Areas Related to Circles

Areas of Sector and Segment of a Circle — Introduction

Chapter 11: Areas Related to Circles

Areas of Sector and Segment of a Circle — Introduction

Welcome to the fascinating world of circles! While you're familiar with finding the area (πr²) and circumference (2πr) of an entire circle, what about its pieces? Think of a pizza. When you take a slice, you're not taking the whole pizza, just a part of it. This slice is a perfect real-world example of a sector. The curved crust of that slice is the arc, and if you draw a straight line from the two corners of the crust, you create a chord. The area between that chord and the crust is a segment.

In this section, we will deconstruct the circle into these fundamental parts. We'll learn how to precisely calculate the area of any "slice" (sector) and any "crust-end" (segment). This isn't just about pizza; these concepts are crucial in design, engineering, and architecture—from designing a circular garden path to calculating the sweep of a radar beam or the glass needed for an arched window.

{{FORMULA: expr=Area of Sector = (θ/360) × πr² | symbols=θ:angle of the sector in degrees, r:radius of the circle}}

Definitions & Formulas

Before we derive the formulas, let's establish our key vocabulary. Understanding these terms is the first step to mastering the calculations.

Remark: In problems, when we simply write ‘sector’ or ‘segment’, we will mean the ‘minor sector’ and the ‘minor segment’ respectively, unless stated otherwise.

{{VISUAL: diagram: A circle with center O. Radii OA and OB are drawn. The area between OA, OB, and arc APB is shaded and labeled "Minor Sector (OAPB)". The angle AOB is labeled θ. The remaining larger area is labeled "Major Sector (OAQB)".}}

Derivation & Logic

How do we get the formula for the area of a sector? It’s not magic; it’s a simple, logical extension of the area of a full circle using the Unitary Method.

1. Start with the whole. A full circle is essentially a sector where the angle at the centre is a full 360°. The area of this "sector" is the area of the entire circle.

   Angle at centre = 360° → Area = πr²

2. Find the value of one unit. Using the Unitary Method, if 360 degrees corresponds to an area of πr², what area corresponds to just 1 degree? We simply divide the total area by 360.

   Area for 1° = (πr²) / 360
   

3. Scale it up to θ. Now that we know the area for a 1° angle, we can find the area for any angle θ by multiplying.

   Area for θ° = ((πr²) / 360) × θ
   

4. Final Formula. Rearranging the terms gives us the familiar formula for the area of a sector.

   Area of Sector = (θ / 360) × πr²
   

The exact same logic applies to finding the length of an arc. The total length of the circle's boundary (circumference) is 2πr for 360°.

• Length for 1° = (2πr) / 360
• Length for θ° = ((2πr) / 360) × θ

This gives us the formula for arc length:

Length of Arc = (θ / 360) × 2πr

{{KEY: type=concept | title=The Logic of Segments | text=The area of a segment is always found indirectly. There is no direct formula. You must first calculate the area of the corresponding sector and then subtract the area of the triangle formed by the two radii and the chord. Area of Segment = Area of Sector – Area of Triangle.}}

Solved Examples

Let's apply these formulas to some problems, starting from easy and moving to more complex scenarios.

Example 1: Basic Sector Area (Easy)

Given: A circle with radius r = 6 cm and the angle of the sector θ = 60°. Use π = 22/7.

To Find: The area of the sector.

Solution:

1. Start with the formula for the area of a sector.

   Area of Sector = (θ / 360) × πr²
   

2. Substitute the given values: θ = 60, r = 6, and π = 22/7.

   Area = (60 / 360) × (22/7) × 6²
   

3. Simplify the fraction and calculate the final area.

   Area = (1 / 6) × (22 / 7) × 36
   

   Area = 1 × (22 / 7) × 6 = 132 / 7
   

   Area ≈ 18.86 cm²
   

Final Answer: The area of the sector is 132/7 cm² or approximately 18.86 cm².

Example 2: Area of a Quadrant from Circumference (Medium)

Given: A circle whose circumference is 22 cm.

To Find: The area of a quadrant of the circle.

Solution:

1. First, understand what a quadrant is. A quadrant is ¼ of a circle, which means the angle θ at the centre is 360° / 4 = 90°.

2. We need the radius r to find the area. We can find it from the given circumference.

   Circumference = 2πr = 22
   

3. Solve for r, using π = 22/7.

   2 × (22 / 7) × r = 22
   

   (44 / 7) × r = 22
   

   r = 22 × (7 / 44) = 7 / 2 cm
   

4. Now, use the sector area formula with θ = 90° and r = 7/2 cm.

   Area of Quadrant = (90 / 360) × πr²
   

   Area = (1 / 4) × (22 / 7) × (7 / 2)²
   

   Area = (1 / 4) × (22 / 7) × (49 / 4)
   

   Area = (1 / 4) × 22 × (7 / 4) = 154 / 16 = 77 / 8 cm²
   

Final Answer: The area of the quadrant is 77/8 cm² or 9.625 cm².

Example 3: Finding the Area of a Segment (Hard)

Given: A chord of a circle of radius 10 cm subtends a right angle (θ = 90°) at the centre. Use π = 3.14.

To Find: The area of the corresponding minor segment.

{{VISUAL: diagram: A circle with center O and radius 10 cm. Radii OA and OB are perpendicular, forming a 90° angle. The chord AB is drawn. The area between chord AB and arc APB is shaded and labeled "Minor Segment".}}

Solution:

1. Recall the relationship: Area of Segment = Area of Sector – Area of Triangle. We need to find both parts.

2. First, calculate the area of the minor sector OAB.

   Area of Sector = (θ / 360) × πr²
   

   Area of Sector = (90 / 360) × 3.14 × 10²
   

   Area of Sector = (1 / 4) × 3.14 × 100 = 314 / 4 = 78.5 cm²
   

3. Next, calculate the area of the triangle OAB. Since ∠AOB = 90°, ΔOAB is a right-angled triangle with the radii OA and OB as its base and height.

