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Circumference of a Circle: Definition and Formula Area of a Circle: Definition and Formula Relationship between Circumference and Area of a Circle


Circles: Circumference and Area



Circumference of a Circle: Definition and Formula

The circumference of a circle is the total distance around its boundary. It is the circular equivalent of the perimeter of a polygon. Imagine 'unrolling' the circle's edge into a straight line; the length of that line is the circumference.


Key Terms Related to a Circle

Circle showing radius r and diameter d

The Constant $\pi$ (Pi)

The definition of $\pi$ is the ratio of a circle's circumference ($C$) to its diameter ($d$).

$\mathbf{\pi = \frac{Circumference}{Diameter} = \frac{C}{d}}$

... (1)

This relationship is constant for all circles. While $\pi$ is irrational and its exact value cannot be written as a simple fraction or terminating decimal, several approximations are commonly used in calculations, depending on the required precision:


Formulas for Circumference

From the definition of $\pi = \frac{C}{d}$ (Equation 1), we can easily derive the formula for the circumference ($C$). Multiply both sides of the equation by $d$:

$\mathbf{C = \pi d}$

[Formula in terms of diameter]          ... (2)

Since the diameter ($d$) is twice the radius ($r$), i.e., $d = 2r$, we can substitute this relationship into formula (2) to get the circumference in terms of the radius:

$C = \pi (2r)$

[Substituting $d=2r$ into (2)]

$\mathbf{C = 2 \pi r}$

[Formula in terms of radius]          ... (3)

These two formulas, $C = \pi d$ and $C = 2 \pi r$, are the standard ways to calculate the circumference of a circle if either the diameter or the radius is known.

The unit of circumference is a linear unit, the same as the unit used for radius or diameter (e.g., cm, m, km).


Conceptual Understanding and Historical Context of $\pi$

The concept of $\pi$ and the relationship between circumference and diameter was known to ancient civilizations. They observed that for any circular object, the distance around it was always a bit more than three times the distance across its middle. Ancient mathematicians in Egypt, Babylon, India, and Greece calculated approximations for $\pi$. Archimedes of Syracuse (around 250 BCE) provided a rigorous method to approximate $\pi$ using the perimeters of inscribed and circumscribed polygons, showing that $\pi$ is between $3\frac{10}{71}$ and $3\frac{1}{7}$ (i.e., between $3.1408$ and $3.1428$). The modern understanding of $\pi$ as a constant ratio is a cornerstone of geometry and trigonometry.

The idea of relating the circle's circumference to polygons can also serve as a conceptual basis for understanding the formula. As you increase the number of sides of a regular polygon inscribed within a circle, its perimeter gets closer and closer to the circle's circumference. Similarly, the apothem (the perpendicular distance from the center to a side) of the polygon gets closer to the circle's radius. In the limit, as the number of sides approaches infinity, the polygon becomes indistinguishable from the circle, its perimeter becomes the circumference $C$, and its apothem becomes the radius $r$. This limiting process is a fundamental idea in calculus.


Examples

Example 1. Find the circumference of a circle with a radius of $14$ cm. (Use $\pi = \frac{22}{7}$)

Answer:

Given:

Radius of the circle, $r = 14$ cm.

Value of $\pi = \frac{22}{7}$.

To Find:

Circumference of the circle, $C$.

Solution:

Using the formula for circumference in terms of radius, $C = 2 \pi r$. Using formula (3) derived above:

"$C = 2 \pi r$"

Substitute the given values for $\pi$ and $r$:

"$C = 2 \times \frac{22}{7} \times 14 \$ \text{cm}$"

[Substituting values]

Simplify the calculation by cancelling the common factor 7:

"$C = 2 \times \frac{22}{\cancel{7}_1} \times \cancel{14}^2 \$ \text{cm}$"

"$C = 2 \times 22 \times 2 \$ \text{cm}$"

"$C = 44 \times 2 \$ \text{cm}$"

"$\mathbf{C = 88 \$\$ cm}$"

Therefore, the circumference of the circle is 88 cm.


Example 2. The diameter of a merry-go-round is $3.5$ metres. Find its circumference. (Use $\pi = \frac{22}{7}$)

Answer:

Given:

Diameter of the merry-go-round, $d = 3.5$ m.

Value of $\pi = \frac{22}{7}$.

To Find:

Circumference of the merry-go-round, $C$.

Solution:

Using the formula for circumference in terms of diameter, $C = \pi d$. Using formula (2) derived above:

"$C = \pi d$"

Substitute the given values for $\pi$ and $d$. It's helpful to write $3.5$ as a fraction $\frac{35}{10} = \frac{7}{2}$ or $3\frac{1}{2}$ when using $\pi = \frac{22}{7}$.

