A circle is a shape consisting of a curved line completely surrounded by an area. The circle contains points in the plane which are at the given distance from a given point, the center, equivalently the curve traced out by a point which moves in a plane so the distance from a given point is constant. The distance between any point and the center of the circle is called the radius (r).

**Area of a Circle**

The area of the circle is computed by the formula:

**π****r ^{2}**

- r is the radius of the circle.
**π**is the ratio of the diameter of a circle to its circumference- The value
**π**is 22/7 (or) 3.14.

**Circumference of a Circle**

The perimeter of the circle is the distance along the boundary of the circle. The perimeter is also called the circumference of a circle. Circumference will be the **π** times of the diameter of the circle.

- Circumference of a Circle is
**2****π****r**.

**Angle of Sector and Arc**

**Sector**

A sector of a circle is defined as the region of a circle enclosed by an arc and two radii(r). The smallest area of a circle is called the minor sector and the largest area of a circle is called the major sector.

### Angle of a Sector

- The angle of a sector is that angle which is enclosed within the two radii of the sector. It consists of an arc.
- A sector that has a central angle of 180° is known as a semicircle.
- A minor arc is smaller than a semicircle. A central angle that is subtended by a minor arc has a measure of less than 180°.
- A major arc is larger than a semicircle. A central angle that is subtended by a major arc has a measure larger than 180°.

**Arc**

- To find the length of an arc of a circle we use the arc length formula:
**length = radius *****θ** - An arc is a part of a curve.
- It is a portion of the circumference of the circle.

### Sample Problems

**Problem 1: Find the area of a sector and arc length of a circle of radius4cm and the central angle is 2**

**π**

**/5.**

**Solution:**

Arc length, l = 4*2π/5

= 8π//5cm.

area of a sector = 1/2θr2

= 1/2 * 2π/5 *42

= 16π/5cm2.

**Circumference of a circle when **the **area is given**

The circumference of the circle can be referred as the linear distance around it. If the circle is opened to form a straight line then, the length of that line will be the s circumference of the circle.

To calculate the circumference of a or a given circle, we need to multiply the diameter of the circle with the π.

**Problem 2: What is the perimeter of the circle whose surface area is 314.159 sq.cm**

**Solution: **

The formula of the surface area of the circle, we know:

A = π x r

^{2}Now, substituting the value:

314.159 = π x r

^{2}314.159 = 3.14 x r

^{2}r

^{2}= 314.159/3.14r

^{2}= 100.05r = √100.05

r = 10 cm

C = 2 * π * r

Substitute value of r

= 2π * 10

= 2 * 3.14 * 10

= 62.8 cm.

**Partial circle area and arc length**

The circle area can be also referred to as the circumference**. **The area of the partial circle which is also called sector can be found by the formula

**[(****π*****r ^{2})/360]θ**

The arc length is given by

**(θ/360)*(2 * ****π ***** r)**

**Problem 3: A circle of radius 4 units, angle of its sector is 45°. Find the area of the sector.**

**Solution: **

Given,

radius r = 4 units

Angle θ = 45°

Area of the sector

= θ/360

^{o }× πr^{2}= 45/360 × 22/7 × 4

^{2 }= 6.28sq.units.

**Area of a Shaded Region**

The shaded region is the region that can be found when one shape is inscribed within the other shape.

- The area of the shaded region can be calculated by subtracting the area of the inscribed shape area from the area of the shape which is inscribed in it.
- For example, let us assume a circle is inscribed in the triangle then the area of the shaded region is the area of the triangle minus the area of the circle inscribed in the triangle(
**shaded area = area of a triangle – area of a circle**).

**Problem 4: Find the area of the shaded region.**

**Solution:**

The area of shaded portion, we have to subtract area of two semicircles from the area of square

Area of shaded portion = Area of square – (Area of semicircle + Area of another semicircle).

= a2 – [(1/2) πr

^{2}) + ((1/2) πr^{2})]= 82 – πr

^{2}= 64 – (22/7)⋅ (7/2)2

= 64- (22/7)⋅ (7/2)⋅ (7/2)

= 64 – 38.5

= 25.5 cm

^{2}

**Problem 5: Find the area of the shaded region.**

**Solution:**

Given radius r = 10 cm

area of shaded region = area of square – area of circle inscribed inside square.

area of square= S

^{2}diameter of the circle will be equal to length of the side of the square so

d = S = 2*10

= 20

area of square =(20)

^{2}= 400cm

^{2}area of circle = πr

^{2}= 3.14 * 10 * 10

= 314 cm

^{2}Area of shaded region = area of square – area of circle inscribed inside square

= 400 – 314

= 86cm

Therefore, area of shaded region is 86cm.

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