   Area of ΔOAB = ½ × base × height
   

   Area of ΔOAB = ½ × OA × OB = ½ × 10 × 10
   

   Area of ΔOAB = 50 cm²
   

4. Finally, subtract the area of the triangle from the area of the sector to get the segment's area.

   Area of Segment = 78.5 – 50
   

   Area of Segment = 28.5 cm²
   

Final Answer: The area of the minor segment is 28.5 cm².

Example 4: The Clock Problem (Tricky)

Given: The length of the minute hand of a clock is 14 cm.

To Find: The area swept by the minute hand in 5 minutes.

{{VISUAL: diagram: A clock face. The minute hand is shown at the '12' and then at the '1'. The area between these two positions is shaded, forming a sector. The length of the hand is labeled "r = 14 cm".}}

Solution:

1. The path swept by the minute hand forms a sector. The length of the hand is the radius r = 14 cm. The challenge is to find the angle θ.

2. The minute hand completes a full circle (360°) in 60 minutes. We can use this to find the angle it sweeps per minute.

   Angle per minute = 360° / 60 minutes = 6° per minute
   

3. Now, calculate the total angle swept in 5 minutes.

   θ = 6°/minute × 5 minutes = 30°
   

4. Now we have r = 14 and θ = 30. We can use the sector area formula. (Use π = 22/7 unless stated otherwise).

   Area = (θ / 360) × πr²
   

   Area = (30 / 360) × (22 / 7) × 14²
   

   Area = (1 / 12) × (22 / 7) × 196
   

   Area = (1 / 12) × 22 × 28 = 616 / 12 = 154 / 3 cm²
   

Final Answer: The area swept by the minute hand is 154/3 cm² or approximately 51.33 cm².

Tips & Tricks

Master these shortcuts to solve problems faster and more efficiently.

Common Mistakes

Avoid these common pitfalls that students often make.

Brain-Teaser Questions

Test your understanding with these slightly more challenging problems.

1. A car has two windscreen wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. (Use π = 22/7)

   💡 Answer: This is a sector problem. For one wiper, r = 25 cm and θ = 115°. Area = (115/360) × (22/7) × 25² = (23/72) × (22/7) × 625 ≈ 627.48 cm². Since there are two wipers, the total area is 2 × 627.48 = 1254.96 cm².

2. The area of a sector of a circle with radius 12 cm is 48π cm². What is the angle θ of this sector?

   💡 Answer: This is a reverse problem. We are given the area and need to find θ. Area = (θ/360) × πr² 48π = (θ/360) × π × 12² 48π = (θ/360) × 144π Divide both sides by π: 48 = (θ/360) × 144 θ = (48 × 360) / 144 = 120°.

3. If the radius of a circle is doubled, and the angle of its sector is also doubled, how many times will the area of the new sector be compared to the original sector?

   💡 Answer: Let original area be A₁ = (θ/360) × πr². The new radius r₂ = 2r and the new angle θ₂ = 2θ. The new area is A₂ = (θ₂/360) × π(r₂)² = (2θ/360) × π(2r)². A₂ = (2θ/360) × π(4r²) = 8 × [(θ/360) × πr²] = 8 × A₁. The new area will be 8 times the original area.

Mini Cheatsheet

Screenshot this table for a quick revision of all the key formulas from this page.

Areas of Sector and Segment of a Circle — Area of Sector and Length of Arc

Chapter 11: Areas Related to Circles

Page 2 of 5: Area of Sector and Length of Arc

Welcome back! In the previous section, we revisited the basic properties of a circle. Now, we're going to zoom in on specific parts of a circle. Have you ever wondered how to calculate the area of just one slice of a pizza, or how far the tip of a clock's minute hand travels in 10 minutes? These questions are all about sectors and arcs.

A circle is a shape of perfect symmetry, and this symmetry provides a powerful and simple way to understand its parts. Every slice, big or small, is just a fraction of the whole. If you can understand the whole circle, you can understand any piece of it. In this section, we'll learn the elegant formulas to precisely measure the area of a sector (the "slice") and the length of an arc (the "crust" of the slice). This is a foundational skill for solving many real-world geometry problems.

{{FORMULA: expr=Area of Sector = (θ/360) × πr² | symbols=θ:angle of the sector in degrees, r:radius of the circle}}

Definitions & Formulas

Before we derive the formulas, let's clearly define the key terms we'll be using. Understanding these variables is the first step to mastering the calculations.

Derivation: The Logic Behind the Formulas

The beauty of these formulas lies in their simplicity, which comes directly from the "unitary method" — a concept you've used for years! We simply treat the sector and arc as fractions of the entire circle.

{{VISUAL: diagram: A circle with center O and radius r. A minor sector OAPB is shaded, with the central angle ∠AOB labeled as θ.}}

Deriving the Area of a Sector

1. A full circular region can be considered a special sector where the angle at the center is 360°. You already know its area.

   Area for an angle of 360° = πr²
   

2. Using the unitary method, let's find the area corresponding to a tiny 1° angle at the center. It would simply be 1/360th of the total area.

   Area for an angle of 1° = (1/360) × πr²
   

3. Now, if we know the area for 1°, we can find the area for any angle θ by multiplying.

   Area for an angle of θ = θ × (1/360) × πr²
   

4. Rearranging this gives us the final, elegant formula for the area of a sector.

   Area of Sector = (θ/360) × πr²
   

Deriving the Length of an Arc

The logic is identical, but this time we start with the total length of the circle's boundary: the circumference.

