In the previous section 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 this section we will see how this knowledge can be put to use.
In the three solved examples, we did the following:
1. In the fig.21.8(a) below, ABC is an equilateral triangle drawn inside a circle of diameter 1 m. The centre of this circle is O
2. A larger circle is drawn with
the same centre.
In the three solved examples, we did the following:
• We calculated the perimeter of
a 3 sided polygon in a circle of diameter 1 m
♦ The result is (3√3)⁄2 m = (3×1.7321)⁄2 m = 2.5981 m
♦ The result is (3√3)⁄2 m = (3×1.7321)⁄2 m = 2.5981 m
• We calculated the perimeter of
a 4 sided polygon in a circle of diameter 1 m
♦ The result is 2√2 m = 2 × 1.4142 m = 2.8284 m
♦ The result is 2√2 m = 2 × 1.4142 m = 2.8284 m
• We calculated the perimeter of
a 6 sided polygon in a circle of diameter 1 m
♦ The result is 3 m
♦ The result is 3 m
So it is clear that there are
mathematical methods to calculate the perimeter of regular polygons with any number of sides 'n' in a circle of diameter 1 m
As n becomes large, the
calculations will become lengthy. We can use a computer to do it because, the actual method for calculating the perimeter is
not important for our present discussion. We need only the final
values. The following table gives those final values:
In the above table, the value
of n up to 50 is used. Still there is no sign of getting a fixed
number. We can try increasing n to numbers more than 50. Even then we will not
get a fixed value. It is a non recurring and non terminating decimal.
But for practical purposes, we can use a rounded value.
| No. of Sides | Perimeter | No. of Sides | Perimeter | |
|---|---|---|---|---|
| 3 | 2.5981 | 15 | 3.1187 | |
| 4 | 2.8284 | 20 | 3.1287 | |
| 5 | 2.9389 | 25 | 3.1333 | |
| 6 | 3.0000 | 30 | 3.1359 | |
| 7 | 3.0372 | 35 | 3.1374 | |
| 8 | 3.0615 | 40 | 3.1384 | |
| 9 | 3.0782 | 45 | 3.1390 | |
| 10 | 3.0902 | 50 | 3.1395 |
• Rounded to 2 decimal places
3.14
• Rounded to 3 decimal places
3.139
• Rounded to 4 decimal places
3.1395
So on . . .
In scientific and engineering
problems, the number of decimal places to be rounded will be
specified.
So we find that diameter of a
circle of 1 m diameter is 3.1395 m approximately.
• There is a special
symbol for this value. The symbol is: π
• It is the Greek letter read as
'pi'.
■ We can write: The diameter
of a circle of 1 m diameter is π
m
So we successfully calculated
the perimeter of a circle of diameter 1 m. Do we need to do the same
process to calculate the perimeter of circles with other diameters? Some examples:
• To find the perimeter of a
circle with 1.5 m diameter, do we have to find this:
♦ The perimeter of a polygon
with a large 'n' drawn inside a circle with 1.5 m diameter?
• To find the perimeter of a
circle with 2 m diameter, do we have to find this:
♦ The perimeter of a polygon
with a large 'n' drawn inside a circle with 2 m diameter?
• To find the perimeter of a
circle with 3.5 m diameter, do we have to find this:
♦ The perimeter of a polygon
with a large 'n' drawn inside a circle with 3.5 m diameter?
Let us see if we can avoid
doing such a calculation:
1. In the fig.21.8(a) below, ABC is an equilateral triangle drawn inside a circle of diameter 1 m. The centre of this circle is O
 |
| Fig.21.8 |
3. Now a triangle PQR is drawn
inside the larger circle by using the following procedure (fig.b):
• OA is extended up to the
larger circle to get point P
• OB is extended up to the
larger circle to get point Q
• OC is extended up to the
larger circle to get point R
4. We want the relation between
the two triangles ABC and PQR
But we have already found the
relation between two such triangles in a previous chapter. (See solved example 19.11 here)
• The relation is this: ΔPQR is a scaled version of ΔABC
5. That means, each side of PQ
can be obtained by multiplying the corresponding side of ABC by a
scale factor 'k'. We can write:
PQ = k × AB, QR = k × BC, PR = k × AC
6. But since it is an equilateral triangle, the sides are equal. Let the length of side be s. Then we can write:
• AB = BC = AC = s
• PQ = QR = PQ = ks
6. But since it is an equilateral triangle, the sides are equal. Let the length of side be s. Then we can write:
• AB = BC = AC = s
• PQ = QR = PQ = ks
7. Let p1 be the perimeter of
ABC. Then we get: p1 = 3s
8. Let p2 be the perimeter of
PQR. Then we get: p2 = PQ + QR + PR
⇒ p2 = (ks + ks + ks)
⇒ p2 = 3ks ⇒ p2 = k × 3s
⇒ p2 = k × p1
⇒ p2 = (ks + ks + ks)
⇒ p2 = 3ks ⇒ p2 = k × 3s
⇒ p2 = k × p1
• That means., perimeter is also
scaled by the same factor k
9. In the solved example 19.11, we saw
another important result:
• The scale factor k is same for
the radii also.
