Areas Related to Circles - Perimeter and Area of Sectors and Segments

The study of two-dimensional planar figures is a cornerstone of Euclidean geometry. While straight-line bounded figures such as triangles, quadrilaterals, and polygons have fixed linear boundaries, curved figures introduce unique geometric dynamics. Among all curved shapes, the circle is the most symmetrical and fundamentally important geometric locus.

In Class 9, you explored the basic theoretical properties of circles, including chords, arcs, subtended angles, and cyclic quadrilaterals. In Class 10, the focus shifts from purely spatial and axiomatic geometry to quantitative measurement—specifically, calculating the lengths of circular arcs, as well as the surface regions enclosed by specified circular boundaries, known as sectors and segments.

Understanding how to calculate the perimeter and area of circular regions is not merely an academic exercise required for board examinations; it forms the quantitative foundation for higher mathematics, physics, architecture, engineering design, and modern manufacturing technology.

1. Fundamental Geometric Review

Before diving into sectors and segments, let us quickly re-establish the foundational metrics of a circle. A circle of radius $r$ centered at point $O$ has:

1. Diameter ($d$): $d = 2r$

2. Circumference ($C$): The total boundary length of the circle. $C = 2\pi r = \pi d$

3. Total Area ($A$): The surface enclosed within the circumference. $A = \pi r^2$

Here, $\pi$ (pi) represents the fundamental irrational constant equal to the ratio of a circle's circumference to its diameter ($\pi = \frac{C}{d}$). For calculation purposes in the CBSE curriculum:

• Use $\pi = \frac{22}{7}$ unless explicitly specified otherwise.
• Use $\pi = 3.14$ when explicitly instructed in the problem statement.

2. Understanding Sectors of a Circle

2.1 What is a Sector?

A sector of a circle is defined as the region bounded by two radii and the arc intercepted by those radii. Think of a sector as a slice cut out of a circular pizza or pie.

          Major Sector
           /   |   \
          /    |    \
         /     O     \
        /    /   \    \
       /   r/     \r   \
      /    / θ deg \    \
     +---A-----------B---+
          Minor Sector

• Minor Sector: The sector formed by the smaller arc (intercepted by a central angle $\theta < 180^\circ$).
• Major Sector: The sector formed by the larger arc (intercepted by a central angle $360^\circ - \theta$).

Note: Unless explicitly mentioned, the term "sector" in NCERT textbook problems refers to the minor sector.

2.2 Arc Length of a Sector ($l$)

An arc is a continuous piece of the circumference of a circle. The length of an arc depends directly on the central angle $\theta$ (in degrees) subtended by it at the center $O$.

Derivation using the Unitary Method:

• A complete central angle of $360^\circ$ corresponds to an arc length equal to the full circumference, $2\pi r$.
• A central angle of $1^\circ$ corresponds to an arc length of: $\frac{2\pi r}{360^\circ}$
• Therefore, a central angle of $\theta$ corresponds to an arc length $l$:

$\text{Length of arc } (l) = \frac{\theta}{360^\circ} \times 2\pi r$

2.3 Area of a Sector ($A_{\text{sector}}$)

Similarly, the area of a sector depends proportionally on its central angle $\theta$.

Derivation using the Unitary Method:

• A complete central angle of $360^\circ$ corresponds to the full area of the circle, $\pi r^2$.
• A central angle of $1^\circ$ corresponds to an area of: $\frac{\pi r^2}{360^\circ}$
• Therefore, the area of a sector with a central angle $\theta$ is:

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

Area of Major Sector:

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

2.4 Perimeter of a Sector ($P_{\text{sector}}$)

Students frequently mistake the arc length for the perimeter of a sector. The perimeter of any closed 2D shape is the total length of its outer boundary. A sector is bounded by one arc and two radii.