1. The full circumference of a circle corresponds to an angle of 360°.

   Length for an angle of 360° = 2πr
   

2. Again, using the unitary method, the length of an arc for a 1° angle is 1/360th of the total circumference.

   Length for an angle of 1° = (1/360) × 2πr
   

3. Therefore, the length of an arc for any angle θ is θ times this base unit.

   Length for an angle of θ = θ × (1/360) × 2πr
   

4. This simplifies to our final formula for the length of an arc.

   Length of Arc (L) = (θ/360) × 2πr
   

{{KEY: type=concept | title=The Proportionality Principle | text=The area of a sector and the length of its corresponding arc are directly proportional to the angle θ. Essentially, the fraction θ/360 tells you what portion of the whole circle you are dealing with. A 60° sector is simply 60/360 = 1/6th of the circle's area and 1/6th of its circumference.}}

Solved Examples

Let's apply these formulas to some problems, starting from easy and moving to more complex scenarios.

Example 1: Direct Application (Easy)

Find the area of a sector of a circle with a radius of 6 cm if the angle of the sector is 60°. (Use π = 22/7)

Given: Radius r = 6 cm, Angle θ = 60°.

To Find: Area of the sector.

Solution:

1. Start with the formula for the area of a sector.

   Area = (θ/360) × πr²
   

2. Substitute the given values of θ, r, and π into the formula.

   Area = (60/360) × (22/7) × 6²
   

3. Simplify the fraction and calculate the result.

   Area = (1/6) × (22/7) × 36
   

4. Perform the final calculation.

   Area = (22/7) × 6 = 132/7 cm²
   

Final Answer: The area of the sector is 132/7 cm², which is approximately 18.86 cm².

Example 2: Working Backwards (Medium)

Find the area of a quadrant of a circle whose circumference is 22 cm. (Use π = 22/7)

Given: Circumference = 22 cm. The sector is a quadrant.

To Find: Area of the quadrant.

Solution:

1. A quadrant of a circle is a sector with a central angle of 90°. So, θ = 90°.

2. We need the radius r to find the area. We can find it from the given circumference.

   Circumference = 2πr
   22 = 2 × (22/7) × r
   

3. Solve for r.

   22 = (44/7) × r
   r = (22 × 7) / 44 = 7/2 cm
   

4. Now use the area of a sector formula with r = 7/2 and θ = 90°.

   Area = (θ/360) × πr²
   Area = (90/360) × (22/7) × (7/2)²
   

5. Simplify and compute the final area.

   Area = (1/4) × (22/7) × (49/4)
   Area = (1/4) × 22 × (7/4) = 154/16 = 77/8 cm²
   

Final Answer: The area of the quadrant is 77/8 cm², or 9.625 cm².

Example 3: Real-World Application (Hard)

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 20 minutes. (Use π = 22/7)

{{VISUAL: diagram: A clock face showing the minute hand of length 14 cm moving from the 12 to the 4, sweeping out a sector. The angle swept, 120°, is marked at the center.}}

Given: Radius (length of minute hand) r = 14 cm. Time = 20 minutes.

To Find: Area swept by the minute hand.

Solution:

1. First, we need to find the angle θ that the minute hand sweeps in 20 minutes. The minute hand completes a full 360° circle in 60 minutes.

   Angle swept in 60 minutes = 360°
   

2. Using the unitary method, find the angle swept in 1 minute.

   Angle swept in 1 minute = 360° / 60 = 6°
   

3. Now, calculate the angle swept in 20 minutes.

   θ = 20 minutes × 6°/minute = 120°
   

4. We now have r = 14 cm and θ = 120°. We can use the area of the sector formula.

   Area = (θ/360) × πr²
   Area = (120/360) × (22/7) × 14²
   

5. Simplify the expression and calculate the final area.

   Area = (1/3) × (22/7) × 196
   Area = (1/3) × 22 × 28 = 616/3 cm²
   

Final Answer: The area swept by the minute hand is 616/3 cm², which is approximately 205.33 cm².

Example 4: Combining Formulas (Tricky)

The length of an arc of a circle is 11 cm, and the radius of the circle is 7 cm. Find the area of the sector corresponding to this arc. (Use π = 22/7)

{{VISUAL: diagram: A sector of a circle with center O, radius labeled as r = 7 cm, and the length of the corresponding arc labeled as L = 11 cm. The central angle is labeled θ.}}

Given: Arc Length L = 11 cm, Radius r = 7 cm.

To Find: Area of the sector.

Solution:

1. The area formula needs the angle θ, which is not given. However, we can find θ using the arc length formula.

   L = (θ/360) × 2πr
   

2. Substitute the given values of L, r, and π to solve for θ.

   11 = (θ/360) × 2 × (22/7) × 7
   11 = (θ/360) × 44
   

3. Isolate θ.

   θ = (11 × 360) / 44
   θ = (1 × 360) / 4 = 90°
   

4. Now that we have θ = 90° and r = 7 cm, we can find the area of the sector.

   Area = (θ/360) × πr²
   Area = (90/360) × (22/7) × 7²
   

5. Calculate the final area.

   Area = (1/4) × (22/7) × 49
   Area = (1/4) × 22 × 7 = 154/4 = 77/2 cm²
   

Final Answer: The area of the sector is 77/2 cm², or 38.5 cm².

Tips & Tricks

Here are a few shortcuts and quick facts to speed up your calculations in exams.

Common Mistakes

Many students make small errors that lead to incorrect answers. Watch out for these common traps!

Brain-Teaser Questions

Test your understanding with these slightly challenging problems.

1. A car has two windshield wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. (Use π = 22/7)

   💡 Answer: The area cleaned by one wiper is the area of a sector with r = 25 cm and θ = 115°. Area of one sector = (115/360) × (22/7) × 25² = (23/72) × (22/7) × 625 ≈ 627.48 cm². Since there are two wipers, the total area is 2 × 627.48 = 1254.96 cm².