"$C = \frac{22}{7} \times 3.5 \$ \text{m}$"

"$C = \frac{22}{7} \times \frac{7}{2} \$ \text{m}$"

Simplify by cancelling common factors:

"$C = \frac{\cancel{22}^{11}}{\cancel{7}_1} \times \frac{\cancel{7}^1}{\cancel{2}_1} \$ \text{m}$"

"$C = 11 \times 1 \$ \text{m}$"

"$\mathbf{C = 11 \$\$ m}$"

Alternatively, using the radius: $r = \frac{d}{2} = \frac{3.5}{2} = 1.75$ m. Then $C = 2 \pi r = 2 \times \frac{22}{7} \times 1.75 = 2 \times \frac{22}{7} \times \frac{175}{100} = 2 \times \frac{22}{7} \times \frac{7}{4} = 2 \times \frac{22}{1} \times \frac{1}{4} = \frac{44}{4} = 11$ m.

The circumference of the merry-go-round is 11 metres (m).


Area of a Circle: Definition and Formula

The area of a circle is the measure of the two-dimensional space or region enclosed by its circumference. It represents the amount of surface the circle covers on a flat plane.


Formula for Area of a Circle

The area ($A$) of a circle is calculated using its radius ($r$) and the constant $\pi$. The formula states that the area is equal to $\pi$ times the square of the radius.

Circle showing the enclosed area

Formula in terms of Radius:

$\mathbf{A = \pi r^2}$

... (1)

where '$r$' is the radius of the circle and '$\pi$' is the constant pi ($\approx 3.14159$).

Formula in terms of Diameter:

Sometimes, the diameter ($d$) is given instead of the radius. Since the diameter is twice the radius ($d = 2r$), the radius can be expressed as half the diameter ($r = \frac{d}{2}$). Substitute this into the area formula (1):

"$A = \pi \left(\frac{d}{2}\right)^2$"

[Substituting $r = d/2$ into (1)]

Square the term in the parenthesis:

"$A = \pi \left(\frac{d^2}{2^2}\right) = \pi \frac{d^2}{4}$"

$\mathbf{A = \frac{\pi d^2}{4}}$

[Formula in terms of diameter]          ... (2)

This formula allows you to calculate the area directly if the diameter is known.

The unit of area is always in square units (e.g., $\text{cm}^2$, $\text{m}^2$, $\text{km}^2$), which is consistent with it being a two-dimensional measurement.


Conceptual Derivation (Rearranging Sectors)

The formula for the area of a circle, $A = \pi r^2$, can be conceptually understood and derived by imagining the circle being divided into many small sectors (like slices of a pizza) and then rearranging these sectors to form a shape whose area is easier to calculate.

Circle divided into many equal sectors

Imagine dividing the circle into a very large number of equal sectors. Now, carefully arrange these sectors alternately, pointing up and down, side by side. For instance, take the first sector pointing upwards, the second pointing downwards next to it, the third upwards, and so on.

Sectors arranged to form a near-rectangle

As the number of sectors increases, the curved edges of the sectors become nearly straight, and the resulting shape closely approximates a rectangle (or a parallelogram, which has the same area formula as a rectangle). Let's analyze the dimensions of this approximate rectangle:

The area of the approximate rectangle is given by Base $\times$ Height:

Area $\approx (\text{Base length}) \times (\text{Height})$

$\mathbf{A \approx (\pi r) \times r = \pi r^2}$

As the number of sectors into which the circle is divided increases and approaches infinity, the rearranged shape becomes a perfect rectangle with dimensions $\pi r$ and $r$. The area of this limiting rectangle is exactly $\pi r^2$. Since the area of the circle is the same as the area of the shape formed by rearranging its parts, the area of the circle is exactly $\pi r^2$.


Examples

Example 1. Calculate the area of a circular garden with a diameter of $21$ metres. (Use $\pi = \frac{22}{7}$)

Answer:

Given:

Diameter of the circular garden, $d = 21$ m.

Value of $\pi = \frac{22}{7}$.

To Find:

Area of the garden, $A$.

