁄_√2
6. So perimeter of the square = 4 × ¹⁄_√2  = ⁴⁄_√2  = ^{(2×√2×√2)}⁄_(√2) = 2√2

Solved example 21.3

Fig.21.7(a) below shows a regular hexagon drawn inside a circle of diameter 1m. 

Fig.21.7

Calculate the perimeter of the regular hexagon.

Solution:

We have only two information:

• The polygon is a regular hexagon

• The diameter of the circle is 1 m

With these, we have to calculate the perimeter of the regular hexagon.
1. Draw the three diagonals of the hexagon
2. The diagonals will interset at the centre O of the circle
3. The diagonals will give 6 equilateral triangles
4. Consider any one of those 6 triangles. Say OAB
5. OB is half of diameter = ¹⁄₂ of 1 m = ¹⁄₂ m = 0.5 m
6. So OA and OB will also be equal to ¹⁄₂ m
7. So perimeter of hexagon = 6 × ¹⁄₂ = 3 m

---

We have calculated the perimeter of a triangle, a square and a hexagon in a circle of 1 m diameter. In this way we can calculate perimeters of polygons with any number of sides.  In the next section we will see how we can put it to use.

PREVIOUS      CONTENTS       NEXT

                        Copyright©2016 High school Maths lessons. blogspot.in - All Rights Reserved