$\text{Perimeter of Sector} = \text{Arc Length } (l) + \text{Radius } (r) + \text{Radius } (r)$ $P_{\text{sector}} = l + 2r = \left(\frac{\theta}{360^\circ} \times 2\pi r\right) + 2r$

2.5 Interrelation Between Sector Area and Arc Length

By dividing the expression for area by the expression for arc length: $\frac{\text{Area}}{\text{Arc Length}} = \frac{\frac{\theta}{360^\circ} \cdot \pi r^2}{\frac{\theta}{360^\circ} \cdot 2\pi r} = \frac{r}{2}$

Rearranging this yields a very useful direct identity (analogous to $A = \frac{1}{2} \times \text{base} \times \text{height}$ for triangles):

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

3. Understanding Segments of a Circle

3.1 What is a Segment?

A segment of a circle is defined as the region bounded by a chord and the corresponding arc connecting the endpoints of the chord.

• Minor Segment: The region bounded by the chord and the minor arc.
• Major Segment: The region bounded by the chord and the major arc.

Unless specified otherwise, the term "segment" refers to the minor segment.

3.2 Area of a Minor Segment

The area of a minor segment is determined by taking the area of the entire minor sector $OAPB$ and subtracting the area of the isosceles triangle $\Delta OAB$ formed by the two radii and the chord.

$\text{Area of Minor Segment} = \text{Area of Sector } OAPB - \text{Area of } \Delta OAB$

$\text{Area of Minor Segment} = \left(\frac{\theta}{360^\circ} \times \pi r^2\right) - \text{Area}(\Delta OAB)$

3.3 Calculating the Area of Triangle $\Delta OAB$

In $\Delta OAB$, two side lengths are equal to the radius ($OA = OB = r$), and the included angle is $\angle AOB = \theta$.

Depending on the value of $\theta$, we compute $\text{Area}(\Delta OAB)$ through different approaches:

Method A: Direct Trigonometric Formula (Recommended for standard calculations)

From trigonometry, the area of any triangle with two sides $r$ and an included angle $\theta$ is given by:

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

Derivation of the Formula:

Drop a perpendicular from center $O$ to chord $AB$ at point $M$. Since $\Delta OAB$ is isosceles ($OA = OB = r$):

1. $OM \perp AB$ bisects the chord $AB$ ($AM = MB = \frac{1}{2} AB$).
2. $OM$ bisects the central angle $\angle AOB$, so $\angle AOM = \angle BOM = \frac{\theta}{2}$.

In right-angled triangle $\Delta OMA$: $\sin\left(\frac{\theta}{2}\right) = \frac{AM}{OA} = \frac{AM}{r} \implies AM = r \sin\left(\frac{\theta}{2}\right) \implies AB = 2r \sin\left(\frac{\theta}{2}\right)$ $\cos\left(\frac{\theta}{2}\right) = \frac{OM}{OA} = \frac{OM}{r} \implies OM = r \cos\left(\frac{\theta}{2}\right)$

Now, applying the basic area formula for a triangle: $\text{Area}(\Delta OAB) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times OM$ $\text{Area}(\Delta OAB) = \frac{1}{2} \times \left[2r \sin\left(\frac{\theta}{2}\right)\right] \times \left[r \cos\left(\frac{\theta}{2}\right)\right] = r^2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)$

Using the trigonometric identity $2\sin(x)\cos(x) = \sin(2x)$, this simplifies to: $\text{Area}(\Delta OAB) = \frac{1}{2} r^2 \sin\theta$

Method B: Standard Case Breakdown (NCERT Conventional Approach)

1. When Central Angle $\theta = 60^\circ$: Since $OA = OB = r$ and $\angle AOB = 60^\circ$, $\Delta OAB$ is an equilateral triangle ($\angle A = \angle B = 60^\circ$). $\text{Area}(\Delta OAB) = \frac{\sqrt{3}}{4} r^2$

2. When Central Angle $\theta = 90^\circ$: $\Delta OAB$ is a right-angled isosceles triangle at $O$. $\text{Area}(\Delta OAB) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times r \times r = \frac{1}{2} r^2$