2. The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. What is the area of the sector?

   💡 Answer: The perimeter of a sector = radius + radius + arc length (r + r + L). 16.4 = 5.2 + 5.2 + L 16.4 = 10.4 + L → L = 6 cm. Now, use the shortcut: Area = ½ × L × r = ½ × 6 × 5.2 = 3 × 5.2 = 15.6 cm².

3. If the radius of a circle is doubled and the angle of its sector is halved, what is the ratio of the area of the new sector to the area of the original sector?

   💡 Answer: Let original radius = r, original angle = θ. Original Area A₁ = (θ/360) × πr². New radius = 2r, new angle = θ/2. New Area A₂ = ((θ/2)/360) × π(2r)² = (θ/720) × π(4r²) = (4/2) × (θ/360) × πr² = 2 × A₁. The ratio A₂ / A₁ is 2/1. The area doubles.

Mini Cheatsheet

Screenshot this table for your last-minute revision!

Areas of Sector and Segment of a Circle — Area of Segment

Page 3 of 5: Areas of Sector and Segment of a Circle — Area of Segment

Welcome back! In the previous section, we mastered calculating the area of a sector, which looks like a slice of pizza. Now, let's focus on a different part of the circle: the segment.

Concept Introduction

Imagine you have a perfect circular pizza slice (a sector). Now, suppose you eat only the pointy, cheesy part, leaving behind just the crusty edge. That leftover piece of pizza, bounded by the arc and the straight-line cut (the chord), is a perfect real-world example of a segment of a circle.

A segment is the region enclosed by a chord and the corresponding arc. Just like sectors, segments come in two sizes: the minor segment (the smaller piece) and the major segment (the larger piece). Our goal is to find the area of this region, and the key lies in a clever subtraction: we take the area of the whole slice (sector) and remove the area of the triangular part.

{{FORMULA: expr=Area of Segment = (θ/360) × πr² - Area of Δ | symbols=θ:angle at center in degrees, r:radius of the circle, Δ:triangle formed by radii and chord}}

Definitions & Formulas

Let's formalize the terms and formulas we'll be using.

The Logic: How to Find the Area of a Segment

Calculating the area of a segment might seem tricky because of its curved edge, but the logic is surprisingly simple. It's just a two-step process of finding two areas you already know and then subtracting them.

{{VISUAL: diagram: a circle with center O. Radii OA and OB form a sector OAPB. A chord AB is drawn. The area of the segment APB is shaded, and labels clearly show that 'Area of Sector OAPB' minus 'Area of Triangle OAB' equals the shaded 'Area of Segment APB'.}}

1. Identify the Full Slice: First, visualize the sector that contains your segment. This is the sector OAPB formed by the radii OA, OB, and the arc APB.

2. Calculate the Sector's Area: Use the formula we learned previously to find the area of this entire sector.

   Area of Sector OAPB = (θ/360) × πr²
   

3. Isolate the Triangle: Now, look at the triangle OAB formed by the two radii (OA and OB) and the chord (AB). This is the part of the sector that is not in the segment.

4. Calculate the Triangle's Area: Find the area of this triangle. The method will depend on the angle θ.

   • If θ = 90°, it's a right-angled triangle. Area = ½ × r × r.
   • If θ = 60°, it's an equilateral triangle. Area = (√3 / 4) × r².
   • For other angles (like 120°), we may need to use trigonometry.

5. Subtract to Find the Segment: Finally, subtract the area of the triangle from the area of the sector. The result is the area of the minor segment.

   Area of Segment APB = Area of Sector OAPB – Area of ΔOAB
   

This simple subtraction is the core concept for this entire topic!

{{KEY: type=concept | title=The Core Formula | text=The area of a minor segment is always the area of its corresponding minor sector minus the area of the triangle formed by the radii and the chord. This relationship is the foundation for all segment-related problems.}}

Solved Examples

Let's apply this logic to some problems, starting from easy and moving to more complex ones. (Unless stated otherwise, use π = 22/7)

Example 1: The Right-Angled Segment (Easy)

Given: A chord of a circle of radius 10 cm subtends a right angle (90°) at the center. (Use π = 3.14)

To Find: The area of the corresponding minor segment.

Solution:

1. First, we calculate the area of the minor sector OAPB. Here, r = 10 cm and θ = 90°.

   Area of Sector = (90/360) × 3.14 × 10²
   

   Area of Sector = (1/4) × 3.14 × 100 = 78.5 cm²
   

2. Next, we find the area of the triangle OAB. Since θ = 90°, ΔOAB is a right-angled triangle with the radii OA and OB as its base and height.

   Area of ΔOAB = ½ × base × height = ½ × 10 × 10
   

   Area of ΔOAB = 50 cm²
   

3. Finally, subtract the area of the triangle from the area of the sector.

   Area of Segment = Area of Sector – Area of ΔOAB
   

   Area of Segment = 78.5 – 50 = 28.5 cm²
   

Final Answer: The area of the minor segment is 28.5 cm².

Example 2: The Equilateral Case (Medium)

Given: In a circle of radius 21 cm, an arc subtends an angle of 60° at the center.

To Find: The area of the segment formed by the corresponding chord.

Solution:

1. Calculate the area of the sector OAPB with r = 21 cm and θ = 60°.

   Area of Sector = (60/360) × (22/7) × 21²
   

   Area of Sector = (1/6) × (22/7) × 21 × 21 = 11 × 21 = 231 cm²
   

2. Now, consider the triangle OAB. We know OA = OB = 21 cm (radii). Since the angle between these equal sides is 60°, ΔOAB is an equilateral triangle. Therefore, all its sides are 21 cm.

3. Calculate the area of the equilateral triangle OAB using the formula Area = (√3 / 4) × side².

   Area of ΔOAB = (√3 / 4) × 21²
   

   Area of ΔOAB = (441√3) / 4 cm²
   

4. Subtract the triangle's area from the sector's area to get the segment's area.

   Area of Segment = 231 – (441√3 / 4) cm²
   

Final Answer: The area of the segment is (231 – 441√3 / 4) cm².

