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### Perimeter and Area of a Circle

♦ Introduction

♦ Perimeter and Area of a Circle : A Review

♦ Perimeter and Area of a Circle : A Review

You are already familiar with some methods of finding perimeters and areas of simple plane figures such as rectangles, squares, parallelograms, triangles and circles.

Many objects that we come across in our daily life are related to the circular shape in some form or the other. Cycle wheels, wheel barrow (thela), dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1).

So, the problem of finding perimeters and areas related to circular figures is of great practical importance.

Here, we shall begin our discussion with a review of the concepts of perimeter (circumference) and area of a circle and apply this knowledge in finding the areas of two special ‘parts’ of a circular region (or briefly of a circle) known as sector and segment.

We shall also see how to find the areas of some combinations of plane figures involving circles or their parts.

Many objects that we come across in our daily life are related to the circular shape in some form or the other. Cycle wheels, wheel barrow (thela), dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1).

So, the problem of finding perimeters and areas related to circular figures is of great practical importance.

Here, we shall begin our discussion with a review of the concepts of perimeter (circumference) and area of a circle and apply this knowledge in finding the areas of two special ‘parts’ of a circular region (or briefly of a circle) known as sector and segment.

We shall also see how to find the areas of some combinations of plane figures involving circles or their parts.

Recall that the distance covered by travelling once around a circle is its perimeter, usually called its circumference. You also know from your earlier classes, that circumference of a circle bears a constant ratio with its diameter.

This constant ratio is denoted by the Greek letter `π` (read as `‘pi’` ). In other words,

`( text ( circumference) )/( text (diameter ) ) = pi`

or, circumference` = π × text ( diameter ) `

`= π × 2r` (where r is the radius of the circle)

`= 2πr`

The great Indian mathematician Aryabhatta (C.E. `476 – 550`) gave an approximate

value of `π`. He stated that `π = (62832)/(20000)` , which is nearly equal to 3.1416. It is alsom interesting to note that using an identity of the great mathematical genius Srinivas Ramanujan of India, mathematicians have been able to calculate the value of `π` correct to million places of decimals.

As you know, `π` is an irrational number and its decimal expansion is non-terminating and non-recurring (non-repeating).

However, for practical purposes, we generally take the value of `π` as `22/7` or `3.14`, approximately.

You may also recall that area of a circle is `πr^2`, where `r` is the radius of the circle. Recall that you have verified it in Class VII, by cutting a circle into a number of sectors and rearranging them as shown in Fig. 12.2

You can see that the shape in Fig. 12.2 (ii) is nearly a rectangle of length . `1/2 xx 2 pi r` and breadth `r`. This suggests that the area of the circle `= 1/2 xx 2 pi r xx r = pi r^2.`

This constant ratio is denoted by the Greek letter `π` (read as `‘pi’` ). In other words,

`( text ( circumference) )/( text (diameter ) ) = pi`

or, circumference` = π × text ( diameter ) `

`= π × 2r` (where r is the radius of the circle)

`= 2πr`

The great Indian mathematician Aryabhatta (C.E. `476 – 550`) gave an approximate

value of `π`. He stated that `π = (62832)/(20000)` , which is nearly equal to 3.1416. It is alsom interesting to note that using an identity of the great mathematical genius Srinivas Ramanujan of India, mathematicians have been able to calculate the value of `π` correct to million places of decimals.

As you know, `π` is an irrational number and its decimal expansion is non-terminating and non-recurring (non-repeating).

However, for practical purposes, we generally take the value of `π` as `22/7` or `3.14`, approximately.

You may also recall that area of a circle is `πr^2`, where `r` is the radius of the circle. Recall that you have verified it in Class VII, by cutting a circle into a number of sectors and rearranging them as shown in Fig. 12.2

You can see that the shape in Fig. 12.2 (ii) is nearly a rectangle of length . `1/2 xx 2 pi r` and breadth `r`. This suggests that the area of the circle `= 1/2 xx 2 pi r xx r = pi r^2.`

Q 3129191911

The cost of fencing a circular field at the rate of Rs. 24 per metre is

Rs. 5280. The field is to be ploughed at the rate of Rs. 0.50 per `m^2`. Find the cost of

ploughing the field (Take` π = 22/7 ` ) .

Class 10 Chapter 12 Example 1

Rs. 5280. The field is to be ploughed at the rate of Rs. 0.50 per `m^2`. Find the cost of

ploughing the field (Take` π = 22/7 ` ) .

Class 10 Chapter 12 Example 1

Length of the fence (in metres) `= ( text ( Total cost) )/(text (Rate)) = 5280/24 = 220`

So, circumference of the field` = 220 m`

Therefore, if r metres is the radius of the field, then

`2 pi r= 220`

or, ` 2 xx 22/7 xx r = 220`

or, `r= (220 xx 7)/(2 xx 22) = 35`

i.e., radius of the field is 35 m.

Therefore, area of the field `= pi r^2 = 22/7 xx 35 xx 35 m^2 = 22 xx 5 xx 35 m^2`

Now, cost of ploughing `1 m^2` of the field = Rs. 0.50

So, total cost of ploughing the field = Rs. 22 × 5 × 35 × 0.50 = Rs. 1925