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Complete Course of Mathematics
Topic 1: Numbers & Numerical Applications Topic 2: Algebra Topic 3: Quantitative Aptitude
Topic 4: Geometry Topic 5: Construction Topic 6: Coordinate Geometry
Topic 7: Mensuration Topic 8: Trigonometry Topic 9: Sets, Relations & Functions
Topic 10: Calculus Topic 11: Mathematical Reasoning Topic 12: Vectors & Three-Dimensional Geometry
Topic 13: Linear Programming Topic 14: Index Numbers & Time-Based Data Topic 15: Financial Mathematics
Topic 16: Statistics & Probability


Content On This Page
Sector of a Circle: Definition and Properties Length of Arc of a Sector Area of a Sector of a Circle
Segment of a Circle: Definition and Properties Area of a Segment of a Circle Perimeter and Area of Sector and Segment (Consolidated Formulas)


Areas Related to Circles: Sectors and Segments



Sector of a Circle: Definition and Properties

In geometry, a sector of a circle is a region bounded by two radii of the circle and the arc connecting their endpoints on the circumference.


Visually, a sector resembles a slice of a circular pie or pizza. It is defined by the central angle formed by the two radii.

Circle with center O, radii OA and OB, arc AB, and central angle theta forming a sector

In the diagram, O is the center of the circle, OA and OB are two radii, and $\overarc{AB}$ is the arc between points A and B. The region OAB enclosed by OA, OB, and $\overarc{AB}$ is a sector.


Components of a Sector


Types of Sectors

For any central angle less than $180^\circ$, there are two sectors associated with it:

Circle showing a minor sector and the corresponding major sector

If the central angle is exactly $180^\circ$, both sectors are identical and are called semicircles.


Properties of Sectors

Understanding these properties and the definition of a sector is essential for calculating its arc length and area.


Length of Arc of a Sector

The length of the arc ($l$) of a sector is the length of the curved part of the sector's boundary. It is a fraction of the total circumference of the circle, and this fraction is determined by the central angle of the sector.


The key relationship used to find the arc length is that the ratio of the sector's central angle to the total angle in a circle ($360^\circ$ or $2\pi$ radians) is equal to the ratio of the sector's arc length to the circle's total circumference ($C = 2\pi r$).

$\mathbf{\frac{Arc \$\$ Length}{Circumference} = \frac{Central \$\$ Angle}{Total \$\$ Angle \$\$ in \$\$ a \$\$ Circle}}$

... (1)


Formula Derivation

Let the radius of the circle be $r$, the circumference be $C = 2\pi r$, the central angle of the sector be $\theta$, and the arc length be $l$.

Using Degrees:

If the central angle $\theta$ is measured in degrees, the total angle in a circle is $360^\circ$. Substitute these values into the proportionality relationship (1):

$\frac{l}{2\pi r} = \frac{\theta}{360^\circ}$

[Using (1) with angle in degrees]

To find $l$, multiply both sides of the equation by $2\pi r$:

"$l = \frac{\theta}{360^\circ} \times 2\pi r$"

Rearranging for clarity:

$\textbf{Length \$\$ of \$\$ Arc} (l) = \mathbf{\frac{\theta}{360^\circ} \times 2\pi r}$

[Formula for arc length, $\theta$ in degrees]          ... (2)

Using Radians:

If the central angle $\theta$ is measured in radians, the total angle in a circle is $2\pi$ radians. Substitute these values into the proportionality relationship (1):

$\frac{l}{2\pi r} = \frac{\theta}{2\pi}$

[Using (1) with angle in radians]

To find $l$, multiply both sides of the equation by $2\pi r$:

"$l = \frac{\theta}{2\pi} \times 2\pi r$"

Simplify the expression by cancelling out $2\pi$ from the numerator and denominator:

"$l = \frac{\theta}{\cancel{2\pi}} \times \cancel{2\pi} r$"

$\textbf{Length \$\$ of \$\$ Arc} (l) = \mathbf{r \theta}$

[Formula for arc length, $\theta$ in radians]          ... (3)

This simpler formula ($l = r\theta$) is one of the main reasons why radians are the standard unit for angles in many areas of mathematics and physics, particularly when dealing with circular motion or sectors.


Examples

Example 1. Find the length of the arc of a sector with a central angle of $60^\circ$ in a circle of radius $21$ cm. (Use $\pi = \frac{22}{7}$)

Answer:

Given:

Central angle, $\theta = 60^\circ$.