10. So, if r1 is the radius of inner circle, and r2 the radius of outer circle, then r2 = k × r1
11. Multiplying both sides of the
above equation by 2. we get:
2 × r2 = 2 × k × r1
⇒ 2 × r2 = k × 2 × r1
⇒ d2 = k × d1
2 × r2 = 2 × k × r1
⇒ 2 × r2 = k × 2 × r1
⇒ d2 = k × d1
• So, the scale factor k is same
for the diameters also.
12. From this we get k = d2⁄d1
So we proved two things:
• Perimeter of the outer triangle = k × perimeter of inner triangle
• k = outer diameter⁄inner diameter
In the above discussion, we
considered equilateral triangles. The above results can be proved for
any regular polygon. For example,
1. In fig.21.9(a) below, we have a regular
pentagon inside a circle of 1 m diameter. A larger circle is drawn with the same centre
 |
| Fig.21.9 |
2. In fig.b, a regular pentagon is drawn in
side the larger circle. This is by extending radial lines through the vertices of the inner polygon.
3. By considering the inner triangles, we can prove that, the
perimeter of the outer pentagon is k times the perimeter of the inner
pentagon
4. Also, k = d2⁄d1,
where d2 is the diameter of the outer circle, and d1 is the diameter
of the inner circle
Now we consider the general
case:
1. We draw a regular polygon of
'n' sides in a circle of diameter 1 m
2. Then we draw a larger circle
with the same centre
3. Next we draw a regular polygon
with the same 'n' sides in the larger circle
4. We can write:
• Perimeter of the outer polygon = k × perimeter of inner polygon
• k = outer diameter⁄inner diameter
5. But we have taken the inner
circle to be of diameter 1 m. In a circle of diameter 1 m, if n is very large,
we have seen that the perimeter of the regular polygon will be equal
to π m
6. So, if n is very large,
perimeter of the outer polygon = k π
7. But, if n is very large, the
perimeter of the outer polygon will be equal to the perimeter of the
outer circle. So we can write:
8. If n is very large, perimeter
of the outer circle = k π
9. We know the value of k from (4): k = outer diameter⁄inner diameter
10. But inner diameter = 1 m
11. So we get: k = outer diameter
12. Thus we can write:
• perimeter of outer circle = outer diameter × π
■ We are given a circle with diameter 'd' m. We are asked to find it's perimeter
Perimeter can be found by following the steps below:
1. Imagine that the given circle of diameter 'd' is placed above the circle of 1 m diameter (fig.21.10.a)
2. It must be placed in such a way that, centres of both are the same
3. Imagine a polygon of 'n' (n has to be very large) sides inside the circle with 1 m diameter (fig.b).
♦ Then the perimeter of that polygon = π m
♦ Then the perimeter of the 1 m dia. circle is also π m. This is the inner perimeter.
(Note that n has to be very large. In the fig., a small n is used to distinguish between the circle and polygon easily)
4. Imagine a polygon with the same n sides in the given circle (fig.c).
5. Outer perimeter = k × inner perimeter
6. The scale factor 'k' in the above step is given by: k = outer diameter⁄inner diameter .
♦ But inner diameter = 1 m
♦ So k = outer diameter = given diameter 'd'
7. From (5) we get: Outer perimeter = k × inner perimeter
⇒ Outer perimeter = d × π
■ The above procedure assumes that the given circle has a diameter 'd' larger than 1 m. Even if 'd' is smaller than 1 m, we will get the same result in (7). The steps are given below:
1. Imagine that the given circle of diameter 'd' is placed above the circle of 1 m diameter (fig.21.11.a). (This time, the given circle falls inside the 1 m diameter circle)
2. It must be placed in such a way that, centres of both are the same
3. Imagine a polygon of 'n' (n has to be very large) sides inside the circle with 1 m diameter (fig.b).
♦ Then the perimeter of that polygon = π m
♦ Then the perimeter of the 1 m dia. circle is also π m. This is the outer perimeter.