Solution:

First, we need the radius ($r$). The radius is half of the diameter:

"$r = \frac{d}{2} = \frac{21}{2} \$ \text{m}$"

[Relationship between radius and diameter]

Now, use the formula for the area of a circle in terms of radius, $A = \pi r^2$. Using formula (1) derived above:

"$A = \pi r^2$"

Substitute the values for $\pi$ and $r$:

"$A = \frac{22}{7} \times \left(\frac{21}{2} \$ \text{m}\right)^2$"

[Substituting values]

Calculate the square of the radius:

"$A = \frac{22}{7} \times \left(\frac{21}{2} \times \frac{21}{2}\right) \$ \text{m}^2$"

"$A = \frac{22}{7} \times \frac{441}{4} \$ \text{m}^2$"

Simplify the calculation by cancelling common factors:

"$A = \frac{\cancel{22}^{11}}{\cancel{7}_1} \times \frac{\cancel{441}^{63}}{\cancel{4}_2} \$ \text{m}^2$"

[Cancel 7 with 441 (441 = 7 * 63), Cancel 22 with 4]

"$A = \frac{11 \times 63}{2} \$ \text{m}^2$"

"$A = \frac{693}{2} \$ \text{m}^2$"

"$\mathbf{A = 346.5 \$\$ m^2}$"

Alternate Solution (using diameter formula):

Using the formula for area in terms of diameter, $A = \frac{\pi d^2}{4}$. Using formula (2) derived above:

"$A = \frac{\pi d^2}{4}$"

Substitute the values for $\pi$ and $d$:

"$A = \frac{1}{4} \times \frac{22}{7} \times (21 \$ \text{m})^2$"

[Substituting values]

"$A = \frac{1}{4} \times \frac{22}{7} \times 21 \$ \text{m} \times 21 \$ \text{m}$"

Simplify the calculation by cancelling common factors:

"$A = \frac{1}{\cancel{4}_2} \times \frac{\cancel{22}^{11}}{\cancel{7}_1} \times \cancel{21}^3 \times 21 \$ \text{m}^2$"

[Cancel 22 with 4 by 2, Cancel 7 with 21 by 7]

"$A = \frac{11 \times 3 \times 21}{2} \$ \text{m}^2$"

"$A = \frac{33 \times 21}{2} \$ \text{m}^2 = \frac{693}{2} \$ \text{m}^2$"

"$\mathbf{A = 346.5 \$\$ m^2}$"

Both methods yield the same result. The area of the circular garden is 346.5 square metres ($\text{m}^2$).


Relationship between Circumference and Area of a Circle

The circumference ($C$) and area ($A$) of a circle are both derived from its radius ($r$) using the constant $\pi$. This means that if you know either the circumference or the area, you can calculate the radius, and subsequently find the other quantity (area from circumference, or circumference from area).


The fundamental formulas connecting these quantities via the radius are:

$\mathbf{Circumference} (C) = \mathbf{2 \pi r}$

... (1)

$\mathbf{Area} (A) = \mathbf{\pi r^2}$

... (2)


Expressing Area in terms of Circumference

If we are given the circumference ($C$) and need to find the area ($A$) without first calculating the radius explicitly, we can derive a formula for $A$ in terms of $C$.

From the circumference formula (1), we can solve for the radius ($r$) in terms of the circumference ($C$) and $\pi$:

"$C = 2 \pi r$"

[From (1)]

Divide both sides by $2\pi$:

$\mathbf{r = \frac{C}{2 \pi}}$

... (3)

Now, substitute this expression for $r$ into the area formula (2):

"$A = \pi r^2$"

[From (2)]

"$A = \pi \left( \frac{C}{2 \pi} \right)^2$"

[Substituting r from (3)]

Calculate the square of the term in the parenthesis:

"$A = \pi \left( \frac{C^2}{(2 \pi)^2} \right) = \pi \left( \frac{C^2}{4 \pi^2} \right)$"

Simplify the expression by cancelling out $\pi$ from the numerator and denominator:

"$A = \frac{\cancel{\pi} C^2}{4 \cancel{\pi^2}^{\pi}}$"

$\mathbf{A = \frac{C^2}{4 \pi}}$

[Area in terms of Circumference]          ... (4)

This formula provides a direct way to compute the area if the circumference is the known value.


Expressing Circumference in terms of Area

If we are given the area ($A$) and need to find the circumference ($C$) without first calculating the radius explicitly, we can derive a formula for $C$ in terms of $A$.

From the area formula (2), we can solve for $r^2$ in terms of $A$ and $\pi$:

"$A = \pi r^2$"

[From (2)]

"$r^2 = \frac{A}{\pi}$"

Take the square root of both sides to find $r$. Since radius must be a positive length, we take the positive square root:

$\mathbf{r = \sqrt{\frac{A}{\pi}}}$

... (5)

Now, substitute this expression for $r$ into the circumference formula (1):

"$C = 2 \pi r$"

[From (1)]

"$C = 2 \pi \sqrt{\frac{A}{\pi}}$"

[Substituting r from (5)]

To simplify the expression, we can bring the term $2\pi$ inside the square root by squaring it $(2\pi)^2 = 4\pi^2$:

"$C = \sqrt{(2 \pi)^2 \times \frac{A}{\pi}}$"

"$C = \sqrt{4 \pi^2 \times \frac{A}{\pi}}$"

Simplify by cancelling out $\pi$ from the numerator and denominator inside the square root:

"$C = \sqrt{4 \cancel{\pi^2}^{\pi} \times \frac{A}{\cancel{\pi}}}$"

"$C = \sqrt{4 \pi A}$"

Take the square root of the perfect square 4:

$\mathbf{C = 2 \sqrt{\pi A}}$

[Circumference in terms of Area]          ... (6)

This formula provides a direct way to compute the circumference if the area is the known value.


