### Theory:

In our daily lives, we came across the circular shape in some form or the other. Such as bangles, Cycle wheels, wheelbarrow (thela), dartboard, round cake, papad, drain cover, various designs, brooches, circular parks, etc. are examples of such objects.

As a result, the problem of determining the perimeters and areas of circular figures is obviously useful. In your earlier classes, you have learned several methods of finding perimeters and areas of simple plane figures such as rectangles, squares, parallelograms, triangles and circles.

In this chapter, we begin with discussing and reviewing the concepts of the 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.

Perimeter and Area of a Circle — A Review:

We know the distance covered by travelling once around a circle is its perimeter; generally, we call it its circumference.

From your earlier classes, we understand that the 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:

\(\frac{\text{Circumference}}{\text{Diameter}} = π\)

\(\text{Circumference = π × Diameter}\)

We know that \(\text{Diameter = 2 × Radius (r) = 2r }\). Here '\(r\)' is the radius of the circle.

Then, \(\text{Circumference = π × 2r}\)

Therefore, \(\text{Circumference = 2πr}\)

We are aware that, usually, we take the value of '\(π\)' as \(\frac{22}{7}\) or \(~3.14\).

In our earlier classes, we learn that the area of the circle is \(πr²\). Here '\(r\)' is the radius of the circle.

Let's do one experiment with circles.

From your earlier classes, we understand that the 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:

\(\frac{\text{Circumference}}{\text{Diameter}} = π\)

\(\text{Circumference = π × Diameter}\)

We know that \(\text{Diameter = 2 × Radius (r) = 2r }\). Here '\(r\)' is the radius of the circle.

Then, \(\text{Circumference = π × 2r}\)

Therefore, \(\text{Circumference = 2πr}\)

We are aware that, usually, we take the value of '\(π\)' as \(\frac{22}{7}\) or \(~3.14\).

In our earlier classes, we learn that the area of the circle is \(πr²\). Here '\(r\)' is the radius of the circle.

Let's do one experiment with circles.

Take a paper and mark it as shown in the below figure. Then cut the paper into pieces and arrange the pieces as shown below.

You can see that the shape in the

**figure - (ii)**is nearly a rectangle of length $\frac{1}{2}\times 2\mathrm{\pi}r$ and breadth \(r\). This suggests that the area of the circle $\frac{1}{2}\times 2\mathrm{\pi}r\times r=\mathrm{\pi}{r}^{2}$ square units.Area of the circular path:

We know that the area of the circular path is formed by two concentric of radii \(r_1\) and \(r_2\):

$\mathrm{\pi}{({r}_{1})}^{2}-\mathrm{\pi}{({r}_{2})}^{2}=\mathrm{\pi}{(({r}_{1})}^{2}-{({r}_{1})}^{2})$, when \(r_1 > r_2\).

Here, \(r_1\) and \(r_2\) are radius of the circles.

In this chapter let's recall and go through same concepts with example for better understanding.

*Watch the below video to know more:*

Reference:

Video credits: TicTacLearn English