Length of an Arc and Area of Sector of a Circle for CBSE Class 10

When you cut a slice of pizza or open a handheld folding fan, you are holding a geometric shape known as a sector of a circle. A sector represents a fractional wedge of a full circular disc, defined by two radial edges and a curved outer arc. From designing windshield wipers and sprinkler irrigation systems to measuring the area swept by the hands of a ticking wall clock, sector geometry is one of the most practical branches of coordinate and planar mensuration.

In CBSE Class 10 Mathematics, Chapter 11 (Areas Related to Circles), deriving and calculating the length of an arc and the area of minor and major sectors are standard 2-mark and 3-mark board examination problems.

What You Will Learn

• Formal definition of a sector and an arc
• Minor sector vs. Major sector
• Derivation and formula for the Length of an Arc: $l = \frac{\theta}{360^\circ} \times 2\pi r$
• Derivation and formula for the Area of a Sector: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$
• The direct relationship between arc length and sector area: $\text{Area} = \frac{1}{2} l r$
• Calculating the area of a Major Sector
• The Clock Face Principle: Angular speed of minute hands ($6^\circ/\text{min}$) and hour hands
• Solved CBSE board examination problems and common traps

1. What is a Sector of a Circle?

Formal Definition

A sector of a circle is the region of the circular plane enclosed by two radii and the corresponding arc connecting their endpoints.

                                  O (Center)
                                 /                         Radius r/   \ Radius r
                               /  θ                                A-------B
                               \     /
                                `---' <-- Curved Arc AB
                            [ MINOR SECTOR ]

1. Minor Sector: The sector corresponding to an angle $\theta < 180^\circ$.
2. Major Sector: The remaining circular region corresponding to angle $(360^\circ - \theta)$.

2. Length of an Arc of a Sector

An arc is a continuous curved piece of the circumference of a circle.

• The complete circumference ($360^\circ$ rotation) has length $2\pi r$.
• Therefore, for an arc subtending an angle of $1^\circ$ at the center, its length is $\frac{2\pi r}{360^\circ}$.
• For an arc subtending an angle of $\theta$ at the center:

Arc Length Formula

$\mathbf{l = \frac{\theta}{360^\circ} \times 2\pi r = \frac{\theta}{180^\circ} \times \pi r}$

3. Area of a Sector of a Circle

Similarly, the total area of a full circular disc ($360^\circ$ rotation) is $\pi r^2$.

• For a sector with central angle of $1^\circ$, the area is $\frac{\pi r^2}{360^\circ}$.
• For a sector with central angle of $\theta$:

Sector Area Formula

$\mathbf{\text{Area of Minor Sector} = \frac{\theta}{360^\circ} \times \pi r^2}$

The Direct Arc-Area Relationship:

Notice that: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{1}{2} \times \left(\frac{\theta}{360^\circ} \times 2\pi r\right) \times r$

$\mathbf{\text{Area of Sector} = \frac{1}{2} \times l \times r}$

This elegant formula mirrors the standard triangle area formula ($\frac{1}{2} \times \text{base} \times \text{height}$), where the curved arc $l$ acts as the base and radius $r$ acts as the height!

4. Area of a Major Sector

The area of the major sector can be found in two equivalent ways:

$\mathbf{\text{Area of Major Sector} = \text{Total Area of Circle} - \text{Area of Minor Sector} = \pi r^2 - \left(\frac{\theta}{360^\circ} \times \pi r^2\right)}$ $\mathbf{\text{Area of Major Sector} = \left(\frac{360^\circ - \theta}{360^\circ}\right) \times \pi r^2}$

5. The Clock Face Principle (CBSE High-Frequency Question)

In many board exam problems, the central angle $\theta$ is not given directly; instead, you are told that the minute hand of a clock moved for a certain number of minutes.

                              Clock Dial (360°)
                                    12
                               11   |    1
                            10      |      2
                           9        O------- 3 (15 min = 90°)
                            8              4
                               7    6    5
                       Minute Hand: 360° / 60 min = 6° per minute!

The Angular Speed of Clock Hands:

1. The Minute Hand:
   • In $60\text{ minutes}$, the minute hand completes a full $360^\circ$ circle.
   • Therefore, in $1\text{ minute}$, the minute hand rotates: $\mathbf{\text{Angle swept in } 1\text{ minute} = \frac{360^\circ}{60} = 6^\circ}$
   • Example: In $5\text{ minutes}$, angle $\theta = 5 \times 6^\circ = \mathbf{30^\circ}$.
   • Example: In $15\text{ minutes}$, angle $\theta = 15 \times 6^\circ = \mathbf{90^\circ}$.
   • Example: In $20\text{ minutes}$, angle $\theta = 20 \times 6^\circ = \mathbf{120^\circ}$.
2. The Hour Hand:
   • In $12\text{ hours}$ ($720\text{ minutes}$), the hour hand rotates $360^\circ$.
   • Angle swept by hour hand in $1\text{ minute} = \frac{360^\circ}{720} = \mathbf{0.5^\circ}$.

6. Solved CBSE Board Examination Problems

Solved Example 1: Area Swept by Minute Hand (NCERT Classic)

Problem: The length of the minute hand of a clock is $14\text{ cm}$. Find the area swept by the minute hand in $5\text{ minutes}$.

Solution:

1. Analyze the Sector Parameters:
   • Radius: $r = 14\text{ cm}$ (the length of the minute hand).
   • Angle swept in $1\text{ minute} = 6^\circ$.
   • Central angle in $5\text{ minutes}$: $\theta = 5 \times 6^\circ = \mathbf{30^\circ}$
2. Apply the Sector Area Formula: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$ $\text{Area} = \frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14$
3. Simplify the Fraction: $\text{Area} = \frac{1}{12} \times 22 \times 2 \times 14 = \frac{1}{12} \times 616 = \frac{154}{3}\text{ cm}^2 = \mathbf{51.33\text{ cm}^2}$
4. Therefore, <u>the area swept by the minute hand in $5\text{ minutes}$ is $\frac{154}{3}\text{ cm}^2$ (or $51.33\text{ cm}^2$)</u>.

Solved Example 2: Arc Length and Major Sector Area

Problem: In a circle of radius $21\text{ cm}$, an arc subtends an angle of $60^\circ$ at the center. Find: (i) the length of the arc, and (ii) the area of the sector formed by the arc.

Solution:

1. Given: $r = 21\text{ cm}$ and $\theta = 60^\circ$.
2. (i) Length of the Arc ($l$): $l = \frac{\theta}{360^\circ} \times 2\pi r = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21$ $l = \frac{1}{6} \times 2 \times 22 \times 3 = \frac{1}{6} \times 132 = \mathbf{22\text{ cm}}$
3. (ii) Area of the Sector: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21$ $\text{Area} = \frac{1}{6} \times 22 \times 3 \times 21 = \frac{1}{6} \times 1386 = \mathbf{231\text{ cm}^2}$ (Alternatively, using $\text{Area} = \frac{1}{2} l r = \frac{1}{2} \times 22 \times 21 = 231\text{ cm}^2$).
4. Therefore, <u>the arc length is $22\text{ cm}$ and the sector area is $231\text{ cm}^2$</u>.

7. Summary and Examination Tips

Exam Tip: In questions where the minute hand sweeps for $t$ minutes, always show the step $\theta = t \times 6^\circ$ explicitly. This single calculation step carries 1 mark in the marking scheme!

Common Mistake: Confusing the perimeter of a sector with arc length. The arc length is only the curved part ($l$). The perimeter of a sector is the curved arc PLUS the two straight radii: $\text{Perimeter} = l + 2r$!