3. When Central Angle $\theta = 120^\circ$: Using the perpendicular construction $OM \perp AB$:

   • $\angle AOM = 60^\circ$
   • $AM = r \sin(60^\circ) = r \frac{\sqrt{3}}{2} \implies AB = 2 \times r \frac{\sqrt{3}}{2} = r\sqrt{3}$
   • $OM = r \cos(60^\circ) = \frac{r}{2}$ $\text{Area}(\Delta OAB) = \frac{1}{2} \times AB \times OM = \frac{1}{2} \times (r\sqrt{3}) \times \left(\frac{r}{2}\right) = \frac{\sqrt{3}}{4} r^2$

3.4 Area of a Major Segment

$\text{Area of Major Segment} = \text{Total Area of Circle} - \text{Area of Minor Segment}$

Alternatively: $\text{Area of Major Segment} = \text{Area of Major Sector} + \text{Area of } \Delta OAB$

3.5 Perimeter of a Segment

The perimeter of a segment consists of the arc length plus the linear length of the subtending chord $AB$.

$\text{Perimeter of Minor Segment} = \text{Arc Length } (l) + \text{Chord Length } (AB)$ $P_{\text{segment}} = \left(\frac{\theta}{360^\circ} \times 2\pi r\right) + 2r \sin\left(\frac{\theta}{2}\right)$

4. Comprehensive Comparative Matrix

5. Real-World Applications

1. Civil & Architectural Engineering (Rotational Irrigation Systems)

Center-pivot irrigation systems feature a long rotating pipe anchored at a central pivot point. As the pipe rotates through a given angle $\theta$, it waters a specific sector of a circular field. Engineers compute the area of this sector to calibrate water volume distribution and crop yields.

2. Automotive & Mechanical Design (Windshield Wipers)

A car's windshield wiper blade is mounted on an arm of radius $R_1$, with the actual rubber blade extending from distance $R_2$ to $R_1$. As it sweeps through an angle $\theta$, the swept region forms the difference between two concentric sectors (an annular sector): $\text{Swept Area} = \frac{\theta}{360^\circ} \pi (R_1^2 - R_2^2)$

3. Navigation & Marine Radar (Lighthouse Beam Range)

Lighthouses project light across a circular arc $\theta$ to warn ships of nearby reefs. The total ocean surface covered by the light beam represents a sector. Determining this area helps navigation authorities assess safety coverage zones near coastal waters.

6. Step-by-Step Solved Textbook Examples

Example 1: Basic Sector Area and Arc Length

Question: Find the length of the arc and the area of a sector of a circle with a radius of $21\text{ cm}$ and a central angle of $60^\circ$. (Use $\pi = \frac{22}{7}$)

Solution:

• Step 1: Identify given values. Radius ($r$) = $21\text{ cm}$ Central angle ($\theta$) = $60^\circ$

• Step 2: Compute Arc Length ($l$). $l = \frac{\theta}{360^\circ} \times 2\pi r$ $l = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21$ $l = \frac{1}{6} \times 2 \times 22 \times 3$ $l = \frac{1}{6} \times 132 = 22\text{ cm}$

• Step 3: Compute Area of Sector ($A$). $A = \frac{\theta}{360^\circ} \times \pi r^2$ $A = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21$ $A = \frac{1}{6} \times 22 \times 3 \times 21$ $A = \frac{1}{6} \times 1386 = 231\text{ cm}^2$

• Final Answer:

  • Arc length = $22\text{ cm}$
  • Sector area = $231\text{ cm}^2$

Example 2: Area of Minor and Major Segment ($\theta = 90^\circ$)

Question: A chord of a circle of radius $10\text{ cm}$ subtends a right angle at the center. Find:

1. The area of the minor segment.
2. The area of the major segment. (Use $\pi = 3.14$)

Solution:

• Step 1: Identify given values. Radius ($r$) = $10\text{ cm}$ Central angle ($\theta$) = $90^\circ$

• Step 2: Calculate Area of Minor Sector. $\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{90^\circ}{360^\circ} \times 3.14 \times (10)^2$ $\text{Area of sector} = \frac{1}{4} \times 3.14 \times 100 = \frac{314}{4} = 78.5\text{ cm}^2$