Example 3: The 120° Angle Problem (Hard)

Given: A chord of a circle of radius 12 cm subtends an angle of 120° at the center. (Use π = 3.14 and √3 = 1.73)

To Find: The area of the corresponding segment of the circle.

Solution:

{{VISUAL: diagram: a circle with center O and radii OA and OB = 12 cm. Angle AOB is 120°. A perpendicular OM is dropped from O to the chord AB. Angle AOM is shown as 60°. The labels OM and AM are marked for calculation using trigonometry.}}

1. First, calculate the area of the sector OAPB where r = 12 cm and θ = 120°.

   Area of Sector = (120/360) × 3.14 × 12²
   

   Area of Sector = (1/3) × 3.14 × 144 = 150.72 cm²
   

2. To find the area of ΔOAB, we need its base and height. Let's draw a perpendicular OM from O to the chord AB. This bisects the angle ∠AOB and the chord AB. So, ∠AOM = 120°/2 = 60° and AM = MB.

3. In the right-angled triangle OMA, we use trigonometry to find the height OM and half-base AM.

   • cos(60°) = OM / OA → ½ = OM / 12 → OM = 6 cm.
   • sin(60°) = AM / OA → √3 / 2 = AM / 12 → AM = 6√3 cm.

4. The full base AB is 2 × AM = 2 × 6√3 = 12√3 cm.

5. Now calculate the area of ΔOAB using Area = ½ × base × height.

   Area of ΔOAB = ½ × AB × OM = ½ × 12√3 × 6
   

   Area of ΔOAB = 36√3 cm²
   

6. Using √3 = 1.73, the area is 36 × 1.73 = 62.28 cm².

7. Finally, find the area of the segment by subtraction.

   Area of Segment = 150.72 – 62.28 = 88.44 cm²
   

Final Answer: The area of the segment is 88.44 cm².

Example 4: Finding the Major Segment (Tricky)

Given: A chord of a circle of radius 15 cm subtends an angle of 60° at the center. (Use π = 3.14 and √3 = 1.73)

To Find: The area of the corresponding major segment.

Solution:

1. The strategy is to find the area of the minor segment first, and then subtract it from the total area of the circle.

2. Calculate the area of the minor sector. Here r = 15 cm and θ = 60°.

   Area of Minor Sector = (60/360) × 3.14 × 15²
   

   Area of Minor Sector = (1/6) × 3.14 × 225 = 117.75 cm²
   

3. Calculate the area of the triangle OAB. Since θ = 60° and OA = OB = 15, it's an equilateral triangle.

   Area of ΔOAB = (√3 / 4) × side² = (√3 / 4) × 15²
   

   Area of ΔOAB = (225√3) / 4 = (225 × 1.73) / 4 = 97.3125 cm²
   

4. Calculate the area of the minor segment.

   Area of Minor Segment = 117.75 – 97.3125 = 20.4375 cm²
   

   {{VISUAL: diagram: a circle showing a minor segment (shaded light blue) and a major segment (shaded light green). Labels indicate how Area of Major Segment = Area of Circle - Area of Minor Segment.}}

5. Now, calculate the total area of the circle.

   Area of Circle = πr² = 3.14 × 15² = 3.14 × 225 = 706.5 cm²
   

6. Finally, subtract the minor segment's area from the circle's area to get the major segment's area.

   Area of Major Segment = 706.5 – 20.4375 = 686.0625 cm²
   

Final Answer: The area of the major segment is 686.0625 cm².

Tips & Tricks

Common Mistakes to Avoid

Brain-Teaser Questions

1. A chord in a circle of radius r subtends an angle θ at the center. If the area of the sector is exactly four times the area of the triangle formed by the radii and the chord, what is the relationship between them?

   💡 Answer: The relationship is (θ/360) × πr² = 4 × Area of ΔOAB. This requires expressing the area of the triangle in terms of r and θ (using ½ r² sinθ), which leads to (πθ / 360) = 2 sinθ.

2. Two equal chords AB and AC of a circle with radius 14 cm subtend angles of 60° each at the center. What is the area of the segment formed by the chord BC? (Hint: What is the angle ∠BOC?)

   💡 Answer: Since ∠AOB = 60° and ∠AOC = 60°, the total angle ∠BOC = 120°. Now it becomes a standard 120° segment problem. Calculate the area of the sector with θ=120° and subtract the area of ΔBOC. Area = (120/360) × (22/7) × 14² - ½ × (14√3) × 7 = (616/3 - 49√3) cm².

3. A square with a side length of 20 cm has a circle inscribed in it. A chord is drawn in the circle that is parallel to one side of the square and is 6 cm away from the center. Find the area of the smaller segment formed by this chord.

   💡 Answer: The radius of the inscribed circle is half the side of the square, so r = 10 cm. We have a chord 6 cm from the center. We can form a right triangle with the radius (hypotenuse=10), distance from center (one side=6), and half the chord length (other side). (Half chord)² = 10² - 6² = 64, so half chord = 8 cm. Now we have ΔOAB with sides 10, 10, and 16. We can find the angle θ using cos(θ/2) = 6/10, or find the triangle area ½ × 16 × 6 = 48 cm². Then find the sector area and subtract. θ ≈ 106.26°. Area of Sector ≈ (106.26/360) × π × 10² ≈ 92.73 cm². Area of segment ≈ 92.73 - 48 = 44.73 cm².

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Areas of Sector and Segment of a Circle — Solved Examples

Areas of Sector and Segment of a Circle — Solved Examples

Welcome back! Now that we've understood the definitions of sectors and segments, it's time to put our knowledge into practice. This section is all about applying the formulas we've derived to solve real problems. We'll move from simple, direct questions to more complex ones, building your confidence step-by-step.