Radius of the circle, $r = 21$ cm.

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

To Find:

Length of the arc, $l$.

Solution:

Since the angle is given in degrees, we use the formula for arc length in degrees:

"$l = \frac{\theta}{360^\circ} \times 2\pi r$"

[Formula (2)]

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

"$l = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21 \$ \text{cm}$"

[Substituting values]

Simplify the fraction involving the angle:

"$l = \frac{\cancel{60}^{1}}{\cancel{360}_6} \times 2 \times \frac{22}{7} \times 21 \$ \text{cm}$"

[Simplifying $\frac{60}{360}$]

"$l = \frac{1}{6} \times 2 \times \frac{22}{7} \times 21 \$ \text{cm}$"

Now perform the multiplication, cancelling common factors:

"$l = \frac{1}{\cancel{6}_3} \times \cancel{2}_1 \times \frac{22}{\cancel{7}_1} \times \cancel{21}^3 \$ \text{cm}$"

[Cancel 6 with 2 and 3, Cancel 7 with 21]

"$l = \frac{1}{3} \times 1 \times 22 \times 3 \$ \text{cm}$"

"$l = \frac{22 \times \cancel{3}}{\cancel{3}} \$ \text{cm}$"

"$\mathbf{l = 22 \$\$ cm}$"

Therefore, the length of the arc of the sector is 22 centimetres (cm).


Example 2. Find the length of the arc of a sector with a central angle of $1.5$ radians in a circle of radius $8$ cm.

Answer:

Given:

Central angle, $\theta = 1.5$ radians.

Radius of the circle, $r = 8$ cm.

To Find:

Length of the arc, $l$.

Solution:

Since the angle is given in radians, we use the simpler formula for arc length in radians:

"$l = r \theta$"

[Formula (3)]

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

"$l = 8 \$ \text{cm} \times 1.5$"

[Substituting values]

Perform the multiplication:

"$8 \times 1.5 = 8 \times \frac{15}{10} = 8 \times \frac{3}{2} = \frac{24}{2} = 12$"

"$\mathbf{l = 12 \$\$ cm}$"

Therefore, the length of the arc is 12 centimetres (cm).


Area of a Sector of a Circle

The area of a sector is the measure of the two-dimensional region enclosed by the two radii and the arc of the sector. Similar to the arc length, the area of a sector is a fraction of the total area of the circle, determined by the size of the central angle.


The key relationship used to find the area of a sector is that the ratio of the sector's area to the total area of the circle ($A_{\text{circle}} = \pi r^2$) is equal to the ratio of the sector's central angle ($\theta$) to the total angle in a circle ($360^\circ$ or $2\pi$ radians).

$\mathbf{\frac{Area \$\$ of \$\$ Sector}{Area \$\$ of \$\$ Circle} = \frac{Central \$\$ Angle}{Total \$\$ Angle \$\$ in \$\$ a \$\$ Circle}}$

... (1)


Formula Derivation

Let the radius of the circle be $r$, the area of the circle be $A_{\text{circle}} = \pi r^2$, the central angle of the sector be $\theta$, and the area of the sector be $A_{\text{sector}}$.

Using Degrees:

If the central angle $\theta$ is measured in degrees, the total angle in a circle is $360^\circ$. Substitute these values into the proportionality relationship (1):

$\frac{A_{\text{sector}}}{\pi r^2} = \frac{\theta}{360^\circ}$

[Using (1) with angle in degrees]

To find $A_{\text{sector}}$, multiply both sides of the equation by $\pi r^2$:

"$A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2$"

Rearranging for clarity:

$\textbf{Area \$\$ of \$\$ Sector} (A_{\text{sector}}) = \mathbf{\frac{\theta}{360^\circ} \times \pi r^2}$

[Formula for sector area, $\theta$ in degrees]          ... (2)

Using Radians:

If the central angle $\theta$ is measured in radians, the total angle in a circle is $2\pi$ radians. Substitute these values into the proportionality relationship (1):

$\frac{A_{\text{sector}}}{\pi r^2} = \frac{\theta}{2\pi}$

[Using (1) with angle in radians]

To find $A_{\text{sector}}$, multiply both sides of the equation by $\pi r^2$:

"$A_{\text{sector}} = \frac{\theta}{2\pi} \times \pi r^2$"

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

"$A_{\text{sector}} = \frac{\theta}{2\cancel{\pi}} \times \cancel{\pi} r^2$"

$\textbf{Area \$\$ of \$\$ Sector} (A_{\text{sector}}) = \mathbf{\frac{1}{2} r^2 \theta}$

[Formula for sector area, $\theta$ in radians]          ... (3)

This formula using radians is particularly elegant.