• Step 3: Calculate Area of Triangle $\Delta OAB$. Since $\theta = 90^\circ$, $\Delta OAB$ is a right-angled triangle: $\text{Area}(\Delta OAB) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times r \times r$ $\text{Area}(\Delta OAB) = \frac{1}{2} \times 10 \times 10 = 50\text{ cm}^2$

• Step 4: Calculate Area of Minor Segment. $\text{Area of minor segment} = \text{Area of sector} - \text{Area}(\Delta OAB)$ $\text{Area of minor segment} = 78.5 - 50 = 28.5\text{ cm}^2$

• Step 5: Calculate Area of Major Segment. $\text{Total area of circle} = \pi r^2 = 3.14 \times 10^2 = 314\text{ cm}^2$ $\text{Area of major segment} = \text{Total area} - \text{Area of minor segment}$ $\text{Area of major segment} = 314 - 28.5 = 285.5\text{ cm}^2$

• Final Answer:

  • Area of minor segment = $28.5\text{ cm}^2$
  • Area of major segment = $285.5\text{ cm}^2$

Example 3: Clock Hands Sweeping Area

Question: The length of the minute hand of a clock is $14\text{ cm}$. Find the area swept by the minute hand in $5\text{ minutes}$. (Use $\pi = \frac{22}{7}$)

Solution:

• Step 1: Determine the angle $\theta$ swept in 5 minutes. A minute hand completes one full rotation ($360^\circ$) in $60\text{ minutes}$. $\text{Angle described in } 1\text{ minute} = \frac{360^\circ}{60} = 6^\circ$ $\text{Angle described in } 5\text{ minutes } (\theta) = 5 \times 6^\circ = 30^\circ$

• Step 2: Apply sector area formula. Here, radius $r = 14\text{ cm}$ and $\theta = 30^\circ$. $\text{Area swept} = \frac{\theta}{360^\circ} \times \pi r^2$ $\text{Area swept} = \frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14$ $\text{Area swept} = \frac{1}{12} \times \frac{22}{7} \times 196$ $\text{Area swept} = \frac{1}{12} \times 22 \times 28 = \frac{616}{12} = \frac{154}{3}\text{ cm}^2 \approx 51.33\text{ cm}^2$

• Final Answer:

  • Area swept = $\frac{154}{3}\text{ cm}^2$ (or $51.33\text{ cm}^2$)

Example 4: Complex Segment Area with Central Angle $\theta = 120^\circ$

Question: A chord of a circle of radius $12\text{ cm}$ subtends an angle of $120^\circ$ at the center. Find the area of the corresponding segment of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

Solution:

              O (120°)
             / | \
       12 cm/  |  \12 cm
           /   |M  \
          A----+----B
            Chord

• Step 1: Calculate Area of Sector $OAPB$. $r = 12\text{ cm}, \quad \theta = 120^\circ$ $\text{Area of sector} = \frac{120^\circ}{360^\circ} \times \pi r^2 = \frac{1}{3} \times 3.14 \times 12 \times 12$ $\text{Area of sector} = 1 \times 3.14 \times 4 \times 12 = 150.72\text{ cm}^2$

• Step 2: Calculate Area of $\Delta OAB$. Draw perpendicular $OM \perp AB$. In $\Delta OMA$, $\angle AOM = \frac{120^\circ}{2} = 60^\circ$. $\cos(60^\circ) = \frac{OM}{OA} \implies \frac{1}{2} = \frac{OM}{12} \implies OM = 6\text{ cm}$ $\sin(60^\circ) = \frac{AM}{OA} \implies \frac{\sqrt{3}}{2} = \frac{AM}{12} \implies AM = 6\sqrt{3}\text{ cm}$ $\text{Chord } AB = 2 \times AM = 12\sqrt{3}\text{ cm}$

  $\text{Area}(\Delta OAB) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times OM$ $\text{Area}(\Delta OAB) = \frac{1}{2} \times 12\sqrt{3} \times 6 = 36\sqrt{3}\text{ cm}^2$

  Substitute $\sqrt{3} = 1.73$: $\text{Area}(\Delta OAB) = 36 \times 1.73 = 62.28\text{ cm}^2$