(Note that n has to be very large. In the fig., a small n is used to distinguish between the circle and polygon easily)
4. Imagine a polygon with the same 'n' sides in the given circle (fig.c).
5. Outer perimeter = k × inner perimeter
6. The scale factor 'k' in the above step is given by: k = outer diameter⁄inner diameter .
♦ But outer diameter = 1 m
♦ So k = 1⁄inner diameter = 1⁄d . given diameter 'd'
7. From (5) we get Outer perimeter = k × inner perimeter
⇒ π = 1⁄d × inner perimeter
⇒ inner perimeter = d × π
This result is same as in (7) of the previous case
■ Once we understand the basics, we need not imagine placing the given circle above another and doing mental calculations. We can directely use the result: p = dπ
Let us see some sample calculations:
1. Perimeter of a circle with diameter 1.25 m = dπ = 1.25 × 3.1395 = 3.9243 m
Analysis:
• This circle is larger than a 1 m dia. circle. So:
• Perimeter of given circle should be greater than Perimeter of 1 m dia. circle (3.1395)
• 3.9243 is indeed greater than 3.1395
2. Perimeter of a circle with diameter 0.80 m = dπ = 0.80 × 3.1395 = 2.5116 m
Analysis:
• This circle is smaller than a 1 m dia. circle. So:
• Perimeter of given circle should be less than Perimeter of a 1 m dia. circle (3.1395)
• 2.5116 is indeed less than 3.1395
• perimeter of outer circle = outer diameter × π
How do we use the above equation in a practical situation?
The following scenario gives the explanation:■ We are given a circle with diameter 'd' m. We are asked to find it's perimeter
Perimeter can be found by following the steps below:
1. Imagine that the given circle of diameter 'd' is placed above the circle of 1 m diameter (fig.21.10.a)
2. It must be placed in such a way that, centres of both are the same
 |
| Fig.21.10 |
♦ Then the perimeter of that polygon = π m
♦ Then the perimeter of the 1 m dia. circle is also π m. This is the inner perimeter.
(Note that n has to be very large. In the fig., a small n is used to distinguish between the circle and polygon easily)
4. Imagine a polygon with the same n sides in the given circle (fig.c).
5. Outer perimeter = k × inner perimeter
6. The scale factor 'k' in the above step is given by: k = outer diameter⁄inner diameter .
♦ But inner diameter = 1 m
♦ So k = outer diameter = given diameter 'd'
7. From (5) we get: Outer perimeter = k × inner perimeter
⇒ Outer perimeter = d × π
■ The above procedure assumes that the given circle has a diameter 'd' larger than 1 m. Even if 'd' is smaller than 1 m, we will get the same result in (7). The steps are given below:
1. Imagine that the given circle of diameter 'd' is placed above the circle of 1 m diameter (fig.21.11.a). (This time, the given circle falls inside the 1 m diameter circle)
2. It must be placed in such a way that, centres of both are the same
 |
| Fig.21.11 |
♦ Then the perimeter of that polygon = π m
♦ Then the perimeter of the 1 m dia. circle is also π m. This is the outer perimeter.
(Note that n has to be very large. In the fig., a small n is used to distinguish between the circle and polygon easily)
4. Imagine a polygon with the same 'n' sides in the given circle (fig.c).
5. Outer perimeter = k × inner perimeter
6. The scale factor 'k' in the above step is given by: k = outer diameter⁄inner diameter .
♦ But outer diameter = 1 m
♦ So k = 1⁄inner diameter = 1⁄d . given diameter 'd'
7. From (5) we get Outer perimeter = k × inner perimeter
⇒ π = 1⁄d × inner perimeter
⇒ inner perimeter = d × π
This result is same as in (7) of the previous case
■ Once we understand the basics, we need not imagine placing the given circle above another and doing mental calculations. We can directely use the result: p = dπ
Let us see some sample calculations:
1. Perimeter of a circle with diameter 1.25 m = dπ = 1.25 × 3.1395 = 3.9243 m
Analysis:
• This circle is larger than a 1 m dia. circle. So:
• Perimeter of given circle should be greater than Perimeter of 1 m dia. circle (3.1395)
• 3.9243 is indeed greater than 3.1395
2. Perimeter of a circle with diameter 0.80 m = dπ = 0.80 × 3.1395 = 2.5116 m
Analysis:
• This circle is smaller than a 1 m dia. circle. So:
• Perimeter of given circle should be less than Perimeter of a 1 m dia. circle (3.1395)
• 2.5116 is indeed less than 3.1395
■ Usually we use radius to
define a circle. So, instead of d, we will write 2 times radius.
If radius is denoted as 'r', we get: p = 2×r×π
⇒ p
= 2πr
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