**What is a Circle?**

A **Circle **is a **round closed figure** where all its boundary points are equidistant from a fixed point, the centre.

**Radius**

The distance from the centre to a point on the boundary is called the **radius of a circle**. It is represented by the letter "\(r\)"

**Diameter**

A straight line which passes through the centre and its endpoints lie on the circle is called the diameter of a circle. It is represented by the letter "\(d\)"

The **diameter of a circle** is twice its radius.

\(\text{Diameter } \left(d\right)\) \(= 2 \times \text{ Radius} \left(r\right)\) |

**Circumference**

The total length of the boundary is called the **circumference of a circle**.

\(\text{Circumference of a Circle}\) \(= 2 \pi r\) |

where \(r\) is the radius of the circle and \( \text{Pi} - \pi\) is the mathematical constant whose value is approximated to \(3.14\) or \( \frac{22}{7} \)

**What is Area of Circle**

The area of a circle is the amount of space enclosed within the boundary of a circle.

In the below figures, the colored region within the boundary of the circle is the area occupied by the circle.

**Area of Circle Example & Formulas**

- Area of circle formula in terms of the radius is:

\( \text{Area of a Circle} = \pi \text{r}^2\) |

- Area of the circle formula in terms of the diameter is:

\(\begin{align} \text{Area of a Circle} = \!\frac{\pi}{4} \!\times\! \text{d}^2\end{align}\) |

\(\text{r}\) is the radius of the circle and \(\text{d}\) is the diameter of the circle

The value of \(\pi \approx \frac{22}{7}\)

Improve your knowledge on * Area of a Circle* with examples from our experts in the Solved Examples section which has few sums based on this formula.

**Derivation of Area of a Circle **

Before we give the formula, let's first understand how the **formula for area of a circle** is derived.

Here's a simple simulation for you to interact with along with explanations.

- As we can see in the simulation above, the circle can be cut into a triangle with the radius being the height of the triangle and the perimeter as its base which is

\[2 \pi r\] - We know that the area of the triangle is found by multiplying its base by the height, and then dividing by 2, which is

\[\begin{align}\left(\frac{1}{2}\right) \times {2} \pi r \times r\end{align}\] - Therefore, the area of the circle is \(\pi {r^2}\)

\(\text{Area of a Circle}\) \(= \pi r^2\) |

where \(r\) is the radius of the circle and the value of \(\pi \approx \frac{22}{7}\)

**Surface Area of Circle Formula**

Surface area is the area occupied by the surface of a 3-D shape. The surface of a sphere will be circular in shape.

A circle is a simple plane 2-Dimensional shape. The total area occupied by the circle is the surface area of the circle

Surface Area Formula of a circle is:

\(\text{Surface Area of a Circle}\) \(= \pi r^2\) |

where \(r\) is the radius of the circle and the value of \(\pi \approx \frac{22}{7}\)

- \(\pi \approx \frac{22}{7}\)
- Radius \(= r\)
- Diameter \(= 2\text{r}\)
- Circumference \(=2 \pi r\) OR \(2\text{d}\)
- Area \(= \pi \text{r}^2\) OR \(\left(\frac{\pi}{4}\right)\text{d}^2\)

**Real-World Examples **

Example 1 |

Rohan and his friends ordered a pizza on Friday night. Each slice was 15 cm in length.

Calculate the area of the pizza that was ordered by Rohan. You can assume that the length of the pizza slice is equal to the pizza’s radius.

**Solution: **

A pizza is circular in shape. So we can use the **formula of a circle** to calculate the area of the pizza.

**Given:**** **Radius 15 cm

**Area of the Circle**

**\[\begin{align} &=\pi r^2 \\ & = \left(\frac{22}{7}\right) \times\, \left({15}\right)^2 \\ & = \left(\frac{22}{7}\right) \times\, \left({225}\right)\text{ cm}^2 \\ & = \text{47.14} \text{ cm}^2 \end{align} \]**

\(\therefore\) Area of the Pizza \(= 44.14 \text { cm}^2 \) |

**Solved Examples **

Example 1 |

Find the circumference and the area of a circle with radius 14 cm.

**Solution:**

**Given:**** **Radius of the circle = 14 cm

**Circumference of the Circle**

\[\begin{align} &= 2\pi r \\ & = 2 \times \left(\frac{22}{7}\right) \times \left({14}\right) \\ & = \text{88} \text{ cm} \end{align} \]

\(\therefore\) Circumference of the Circle \(= 88 \text { cm}\) |

**Area of a Circle **

\[\begin{align} &= \pi r^2 \\ &= \left(\frac{22}{7}\right) \times \left({14}\right) \times \left({14}\right) \\ &= \text{616} \text{ cm}^2 \end{align} \]

\(\therefore\) Area of the Circle \(= 616 \text { cm}^2\) |

Example 2 |

The ratio of the area of 2 circles is 4:9. Find the ratio of their radii.

**Solution:**

Let us assume the following.

- Radius of the 1st circle = \(r_1\)
- Area of the 1st circle = \(A_1\)
- Radius of the 2nd circle = \(r_2\)
- Area of the 2nd circle = \(A_2\)

**Given:**** **\(A_1\) **:** \(A_2\) = 4 **:** 9

**Area of a Circle \(=\pi r^2\)**

\[\begin{align} &⇒ \pi (r_1)^2 : \pi (r_2)^2 = 4:9 \\ &⇒ (r_1)^2 : (r_2)^2 =4:9 \\ &⇒ (r_1 : r_2)^2 = 4:9 \\ &⇒ (r_1 : r_2) = 2:3 \end{align} \]

\(\therefore\) Ratio of the Radii \(= 2:3 \) |

Example 3 |

A race-track is in the form of a circular ring. The inner radius of the track is 58 m and the outer radius is 63 m. Find the area of the race-track.

