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**Major Arc|Definition & Meaning**

**Definition**

The** lengthier of the two arcs** joining any two points on a circle’s perimeter is the **major arc**, while the shorter one is the **minor arc**.

Figure 1 – The major and minor arcs between points A and B on a circle.

## What Is an Arc?

An **arc** is a line **curved around** a single point. In geometry, the term refers to a **part of** a circle’s **circumference** (i.e., its boundary). Therefore, the length of any arc of a circle is always **smaller** than its circumference.

Forming an arc requires** two** **distinct points** on the circle. Let us call these two points A and B. The part of the circle between these two points is arc AB. We can calculate the **arc’s length** with the formula:

l = r x θ

Given that O is the **center** of the circle, then **l** is the **arc length**, $\boldsymbol{\mathsf{\theta}}$ is the **angle** in radians between the two radii (OA and OB) bounding it, and **r** is the value of the **circle’s radius**. We illustrate these parts of the equation in** Figure 2** below.

**Why Do the Major and Minor Arcs Exist?**

Because there are two ways to traverse a circle (clockwise or anti-clockwise), there are **two ways of describing the resulting arc AB**. You could go **clockwise** or **anti-clockwise** from A to B, resulting in **distinct arcs** with different lengths.

Usually, one of these two arcs is **longer** than the other one. Whatever the case, the **sum** of the two arcs’ lengths always **equals the circumference**.

**Major and Minor Sectors and Segments**

**Sectors** and **segments** represent the** inner portions** of the circle in addition to the **outer boundary** (the arc). Again, since the circle splits into two parts, we have the concept of **major **and** minor **sectors and segments. If the arc AB was a **major arc**, the **corresponding** segment and sector are the **major segment** and **sector**, and vice versa.

Figure 2 – The major and minor arcs reuslting in major and minor segments and sectors.

Note:

area of major segment + area of minor segment = area of the circle

area of major sector + area of minor sector = area of the circle

The segment AB of a circle **splits it into two** parts **about a chord** joining points A and B. The result looks like two different semicircles. For example, when you cut a cake in a straight line such that the two halves are not equal, you have two segments of the cake.

area of segment (θ in radians) = $\displaystyle \mathsf{\left( \frac{\theta-\sin(\theta)}{2} \right)}$ x r^{2}

On the other hand, the sector is the region **enclosed within the arc** and the **bounding radii** OA and OB. For example, when you cut a pizza into many slices, you normally cut it into sectors.

area of sector (θ in radians) = $\mathsf{\dfrac{\theta}{2}}$ x r^{2}

If the **angle is in degrees**, replace θ in the formula with the following (except in sin(θ), keep your calculator in degree mode for that):

\[ \mathsf{\theta \textsf{ (in rad)} = \dfrac{\theta \textsf{ (in deg)} \times \pi}{180}} \]

**The Semicircle – A Special Case**

The circle divides into **two semicircles** when the major and minor **arc lengths are equal**. Therefore, there is no distinction between major and minor segments and sectors in a semicircle. All of them occupy the **same area**. Further, the sector and segment of a semicircle physically represent the same region (this is not the case in any other scenario).

**A Few Solved Examples to Clear Things Up**

**Example 1**

Consider the unit circle (radius = r = 1 unit) centered at the origin, as shown in the figure below.

Find the clockwise arc between the point y = 0.5 in the second quadrant and the point x = 0.3 in the fourth quadrant. Is this arc major or minor?

**Solution**

Point in the second quadrant: **(0.3. -0.954)**

Point in the fourth quadrant: **(-0.866, 0.5)**

Traversing the circle clockwise from A to B, it turns out to be a major arc.

**Example 2**

Consider the same circle from Example 1. Now find the arc in the anti-clockwise direction between the points y = -0.8 in the third quadrant and x = 0.6 in the first quadrant. Is this arc major or minor?

**Solution**

Point in the third quadrant:** (-0.6. -0.8)**

Point in the first quadrant: **(0.6, 0.8)**

Since the result is a pair of equivalent semicircles, there is no distinction between the major and minor arcs.

**Example 3**

Find the areas of the minor sectors and segments defined by the arcs in Examples 1 and 2. Further, what is the major arc’s length in Example 1?

**Solution**

First, we consider the results of **Example 2** as they involve semicircles. We can see that:

area of the first sector = area of the first segment = area of the first semicircle

area of the second sector = area of the second segment = area of the second semicircle

**area of the first semicircle = area of the second semicircle**

The area of a semicircle must be **half** the area of the circle, so we have the following:

area of circle = A_{circle} = $\pi$r^{2} = 3.14159 x (1)^{2} = 3.14159 square units

**area of each semicircle = 0.5 x A _{circle} = 0.5(3.14159) = 1.57079 square units**

We can see that the sum of the two semicircles’ area adds to the area of the full circle, except for some rounding errors.

Now, consider the case in **Example 1**. The formulae for the areas of sector and segments use the angle between the two bounding radii (θ), which we do not have.

**Finding the Angle Between the Bounding Radii**

To find it, we must find the length of the segment AB. Using the **distance formula** (x_{a} = -0.866, x_{b} = 0.3, y_{a} = 0.5, y_{b} = -0.954):

\[ \mathsf{|\overline{AB}| = \sqrt{(x_a-x_b)^2+(y_a-y_b)^2} = \sqrt{3.474} = 1.864} \]

Now we have the length of three sides of the triangle OAB, such that AB = 1.864 and OA = OB = 1. We will use the** law of cosines**. Given three sides, a, b, and c, of a triangle, we can find the angle $\theta$ between a and b:

Figure 7 – Finding the angle θ between the bounding radii of the minor segment OAB.

\[ \mathsf{\theta = \cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right)} \]

In our case, a = OA = 1, b = OB = 1, and c = AB = 1.864. Plugging these in:

\[ \mathsf{\theta = \cos^{-1}\left( \frac{-1.4745}{2} \right) = 137.497^\circ = 137.5^\circ} \]

**Length of the Major Arc and Area of the Minor Sector and Segment OAB**

Given that the minor arc angle is 137.5$^\circ$, the major arc’s angle must be 360 – 137.5 = 222.5$^\circ$. Therefore, with r = 1, the length of the major arc is:

l = r x $\displaystyle\mathsf{\frac{\theta \times \pi}{180}}$

**l = 1 x (222.5 x 3.14159) / 180 = 3.8834 units**

Now, plugging in $\theta$ = 137.5$^\circ$ and r = 1 into the formulae for sector and segment areas (after converting from degrees to radians), we get:

**area of sector OAB = 1.2 square units**

**area of segment AB = 0.862 square units**

*All images/mathematical drawings were created with GeoGebra.*