Example

Example 1. The circumference of a circular sheet is $154$ m. Find its radius and also find its area. (Use $\pi = \frac{22}{7}$)

Answer:

Given:

Circumference of the circular sheet, $C = 154$ m.

Value of $\pi = \frac{22}{7}$.

To Find:

1. Radius of the sheet, $r$.

2. Area of the sheet, $A$.

Solution:

Part 1: Finding the radius ($r$)

We can use the formula relating circumference and radius, $C = 2 \pi r$.

"$C = 2 \pi r$"

[From (1)]

Substitute the given values for $C$ and $\pi$:

"$154 \$ \text{m} = 2 \times \frac{22}{7} \times r$"

[Substituting values]

"$154 = \frac{44}{7} \times r$"

To find $r$, multiply both sides of the equation by the reciprocal of $\frac{44}{7}$, which is $\frac{7}{44}$:

"$r = 154 \times \frac{7}{44} \$ \text{m}$"

Simplify the fraction. Both 154 and 44 are divisible by 22 ($154 \div 22 = 7$ and $44 \div 22 = 2$):

"$r = \frac{\cancel{154}^7}{\cancel{44}_2} \times 7 \$ \text{m}$"

"$r = \frac{7 \times 7}{2} \$ \text{m} = \frac{49}{2} \$ \text{m}$"

"$\mathbf{r = 24.5 \$\$ m}$"

The radius of the circular sheet is 24.5 metres (m).

Part 2: Finding the area ($A$)

Now that we have the radius ($r = 24.5$ m), we can use the formula for the area of a circle, $A = \pi r^2$. Using formula (2):

"$A = \pi r^2$"

[From (2)]

Substitute the values for $\pi$ and $r$. It's easier to use $r = \frac{49}{2}$ m for calculations involving $\pi = \frac{22}{7}$:

"$A = \frac{22}{7} \times \left(\frac{49}{2} \$ \text{m}\right)^2$"

[Substituting values]

"$A = \frac{22}{7} \times \frac{49}{2} \$ \text{m} \times \frac{49}{2} \$ \text{m}$"

Simplify the calculation by cancelling common factors:

"$A = \frac{\cancel{22}^{11}}{\cancel{7}_1} \times \frac{\cancel{49}^7}{\cancel{2}_1} \times \frac{49}{2} \$ \text{m}^2$"

[Cancel 22 with 2, Cancel 7 with 49]

"$A = 11 \times 7 \times \frac{49}{2} \$ \text{m}^2$"

"$A = 77 \times \frac{49}{2} \$ \text{m}^2$"

Perform the multiplication and division:

"$77 \times 49 = 3773$"

"$A = \frac{3773}{2} \$ \text{m}^2$"

"$\mathbf{A = 1886.5 \$\$ m^2}$"

Alternate Method for Part 2 (using formula 4):

We can also calculate the area directly from the circumference ($C = 154$ m) using the formula $A = \frac{C^2}{4 \pi}$. Using formula (4):

"$A = \frac{C^2}{4 \pi}$"

Substitute the given values for $C$ and $\pi$:

"$A = \frac{(154 \$ \text{m})^2}{4 \times \frac{22}{7}}$"

[Substituting values]

"$A = \frac{154 \times 154 \$ \text{m}^2}{\frac{88}{7}}$"

To divide by a fraction, multiply by its reciprocal:

"$A = \frac{154 \times 154 \times 7}{88} \$ \text{m}^2$"

Simplify by dividing 154 and 88 by their greatest common divisor, which is 22:

"$A = \frac{\cancel{154}^7 \times 154 \times 7}{\cancel{88}_4} \$ \text{m}^2$"

[Divide 154 and 88 by 22]

Simplify 154 and 4 by dividing by 2:

"$A = \frac{7 \times \cancel{154}^{77} \times 7}{\cancel{4}_2} \$ \text{m}^2$"

[Divide 154 and 4 by 2]

"$A = \frac{7 \times 77 \times 7}{2} \$ \text{m}^2$"

"$A = \frac{49 \times 77}{2} \$ \text{m}^2$"

"$A = \frac{3773}{2} \$ \text{m}^2$"

"$\mathbf{A = 1886.5 \$\$ m^2}$"

Both methods give the same result for the area. The area of the circular sheet is 1886.5 square metres ($\text{m}^2$).