**Solution:**

**Given:**

\(R\) = 63 cm

\(r\) = 56 cm

**Let, Area of outer circle be **\(A_1\) and **Area of inner circle be **\(A_2\)

**Area of race track**

\[\begin{align} &= (\text A_1) - (\text A_2) \\ & = \pi(R)^2 - \pi (r)^2 \\ & = \pi (63^2 -56^2) \\ & = \left(\frac{22}{7}\right) \times 833 \\ & = 2618 \text { cm}^2 \end{align} \]

\(\therefore\) Area of the Race-track \(= 2618 \text { cm}^2\) |

Example 4 |

A wire is in the shape of an equilateral triangle. Each side of the triangle measures 7 cm. The wire is bent into the shape of a circle. Find the area of the circle that is formed.

**Solution:**

**Perimeter of the Equilateral Triangle**

\[\begin{align} &= \text{3} \times \text{Side} \\ & = 3 \times 7 \\ & = 21 \text { cm} \end{align} \]

\(\therefore\) Perimeter of the Triangle \(= 21\text { cm}\) |

Since, Perimeter of the Equilateral Triangle = Circumference of the Circle formed;

**Circumference of a Circle \( =2 \pi r\) **

\[\begin{align} 2 \pi r &= 21 \text { cm} \\ r&=3.34 \text { cm} \end{align} \]

Let us keep the above result in that form.

Therefore, **Radius of the Circle is 3.34 cm**

**Area of a Circle \(=\pi r^2\)**

\[\begin{align} &= \left(\frac{22}{7}\right) \times (3.34)^2 \\ &=35.06\text{ cm}^2 \end{align} \]

\(\therefore\) Area of the Circle \(= 35.06 \text { cm}^2\) |

Example 5 |

The time shown in a circular-clock is 3:00 pm. The length of the minute hand is 21 cm. Find the distance travelled by the tip of the minute hand when the time is 3:30 pm.

**Solution:**

When the minute-hand is at 3:30 pm, it covers half of the circle.

So, the distance travelled by the minute hand is actually half of the circumference.

Distance \(= \pi r\) (where \(r\) is the length of the minute hand)

\[\begin{align} &= \left(\frac{22}{7}\right) \times (21) \\ & =66\text{ cm} \end{align} \]

\(\therefore\) Distance travelled \(= 66 \text { cm}\) |

- How would you find the area of a sector, of the circle?
- To warn ships of underwater rocks, a lighthouse spreads a red-coloured light over a sector of angle 80°. The light reaches a distance of 16.5 km. Find the area of sea over which the ships are warned.
- Tom is riding a cycle to his friend's house. The wheel of his cycle has a radius of 20 cm and his friend lives 660 m away. How many times does his front wheel rotate when he goes and comes back from his friend's house?

**Practice Questions**

**Here are few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.**

1. | The radius of the circles is given as 1, 2, 3, 4 and 5 cm. Find the area of the shaded region. |

**Maths Olympiad Sample Papers**

IMO (International Maths Olympiad) is a competitive exam in Mathematics conducted annually for school students. It encourages children to develop their math solving skills from a competition perspective.

You can download the FREE grade-wise sample papers from below:

- IMO Sample Paper Class 1
- IMO Sample Paper Class 2
- IMO Sample Paper Class 3
- IMO Sample Paper Class 4
- IMO Sample Paper Class 5
- IMO Sample Paper Class 6
- IMO Sample Paper Class 7
- IMO Sample Paper Class 8
- IMO Sample Paper Class 9
- IMO Sample Paper Class 10

To know more about the Maths Olympiad you can **click here**

**Frequently Asked Questions (FAQs) **

## 1. How do you find the area of a circle?

The area of a circle is \(\pi\) multiplied by the square of the radius. It can be found using the formula given below:

Area of a Circle | \(\pi r^2\) |
---|

Where \(r\) is the radius of the circle and \(\pi \approx 3.14\)

## 2. What is the formula of area of the circle?

Formula for area of the circle is given below:

Area of a Circle when the radius \(\left(r\right)\) is known | \(\pi r^2\) |
---|---|

Area of a Circle when the diameter \(\left(d\right)\) is known | \(\begin{align} \frac{\pi}{4} \!\times\! \text{d}^2\end{align}\) |

## 3. How do you find the area and circumference of a circle?

The area and circumference of a circle can be calculated using the below formulas:

Circumference | \(2 \pi r\) |
---|---|

Area of a Circle | \(\pi r^2\) |

## 4. Why is the Area of a Circle \(\pi r^2\)?

- A circle can be divided into many small sectors which can then be rearranged accordingly to form a parallelogram.
- When the circle is divided into even smaller sectors, it gradually becomes the shape of a rectangle.
- We can clearly see that one of the sides of the rectangle will be the radius and the other will be half the length of the circumference, i.e, \(\pi r\)
- As we know that the area of a rectangle is its length multiplied to the breadth which is \(\pi r\) multiplied to \(r\).
- Therefore, the area of the circle is \(\pi r^2\)