Think of a pizza. The entire pizza is a circular region. When you take a slice, you're taking a sector. The crust of that slice is the arc, and the straight-line cut across the crust (if you were to cut it off) would form a chord. The area of the topping on the crusty edge, separate from the main slice, would be the segment. Mastering these calculations is key to solving a wide variety of geometry problems, from designing a garden to calculating the area a wiper clears on a car's windshield.

{{FORMULA: expr=(θ/360) × πr² | symbols=θ:angle of the sector in degrees, r:radius of the circle}}

Definitions & Formulas

Let's quickly recap the essential terms and formulas we'll be using throughout this section.

Logic Behind the Formulas

The formulas for the area of a sector and the length of an arc are derived using a simple and powerful idea: the Unitary Method. We treat the entire circle as a sector with a central angle of 360°.

1. The total area of a circle is πr². This corresponds to the total angle at the center, which is 360°.

2. Using the unitary method, if 360° corresponds to an area of πr², then an angle of 1° must correspond to a much smaller area.

   Area for 1° = (πr²) / 360
   

3. Therefore, for any given angle θ (in degrees), the area of the corresponding sector is simply θ times the area for 1°.

   Area of Sector = θ × (πr² / 360)
   

4. Rearranging this gives us the final formula.

   Area of the sector of angle θ = (θ/360) × πr²
   

5. The exact same logic applies to the length of an arc. The total length (circumference) is 2πr for 360°. So, for an angle θ, the length of the arc is:

   Length of an arc of angle θ = (θ/360) × 2πr
   

{{VISUAL: diagram: A circle with center O and radius r. A sector OAPB is shaded, with the angle AOB labeled as θ. The derivation steps are shown pointing to the respective parts of the circle.}}

Solved Examples

Let's work through some examples, starting from easy and gradually increasing in difficulty.

Example 1: Basic Sector Area (Easy)

Find the area of a sector of a circle with a radius of 6 cm if the angle of the sector is 60°. (Use π = 22/7)

Given: Radius r = 6 cm, Angle θ = 60°

To Find: Area of the sector.

Solution:

1. Start with the formula for the area of a sector.

   Area = (θ/360) × πr²
   

2. Substitute the given values of θ, r, and π into the formula.

   Area = (60/360) × (22/7) × 6²
   

3. Simplify the fraction and calculate the square. 60/360 simplifies to 1/6.

   Area = (1/6) × (22/7) × 36
   

4. Perform the final calculation. The 6 in the denominator cancels with 36 to leave 6.

   Area = (22/7) × 6 = 132/7
   

Final Answer:

Area of the sector = 132/7 cm² (approx. 18.86 cm²)

Example 2: The Clock Problem (Medium)

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. (Use π = 22/7)

Given: Radius (length of minute hand) r = 14 cm, Time t = 5 minutes.

To Find: Area swept by the minute hand.

Solution:

1. First, we need to find the angle θ swept by the minute hand in 5 minutes. The minute hand completes a full 360° circle in 60 minutes.

2. Using the unitary method, we can find the angle swept per minute.

   Angle swept in 60 minutes = 360°
   Angle swept in 1 minute = 360° / 60 = 6°
   

3. Now, calculate the angle swept in 5 minutes.

   θ = 5 minutes × 6°/minute = 30°
   

4. Now that we have r = 14 cm and θ = 30°, we can use the area of a sector formula.

   Area = (θ/360) × πr²
   

5. Substitute the values into the formula.

   Area = (30/360) × (22/7) × 14 × 14
   

6. Simplify the expression. 30/360 is 1/12. The 7 cancels with one 14 to leave 2.

   Area = (1/12) × 22 × 2 × 14 = (1/12) × 616
   

7. Calculate the final area.

   Area = 616 / 12 = 154 / 3
   

Final Answer:

Area swept = 154/3 cm² (approx. 51.33 cm²)

Example 3: Minor Segment with a Right Angle (Hard)

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. (Use π = 3.14)

Given: Radius r = 10 cm, Angle θ = 90°.

To Find: Area of the minor segment.

Solution:

1. The formula for the area of a segment is Area of Sector - Area of Triangle. Let's find each part separately.

   {{VISUAL: diagram: A circle with center O and radius 10cm. A chord AB is drawn such that angle AOB is 90°. The segment APB is shaded. Triangle OAB is clearly a right-angled triangle.}}

2. First, calculate the area of the sector OAPB.

   Area of Sector = (θ/360) × πr² = (90/360) × 3.14 × 10²
   

3. Simplify and compute the value. 90/360 is 1/4.

   Area of Sector = (1/4) × 3.14 × 100 = 314 / 4 = 78.5 cm²
   

4. Next, calculate the area of the triangle ΔOAB. Since ∠AOB = 90°, ΔOAB is a right-angled triangle with base OA and height OB (both are radii).

   Area of ΔOAB = ½ × base × height = ½ × r × r
   

5. Substitute the value of the radius.

   Area of ΔOAB = ½ × 10 × 10 = 50 cm²
   

6. Finally, subtract the area of the triangle from the area of the sector to get the area of the segment.

   Area of Segment = 78.5 - 50 = 28.5 cm²
   

Final Answer:

Area of the minor segment = 28.5 cm²

{{KEY: type=concept | title=Finding the Area of the Triangle | text=The most crucial step in finding a segment's area is correctly calculating the area of the triangle formed by the two radii and the chord. Identify the type of triangle: it's isosceles (two sides are radii). If θ=90°, it's a right-angled triangle. If θ=60°, it's equilateral. For other angles like 120°, you'll need to use trigonometry.}}

Example 4: Segment with a 120° Angle (Tricky)

A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)

Given: Radius r = 12 cm, Angle θ = 120°.

