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

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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.

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