What is a Circle?

What is Area of Circle

Area of Circle Example & Formulas

Derivation of Area of a Circle 

Surface Area of Circle Formula

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 \(= 44.14 \text { cm}^2 \)

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 \(= 66 \text { cm}\)