Formula using Arc Length:

We can also express the area of a sector in terms of its arc length ($l$) and radius ($r$). From the formula for arc length in radians, $l = r\theta$ (or $\theta = l/r$), assuming $\theta$ is in radians. Substitute this expression for $\theta$ into the sector area formula in radians (3):

"$A_{\text{sector}} = \frac{1}{2} r^2 \theta$"

[From (3)]

"$A_{\text{sector}} = \frac{1}{2} r^2 \left(\frac{l}{r}\right)$"

[Substitute $\theta = l/r$]

Simplify the expression:

"$A_{\text{sector}} = \frac{1}{2} r^{\cancel{2}} \frac{l}{\cancel{r}}$"

$\textbf{Area \$\$ of \$\$ Sector} (A_{\text{sector}}) = \mathbf{\frac{1}{2} l r}$

[Formula for sector area, using arc length and radius]          ... (4)

This formula resembles the area of a triangle ($\frac{1}{2} \times \text{base} \times \text{height}$), where the arc length $l$ serves as the 'base' and the radius $r$ as the 'height'.


Examples

Example 1. Find the area of a sector of a circle with radius $4$ cm and an angle of $30^\circ$. Also, find the area of the corresponding major sector. (Use $\pi = 3.14$)

Answer:

Given:

Radius of the circle, $r = 4$ cm.

Central angle of the minor sector, $\theta = 30^\circ$.

Value of $\pi = 3.14$.

To Find:

1. Area of the minor sector ($A_{\text{minor}}$).

2. Area of the corresponding major sector ($A_{\text{major}}$).

Solution:

Part 1: Area of the minor sector ($A_{\text{minor}}$)

Since the angle is given in degrees, we use the formula for the area of a sector in degrees:

"$A_{\text{minor}} = \frac{\theta}{360^\circ} \times \pi r^2$"

[Formula (2)]

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

"$A_{\text{minor}} = \frac{30^\circ}{360^\circ} \times 3.14 \times (4 \$ \text{cm})^2$"

[Substituting values]

Simplify the fraction involving the angle:

"$A_{\text{minor}} = \frac{\cancel{30}^{1}}{\cancel{360}_{12}} \times 3.14 \times 16 \$ \text{cm}^2$"

[Simplifying $\frac{30}{360}$ and $4^2$]

"$A_{\text{minor}} = \frac{1}{12} \times 3.14 \times 16 \$ \text{cm}^2$"

Perform the multiplication and simplification:

"$A_{\text{minor}} = \frac{3.14 \times 16}{12} \$ \text{cm}^2$"

"$A_{\text{minor}} = \frac{3.14 \times \cancel{16}^4}{\cancel{12}_3} \$ \text{cm}^2$"

[Simplifying $\frac{16}{12}$ by dividing by 4]

"$A_{\text{minor}} = \frac{3.14 \times 4}{3} \$ \text{cm}^2 = \frac{12.56}{3} \$ \text{cm}^2$"

Calculate the decimal value:

"$12.56 \div 3 \approx 4.1866...$"

"$\mathbf{A_{\text{minor}} \approx 4.19 \$\$ cm^2}$"

[Rounding to two decimal places]

The area of the minor sector is approximately 4.19 square centimetres ($\text{cm}^2$).

Part 2: Area of the corresponding major sector ($A_{\text{major}}$)

Method 1: Using the angle of the major sector

The central angle of the major sector is the remaining angle in the circle:

Angle of Major Sector $= 360^\circ - \theta = 360^\circ - 30^\circ = 330^\circ$

Using the formula for sector area with this angle:

"$A_{\text{major}} = \frac{330^\circ}{360^\circ} \times \pi r^2$"

[Using Formula (2)]

Substitute the values:

"$A_{\text{major}} = \frac{\cancel{330}^{11}}{\cancel{360}_{12}} \times 3.14 \times (4 \$ \text{cm})^2$"

[Simplifying $\frac{330}{360}$ and $4^2$]

"$A_{\text{major}} = \frac{11}{12} \times 3.14 \times 16 \$ \text{cm}^2$"

"$A_{\text{major}} = \frac{11 \times 3.14 \times \cancel{16}^4}{\cancel{12}_3} \$ \text{cm}^2$"