• Step 3: Calculate Area of Minor Segment. $\text{Area of segment} = \text{Area of sector} - \text{Area of } \Delta OAB$ $\text{Area of segment} = 150.72 - 62.28 = 88.44\text{ cm}^2$

• Final Answer:

  • Area of the segment = $88.44\text{ cm}^2$

7. Common Student Mistakes to Avoid

Mistake 1: Confusing Arc Length with Sector Perimeter

• Incorrect: Writing $\text{Perimeter of Sector} = \frac{\theta}{360^\circ} \times 2\pi r$.
• Correct: The perimeter includes the two enclosing radii boundaries. $\text{Perimeter of Sector} = \frac{\theta}{360^\circ} \times 2\pi r + 2r$

Mistake 2: Mixing up Minor Segment and Minor Sector Formulae

• Incorrect: Directly applying $\frac{\theta}{360^\circ} \times \pi r^2$ when asked for the area of a segment.
• Correct: A segment is bounded by a chord. You must calculate the sector area and subtract the central triangle area: $\text{Area of Segment} = \text{Area of Sector} - \text{Area of Triangle}$

Mistake 3: Misapplying Values of $\pi$ and $\sqrt{3}$

• Incorrect: Using $\pi = \frac{22}{7}$ when the question explicitly specifies Use $\pi = 3.14$. This results in minor rounding discrepancies that lead to lost marks in CBSE marking schemes.
• Correct: Always check the instruction at the end of the question prompt for explicit values of $\pi$ and $\sqrt{3}$.

Mistake 4: Calculation Errors in Triangle Area for $\theta = 120^\circ$

• Incorrect: Assuming $\Delta OAB$ is equilateral when $\theta = 120^\circ$ and using $\frac{\sqrt{3}}{4} r^2$.
• Correct: For $\theta = 120^\circ$, the triangle area is $\frac{\sqrt{3}}{4} r^2$ algebraically by coincidence of values, but the chord length is $r\sqrt{3}$ and height is $\frac{r}{2}$. Ensure you show the trigonometric derivation ($OM \perp AB$) step-by-step in board exams to earn full working marks.

8. Self-Assessment Practice Questions

Practice Question 1

Find the area of the region enclosed between two concentric circles of radii $7\text{ cm}$ and $14\text{ cm}$, within a central sector angle of $40^\circ$.

??? note "Click to view Solution" Solution:

* **Step 1: Understand the shape.**
  The region is the area of the sector of the outer circle minus the area of the sector of the inner circle.
  Outer radius $R = 14\text{ cm}$, Inner radius $r = 7\text{ cm}$, Central angle $\theta = 40^\circ$.

* **Step 2: Formula setup.**
  $$\text{Required Area} = \frac{\theta}{360^\circ} \pi R^2 - \frac{\theta}{360^\circ} \pi r^2 = \frac{\theta}{360^\circ} \pi (R^2 - r^2)$$

* **Step 3: Execute calculations.**
  $$\text{Required Area} = \frac{40^\circ}{360^\circ} \times \frac{22}{7} \times (14^2 - 7^2)$$
  $$\text{Required Area} = \frac{1}{9} \times \frac{22}{7} \times (196 - 49)$$
  $$\text{Required Area} = \frac{1}{9} \times \frac{22}{7} \times 147$$
  $$\text{Required Area} = \frac{1}{9} \times 22 \times 21 = \frac{462}{9} = \frac{154}{3}\text{ cm}^2 \approx 51.33\text{ cm}^2$$

* **Final Answer:** **$\frac{154}{3}\text{ cm}^2$** or **$51.33\text{ cm}^2$**

Practice Question 2

A chord of a circle of radius $15\text{ cm}$ subtends an angle of $60^\circ$ at the center. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

??? note "Click to view Solution" Solution:

* **Step 1: Calculate Area of Minor Sector.**
  $$r = 15\text{ cm}, \quad \theta = 60^\circ$$
  $$\text{Area of minor sector} = \frac{60^\circ}{360^\circ} \times 3.14 \times 15 \times 15$$
  $$\text{Area of minor sector} = \frac{1}{6} \times 3.14 \times 225 = \frac{706.5}{6} = 117.75\text{ cm}^2$$