To Find: Area of the segment.

Solution:

1. As before, we need to find Area of Sector - Area of ΔOAB. Let's start with the sector.

   Area of Sector = (120/360) × 3.14 × 12² = (1/3) × 3.14 × 144
   

2. Calculate the sector area.

   Area of Sector = 3.14 × 48 = 150.72 cm²
   

3. Now, we must find the area of ΔOAB. This is an isosceles triangle with OA = OB = 12 cm. We can find its area by dropping a perpendicular OM from O to the chord AB.

   {{VISUAL: diagram: A circle with center O and radius 12cm. A chord AB creates an angle AOB of 120°. A perpendicular OM is dropped from O to AB. Triangle OMA is shown as a right-angled triangle with angle AOM = 60°.}}

4. In an isosceles triangle, the perpendicular from the vertex bisects the angle and the base. So, ∠AOM = 120°/2 = 60° and AM = MB.

5. In the right-angled triangle ΔOMA, we use trigonometry to find the height OM and the base AM.

   OM/OA = cos(60°)  =>  OM = 12 × (½) = 6 cm
   AM/OA = sin(60°)  =>  AM = 12 × (√3 / 2) = 6√3 cm
   

6. The full base of the triangle is AB = 2 × AM.

   AB = 2 × 6√3 = 12√3 cm
   

7. Now calculate the area of ΔOAB using the formula ½ × base × height.

   Area of ΔOAB = ½ × AB × OM = ½ × 12√3 × 6 = 36√3 cm²
   

8. Substitute √3 = 1.73.

   Area of ΔOAB = 36 × 1.73 = 62.28 cm²
   

9. Finally, calculate the area of the segment.

   Area of Segment = Area of Sector - Area of ΔOAB = 150.72 - 62.28
   

Final Answer:

Area of the segment = 88.44 cm²

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m long rope. What is the area of the part of the field in which the horse can graze?

   💡 Answer: The horse can graze in a quadrant (a sector with a 90° angle) of a circle with a radius of 5 m. Area = (90/360) × πr² = (1/4) × 3.14 × 5² = 19.625 m².

2. The area of a sector is 1/12th the area of the circle. What is the central angle of the sector?

   💡 Answer: If the area of the sector is 1/12th the area of the circle, then the angle must be 1/12th of the total angle (360°). Angle = (1/12) × 360° = 30°.

3. In a circle of radius 35 cm, an arc subtends an angle of 72° at the centre. Find the area of the sector and the length of the arc. (Use π = 22/7)

   💡 Answer: Area of Sector = (72/360) × (22/7) × 35² = (1/5) × 22 × 5 × 35 = 770 cm². Length of Arc = (72/360) × 2 × (22/7) × 35 = (1/5) × 2 × 22 × 5 = 44 cm.

Mini Cheatsheet

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Summary & Quick Revision

Chapter 11: Areas Related to Circles

Page 5/5: Summary & Quick Revision

{{FORMULA: expr=(θ/360) × πr² | symbols=θ:angle of the sector in degrees, r:radius of the circle, π:Pi (≈ 22/7 or 3.14)}}

Concept Introduction

Welcome to the final revision page for "Areas Related to Circles"! Throughout this chapter, we've explored how to break down a circle into its fundamental parts: sectors, segments, and arcs. Think of a pizza. The entire pizza is the circle. When you take a slice, you're holding a sector. The curved crust of your slice is the arc. If you were to draw a straight line from one end of the crust to the other, the smaller piece containing just the toppings and crust is a segment.

Understanding these components allows us to calculate specific areas and lengths in countless real-world scenarios, from designing a garden path to calculating the area swept by a radar. This summary will reinforce the key formulas and problem-solving techniques you've learned, ensuring you can tackle any question with confidence. Let's consolidate our knowledge!

Definitions & Formulas

Here are the essential terms and the formulas that connect them. Remember, θ is always measured in degrees for these formulas.

Logic Behind the Formulas

The formulas for arc length and sector area are not magic; they are derived from a simple, proportional relationship with the entire circle.

1. A complete circle represents an angle of 360°. The total area of this circle is πr².

2. If we consider a tiny slice of the circle with an angle of just 1°, its area would be a fraction of the total area.

   Area of 1° sector = (1/360) × πr²
   

3. Now, to find the area of a sector with any angle θ, we simply multiply the area of the 1° sector by θ.

   Area of sector with angle θ = θ × (1/360) × πr² = (θ/360) × πr²
   

4. The same logic applies to the length of an arc. The total length around a circle (its circumference) is 2πr for 360°.

   {{VISUAL: diagram: A circle with center O and radius r. A sector AOB is shaded, with the central angle marked as θ. The arc length between A and B is labeled 'l'.}}

5. The length of an arc for a 1° sector would be a tiny fraction of the total circumference.

   Length of 1° arc = (1/360) × 2πr
   

6. Therefore, for a sector with an angle θ, the arc length l is θ times the length of the 1° arc.

   Length of arc with angle θ = θ × (1/360) × 2πr = (θ/360) × 2πr
   

Solved Examples

Let's work through some problems, starting from easy and moving to tricky. (Use π = 22/7 unless stated otherwise)

Example 1: Direct Formula Application (Easy)

Given: A sector of a circle with radius r = 21 cm and angle θ = 60°.

To Find: The area of the sector.

Solution:

1. Start with the formula for the area of a sector.

   Area = (θ/360) × πr²
   

2. Substitute the given values of θ, r, and π.

   Area = (60/360) × (22/7) × (21)²
   

3. Simplify the fraction and calculate the square of the radius.

   Area = (1/6) × (22/7) × (21 × 21)
   

4. Perform the final calculation. Notice that 6 × 7 = 42, which cancels out neatly with 21 × 2.

   Area = (1/6) × 22 × 3 × 21 = 11 × 21 = 231 cm²
   

Final Answer:

The area of the sector is 231 cm².