[Simplifying $\frac{16}{12}$ by dividing by 4]

"$A_{\text{major}} = \frac{11 \times 12.56}{3} \$ \text{cm}^2 = \frac{138.16}{3} \$ \text{cm}^2$"

"$138.16 \div 3 \approx 46.0533...$"

"$\mathbf{A_{\text{major}} \approx 46.05 \$\$ cm^2}$"

[Rounding to two decimal places]

Method 2: Subtracting minor sector area from total circle area

Calculate the total area of the circle:

Area of Circle ($A_{\text{circle}}$) $= \pi r^2 = 3.14 \times (4 \$ \text{cm})^2$"

"$A_{\text{circle}} = 3.14 \times 16 \$ \text{cm}^2$"

"$A_{\text{circle}} = 50.24 \$ \text{cm}^2$"

The area of the major sector is the total area minus the minor sector area:

"$A_{\text{major}} = A_{\text{circle}} - A_{\text{minor}}$"

"$\approx 50.24 \$ \text{cm}^2 - 4.1867 \$ \text{cm}^2$"

[Using the more precise value for $A_{\text{minor}}$]

"$\approx 46.0533 \$ \text{cm}^2$"

"$\mathbf{A_{\text{major}} \approx 46.05 \$\$ cm^2}$"

[Rounding to two decimal places]

Both methods give approximately the same result. The area of the corresponding major sector is approximately 46.05 square centimetres ($\text{cm}^2$).



Segment of a Circle: Definition and Properties

A segment of a circle is a region of the circle bounded by a chord and the arc that the chord subtends. In simpler terms, it's the area cut from a circle by a straight line (the chord).


Think of a circular pizza or cake. If you make a straight cut across the cake (this is the chord), the smaller piece you separate (bounded by the cut and the curved crust) is a segment of the circle.

Circle showing a chord AB and the arc AB enclosing a segment

In the diagram, AB is a chord. The region enclosed by the chord AB and the arc $\overarc{AB}$ is a segment of the circle.


Components of a Segment


Types of Segments

Any chord that is not a diameter divides the circle into two distinct segments:

Circle showing a minor segment and the corresponding major segment divided by a chord

If the chord is a diameter, it passes through the center and divides the circle into two equal parts. Each of these equal segments is called a semicircle.


Relationship with Sector and Triangle

A segment is closely related to a sector that shares the same arc and endpoints of the chord. Consider a circle with center O, and a chord AB. The radii OA and OB form a sector OAB.

Diagram showing sector OAB and triangle OAB used to find segment area

The segment bounded by chord AB and arc AB is the region between the chord AB and the arc AB. This region can be seen as the difference between the sector OAB and the triangle OAB formed by the two radii and the chord.

This relationship is crucial for calculating the area of a segment, as shown in the next section.


Area of a Segment of a Circle

The area of a segment of a circle is calculated by subtracting the area of the triangle formed by the two radii and the chord from the area of the corresponding sector. This method applies to both minor and major segments, though it's usually easier to calculate the minor segment and subtract from the total circle area for the major segment.


Formula Derivation (for Minor Segment)

Consider the minor segment formed by chord AB and minor arc AB in a circle with center O and radius $r$. Let the central angle subtended by the arc (and chord) at the center be $\theta$.

Diagram showing sector OAB and triangle OAB used to find segment area

From the previous section, we know that the area of the minor segment is:

$\mathbf{A_{\text{segment}} = Area(Minor \$\$ Sector \$\$ OAB) - Area(\triangle OAB)}$

... (1)

We need the formulas for the area of the sector and the area of the triangle $\triangle \text{OAB}$ in terms of $r$ and $\theta$.

Area of Sector OAB:

The area of a sector with radius $r$ and central angle $\theta$ is:

Area of Triangle OAB:

In $\triangle \text{OAB}$, we have two sides equal to the radius ($OA=OB=r$), and the included angle is $\theta$. The area of a triangle given two sides and the included angle is $\frac{1}{2} \times \text{side}_1 \times \text{side}_2 \times \sin(\text{included angle})$.

Area($\triangle$ OAB) $= \frac{1}{2} \times OA \times OB \times \sin(\theta)$

$\mathbf{Area(\triangle OAB) = \frac{1}{2} r^2 \sin(\theta)}$

... (3)

This formula for the area of the triangle works whether $\theta$ is in degrees or radians, as long as the $\sin(\theta)$ function is evaluated using the correct angle unit.