* **Step 2: Calculate Area of Equilateral Triangle $\Delta OAB$.**
  Since $\theta = 60^\circ$, $\Delta OAB$ is equilateral.
  $$\text{Area}(\Delta OAB) = \frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times (15)^2 = \frac{1.73 \times 225}{4} = \frac{389.25}{4} = 97.3125\text{ cm}^2$$

* **Step 3: Calculate Area of Minor Segment.**
  $$\text{Area of minor segment} = 117.75 - 97.3125 = 20.4375\text{ cm}^2$$

* **Step 4: Calculate Area of Major Segment.**
  $$\text{Total area of circle} = \pi r^2 = 3.14 \times (15)^2 = 3.14 \times 225 = 706.5\text{ cm}^2$$
  $$\text{Area of major segment} = 706.5 - 20.4375 = 686.0625\text{ cm}^2$$

* **Final Answer:**
  * Area of minor segment = **$20.4375\text{ cm}^2$**
  * Area of major segment = **$686.0625\text{ cm}^2$**

Practice Question 3

An umbrella has 8 ribs which are equally spaced. Assuming the umbrella to be a flat circle of radius $45\text{ cm}$, find the area between two consecutive ribs of the umbrella.

??? note "Click to view Solution" Solution:

* **Step 1: Calculate central angle between two consecutive ribs.**
  Total central angle = $360^\circ$.
  Number of equal sectors = $8$.
  $$\theta = \frac{360^\circ}{8} = 45^\circ$$

* **Step 2: Calculate Area of Sector.**
  Radius $r = 45\text{ cm}$.
  $$\text{Area between two ribs} = \frac{\theta}{360^\circ} \times \pi r^2$$
  $$\text{Area} = \frac{45^\circ}{360^\circ} \times \frac{22}{7} \times 45 \times 45$$
  $$\text{Area} = \frac{1}{8} \times \frac{22}{7} \times 2025$$
  $$\text{Area} = \frac{11 \times 2025}{28} = \frac{22275}{28}\text{ cm}^2 \approx 795.54\text{ cm}^2$$

* **Final Answer:** **$\frac{22275}{28}\text{ cm}^2$** (or **$795.54\text{ cm}^2$**)

9. Board Exam FAQs & Quick Revision

Q1: How do I calculate the area of a sector if the arc length $l$ and radius $r$ are given, but the angle $\theta$ is unknown?

Answer: Use the direct relationship formula: $\text{Area of Sector} = \frac{1}{2} \times l \times r$ This saves time by eliminating the need to calculate $\theta$ first.

Q2: What is the perimeter of a semi-circular protractor of radius $r$?

Answer: A semi-circle is a sector with central angle $\theta = 180^\circ$. Its boundary consists of the semi-circular arc plus the diameter ($2r$): $\text{Perimeter} = \pi r + 2r = r(\pi + 2)$ For instance, if $r = 7\text{ cm}$: $\text{Perimeter} = \left(\frac{22}{7} \times 7\right) + (2 \times 7) = 22 + 14 = 36\text{ cm}$

Q3: If the perimeter and the area of a circle are numerically equal, what is the radius of the circle?

Answer: Set the area expression equal to the circumference expression: $\pi r^2 = 2\pi r$ Since radius $r \neq 0$, divide both sides by $\pi r$: $r = 2\text{ units}$

Q4: How does doubling the radius of a circle affect the area of a sector with a fixed central angle $\theta$?

Answer: Since Sector Area $A = \left(\frac{\theta}{360^\circ}\right) \pi r^2$, the area is proportional to the square of the radius ($A \propto r^2$). If the radius is doubled ($r' = 2r$): $A' = \left(\frac{\theta}{360^\circ}\right) \pi (2r)^2 = 4 \times \left[\left(\frac{\theta}{360^\circ}\right) \pi r^2\right] = 4A$ Thus, doubling the radius quadruples the area of the sector.