Example 2: Finding the Area of a Segment (Medium)

Given: A chord of a circle with radius r = 14 cm subtends an angle θ = 90° at the center.

To Find: The area of the corresponding minor segment.

Solution:

1. The area of a segment is the area of the sector minus the area of the triangle formed by the two radii and the chord.

   Area of Segment = Area of Sector - Area of Triangle OAB
   

   {{VISUAL: diagram: A circle with center O and radius 14 cm. Radii OA and OB form a 90° angle. The chord AB is drawn. The minor segment (the area between chord AB and arc AB) is shaded.}}

2. First, calculate the area of the sector OAB.

   Area of Sector = (90/360) × (22/7) × 14² = (1/4) × (22/7) × 196
   

   Area of Sector = (1/4) × 22 × 28 = 22 × 7 = 154 cm²
   

3. Next, calculate the area of the triangle OAB. Since the angle at the center is 90°, triangle OAB is a right-angled triangle with the radii OA and OB as its base and height.

   Area of ΔOAB = ½ × base × height = ½ × 14 × 14
   

   Area of ΔOAB = ½ × 196 = 98 cm²
   

4. Finally, subtract the triangle's area from the sector's area.

   Area of Segment = 154 - 98 = 56 cm²
   

Final Answer:

The area of the minor segment is 56 cm².

Example 3: Real-World Application (Hard)

Given: The minute hand of a wall clock is 7 cm long.

To Find: The area swept by the minute hand in 20 minutes.

Solution:

1. The area swept by the minute hand is a sector. We have the radius r = 7 cm, but we need to find the angle θ it moves in 20 minutes.

2. The minute hand completes a full circle (360°) in 60 minutes. We can find the angle it moves per minute.

   Angle per minute = 360° / 60 = 6°
   

3. Now, calculate the total angle swept in 20 minutes.

   θ = 20 minutes × 6°/minute = 120°
   

4. Now we have r = 7 cm and θ = 120°. We can use the area of a sector formula.

   Area = (θ/360) × πr² = (120/360) × (22/7) × 7²
   

5. Simplify the expression.

   Area = (1/3) × (22/7) × 49 = (1/3) × 22 × 7
   

   Area = 154 / 3 ≈ 51.33 cm²
   

Final Answer:

The area swept by the minute hand in 20 minutes is 154/3 cm² (or approx. 51.33 cm²).

Example 4: The Tricky Multiple Choice

Given: A question asks for the area of a sector with angle p (in degrees) and radius R. The options are: (A) (p/180) × 2πR (B) (p/180) × πR² (C) (p/360) × 2πR (D) (p/720) × 2πR²

To Find: The correct option.

Solution:

1. Write down the standard formula for the area of a sector, using the variables p for the angle and R for the radius.

   Area = (p/360) × πR²
   

2. Now, evaluate each option to see which one is mathematically equivalent to our standard formula.

   • Option (A) is the formula for arc length, but with 180 instead of 360. Incorrect.
   • Option (B) is the area formula, but with 180 instead of 360. Incorrect.
   • Option (C) is the correct formula for the length of an arc. Incorrect.

3. Let's examine Option (D) carefully. It looks different, but we can simplify it.

   Option (D) = (p/720) × 2πR²
   

4. We can cancel the 2 in the numerator with the 720 in the denominator.

   (p / (2 × 360)) × 2πR² = (p/360) × πR²
   

5. This simplified form matches our standard formula exactly. This type of manipulation is a common trick in multiple-choice questions to test your algebraic confidence.

Final Answer:

Option (D) is the correct answer.

{{KEY: type=concept | title=Formula Manipulation | text=Examiners often disguise correct formulas by multiplying the numerator and denominator by the same number (in the last example, by 2). Always simplify the options provided before concluding they are incorrect.}}

Tips & Tricks

Here are a few shortcuts to speed up your calculations.

Common Mistakes to Avoid

Pay close attention to these common pitfalls. A small mistake can lead to a completely different answer.

{{VISUAL: diagram: A side-by-side comparison. Left side shows a full sector shaded, labeled 'Area of Sector'. Right side shows the same sector, but with the inner triangle un-shaded and only the segment shaded, labeled 'Area of Segment'.}}

Brain-Teaser Questions

1. The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. What is the area of the sector?

   💡 Answer: The perimeter of a sector = 2r + l (two radii + arc length). Given P = 16.4 cm and r = 5.2 cm. 16.4 = 2(5.2) + l → 16.4 = 10.4 + l → l = 6 cm. Now use the shortcut formula: Area = ½ × l × r Area = ½ × 6 × 5.2 = 3 × 5.2 = 15.6 cm².

2. A car has two windshield wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. (Use π = 22/7)

   💡 Answer: The area cleaned by one wiper is the area of a sector with r = 25 cm and θ = 115°. Area of one sector = (115/360) × (22/7) × 25². Area = (23/72) × (22/7) × 625. Area = (23 × 11 × 625) / (36 × 7) = 158125 / 252 cm². Since there are two wipers, the total area is 2 × (158125 / 252) = 158125 / 126 ≈ 1254.96 cm².

3. The arc length of a sector is l. If the radius of the circle is doubled, keeping the angle of the sector the same, what is the new arc length? And what is the effect on the area of the sector?

   💡 Answer: Arc length l = (θ/360) × 2πr. If r becomes 2r, the new length l' = (θ/360) × 2π(2r) = 2 × [(θ/360) × 2πr] = 2l. The arc length doubles. Area A = (θ/360) × πr². If r becomes 2r, the new area A' = (θ/360) × π(2r)² = (θ/360) × π(4r²) = 4 × [(θ/360) × πr²] = 4A. The area becomes four times larger.

Mini Cheatsheet

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