Combining to find Segment Area:

Substitute the formulas for the area of the sector and the area of the triangle into equation (1).

Area of Major Segment:

The area of the major segment is the area of the entire circle minus the area of the corresponding minor segment:

$\mathbf{A_{\text{major segment}} = Area(Circle) - Area(Minor \$\$ Segment)}$

... (5)

$\mathbf{A_{\text{major segment}} = \pi r^2 - A_{\text{minor segment}}}$

[Using Formula (5)]


Example

Example 1. Find the area of the segment corresponding to a chord of a circle of radius $15$ cm which subtends an angle of $120^\circ$ at the center. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

Answer:

Given:

Radius of the circle, $r = 15$ cm.

Central angle subtended by the chord, $\theta = 120^\circ$.

Approximate values: $\pi \approx 3.14$, $\sqrt{3} \approx 1.73$.

To Find:

Area of the corresponding minor segment.

Solution:

The area of the minor segment is the difference between the area of the sector formed by the angle $\theta$ and the area of the triangle formed by the two radii and the chord.

Step 1: Calculate the Area of the Sector

Using the formula for the area of a sector with angle in degrees:

"$A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2$"

[Using Formula (2a)]

Substitute the given values:

"$A_{\text{sector}} = \frac{120^\circ}{360^\circ} \times 3.14 \times (15 \$ \text{cm})^2$"

[Substituting values]

Simplify the fraction $\frac{120}{360} = \frac{1}{3}$ and calculate $(15)^2 = 225$:

"$A_{\text{sector}} = \frac{1}{3} \times 3.14 \times 225 \$ \text{cm}^2$"

"$A_{\text{sector}} = 3.14 \times \frac{225}{3} \$ \text{cm}^2 = 3.14 \times 75 \$ \text{cm}^2$"

[Simplifying $\frac{225}{3}$]

Calculate the product:

$\begin{array}{cc}& & 3\ . & 1 & 4 \\ \times & & & 7 & 5 \\ \hline && 1\ 5 & 7 & 0 \\ & 219 & 8 & \times \\ \hline & 235\ . & 5 & 0 \\ \hline \end{array}$

"$A_{\text{sector}} = 235.50 \$ \text{cm}^2$"

Step 2: Calculate the Area of the Triangle (OAB)

Using the formula for the area of $\triangle \text{OAB}$:

"$A_{\triangle OAB} = \frac{1}{2} r^2 \sin(\theta)$"

[Using Formula (3)]

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

"$A_{\triangle OAB} = \frac{1}{2} \times (15 \$ \text{cm})^2 \times \sin(120^\circ)$"

[Substituting values]

Calculate $(15)^2 = 225$ and find the value of $\sin(120^\circ)$. We know that $\sin(120^\circ)$ lies in the second quadrant, where sine is positive. $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}$.

"$A_{\triangle OAB} = \frac{1}{2} \times 225 \$ \text{cm}^2 \times \frac{\sqrt{3}}{2}$"

[Using $\sin(120^\circ) = \frac{\sqrt{3}}{2}$]

"$A_{\triangle OAB} = \frac{225 \sqrt{3}}{4} \$ \text{cm}^2$"

Substitute the approximate value of $\sqrt{3} = 1.73$:

"$A_{\triangle OAB} = \frac{225 \times 1.73}{4} \$ \text{cm}^2$"

Calculate the numerator: $225 \times 1.73$

$\begin{array}{cc}& & 2 & 2 & 5 \\ \times & & 1\ . & 7 & 3 \\ \hline && 6 & 7 & 5 \\ & 157 & 5 & \times \\ 225 & \times & \times \\ \hline 389\ . & 2 & 5 \\ \hline \end{array}$

"$A_{\triangle OAB} = \frac{389.25}{4} \$ \text{cm}^2$"

Perform the division:

$\begin{array}{r} 97\ . \ 3\ 1\ 2\ 5\phantom{)} \\ 4{\overline{\smash{\big)}\,389\ . \ 2500\phantom{)}}} \\ \underline{-~\phantom{(}36\phantom{.0000)}} \\ 29\phantom{.0000)} \\ \underline{-~\phantom{(}28\phantom{.0000)}} \\ 1\ 2\phantom{000)} \\ \underline{-~\phantom{()1}2\phantom{000)}} \\ 0\ 5\phantom{00)} \\ \underline{-~\phantom{()0}4\phantom{00)}} \\ 10\phantom{0)} \\ \underline{-~\phantom{()0}8\phantom{0)}} \\ 20\phantom{)} \\ \underline{-~\phantom{()}20\phantom{)}}\\ 0\phantom{)} \end{array}$

"$A_{\triangle OAB} = 97.3125 \$ \text{cm}^2$"

Step 3: Calculate the Area of the Segment

Using formula (1):

"$A_{\text{segment}} = A_{\text{sector}} - A_{\triangle OAB}$"

Substitute the calculated areas:

"$A_{\text{segment}} = 235.50 \$ \text{cm}^2 - 97.3125 \$ \text{cm}^2$"

[Substituting areas]

Perform the subtraction:

$\begin{array}{ccccccc} & 2 & 3 & 5\ . & 5 & 0 & 0 & 0 \\ - & & 9 & 7\ . & 3 & 1 & 2 & 5 \\ \hline & 1 & 3 & 8\ . & 1 & 8 & 7 & 5 \\ \hline \end{array}$

"$A_{\text{segment}} = 138.1875 \$ \text{cm}^2$"

Rounding to two decimal places as the input values for $\pi$ and $\sqrt{3}$ suggest approximate calculation:

"$\mathbf{A_{\text{segment}} \approx 138.19 \$\$ cm^2}$"

Therefore, the area of the segment is approximately 138.19 square centimetres ($\text{cm}^2$).


Perimeter and Area of Sector and Segment (Consolidated Formulas)

This section provides a summary of the key formulas for calculating the perimeter and area of sectors and segments of a circle with radius $r$ and central angle $\theta$. It's important to use consistent units for the angle ($\theta$) – either degrees or radians – when applying these formulas.


Consolidated Formulas

Quantity Formula ( $\theta$ in degrees) Formula ( $\theta$ in radians) Notes / Additional Information
Arc Length ($l$)

(Length of the curved boundary)

$l = \frac{\theta}{360^\circ} \times 2\pi r$ $l = r \theta$ $\theta$ must be in the specified unit (degrees or radians) for each formula.
Area of Sector

(Region bounded by two radii and the arc)

$A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2$ $A_{\text{sector}} = \frac{1}{2} r^2 \theta$ Both degree and radian formulas are consistent. The area can also be found using the arc length: $A_{\text{sector}} = \frac{1}{2} l r$.
Area of Triangle formed by Radii and Chord ($\triangle$ OAB) $A_{\triangle} = \frac{1}{2} r^2 \sin(\theta)$ This triangle is formed by the two radii OA, OB and the chord AB. Formula is the same for $\theta$ in degrees or radians, but ensure $\sin(\theta)$ is evaluated using the correct mode for $\theta$.
Area of Minor Segment

(Region bounded by chord and minor arc)

$A_{\text{minor seg}} = A_{\text{sector}} - A_{\triangle}$

$A_{\text{minor seg}} = \left( \frac{\theta}{360^\circ} \times \pi r^2 \right) - \left( \frac{1}{2} r^2 \sin(\theta) \right)$

$A_{\text{minor seg}} = A_{\text{sector}} - A_{\triangle}$

$A_{\text{minor seg}} = \frac{1}{2} r^2 \theta - \frac{1}{2} r^2 \sin(\theta)$
$A_{\text{minor seg}} = \frac{1}{2} r^2 (\theta - \sin\theta)$

Difference between the sector area and the triangle area. Use the $\theta$ in radians formula for the $\frac{1}{2}r^2(\theta - \sin\theta)$ form.
Area of Major Segment

(Region bounded by chord and major arc)

$A_{\text{major seg}} = \text{Area of Circle} - A_{\text{minor seg}}$

$A_{\text{major seg}} = \pi r^2 - A_{\text{minor seg}}$

Sum of Area(Major Sector) + Area(Triangle). Usually easier to calculate via subtraction.
Perimeter of Sector

(Boundary length of sector)

$P_{\text{sector}} = \text{Length of Arc} (l) + 2 \times \text{Radius} (r)$

$P_{\text{sector}} = l + 2r$

Boundary consists of the arc and the two radii. Calculate arc length using appropriate formula ($\theta$ in deg or rad).
Perimeter of Segment

(Boundary length of segment)

$P_{\text{segment}} = \text{Length of Arc} (l) + \text{Length of Chord AB}$ Boundary consists of the arc and the chord. Chord length AB = $2r \sin(\theta/2)$. Calculate arc length $l$ using appropriate formula. Use consistent units for $\theta$ for both $l$ and $\sin(\theta/2)$.

Important Reminders: