Areas Related To Circles

Theory:

A sector is a region of a circle bounded by two radii and the corresponding arc.

The area of a sector depends on the central angle \(\theta\) and radius \(r\).

Formula:

\[ \text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2 \]

Solution Roadmap:

Identify radius and central angle
Substitute values into sector area formula
Simplify step-by-step carefully
Write final answer with correct unit

Solution:

Radius of the circle, \(r = 6\ cm\)
Angle of the sector, \(\theta = 60^\circ\)

Step 1: Write formula

\[ A = \frac{\theta}{360^\circ} \times \pi r^2 \]

Step 2: Substitute values

\[ A = \frac{60}{360} \times \pi \times 6^2 \]

Step 3: Simplify fraction

\[ \frac{60}{360} = \frac{1}{6} \]

\[ A = \frac{1}{6} \times \pi \times 36 \]

Step 4: Multiply

\[ A = 6\pi \]

Step 5: Use \(\pi = \frac{22}{7}\)

\[ A = 6 \times \frac{22}{7} \]

\[ A = \frac{132}{7} \]

Step 6: Decimal value

\[ A \approx 18.857\ cm^2 \]

Final Answer: \(A \approx 18.857\ cm^2\)

Exam Significance:

Very common 2–3 mark direct formula-based question in CBSE Board exams
Tests clarity of formula application and simplification
Frequently used as a base for mixed problems (sector + segment)
Important for competitive exams like NTSE, Olympiads, and foundation JEE

Theory:

The circumference of a circle is given by:

\[ C = 2\pi r \]

A quadrant is one-fourth of a circle, so its central angle is:

\[ \theta = 90^\circ \]

Area of a sector:

\[ A = \frac{\theta}{360^\circ} \times \pi r^2 \]

Solution Roadmap:

Use circumference formula to find radius
Substitute radius into sector (quadrant) formula
Simplify step-by-step
Write final answer with unit

Solution:

Circumference of the circle, \(C = 22\ cm\)

Step 1: Use circumference formula

\[ 2\pi r = 22 \]

Step 2: Solve for radius

\[ r = \frac{22}{2\pi} \]

\[ r = \frac{11}{\pi} \]

Step 3: Substitute \(\pi = \frac{22}{7}\)

\[ r = \frac{11}{22/7} \]

\[ r = \frac{11 \times 7}{22} \]

\[ r = \frac{77}{22} \]

\[ r = 3.5\ cm \]

Step 4: Area of quadrant

\[ A = \frac{90^\circ}{360^\circ} \times \pi r^2 \]

\[ A = \frac{1}{4} \times \pi \times (3.5)^2 \]

Step 5: Square the radius

\[ (3.5)^2 = 12.25 \]

\[ A = \frac{1}{4} \times \pi \times 12.25 \]

Step 6: Substitute \(\pi = \frac{22}{7}\)

\[ A = \frac{1}{4} \times \frac{22}{7} \times 12.25 \]

Step 7: Simplify

\[ A = \frac{22 \times 12.25}{28} \]

\[ A = \frac{269.5}{28} \]

\[ A = 9.625\ cm^2 \]

Final Answer: \(A = 9.625\ cm^2\)

Exam Significance:

Tests ability to link circumference with area (multi-step concept)
Common in CBSE as 2–4 mark structured question
Important for problems involving sector transformations
Useful for competitive exams where formula chaining is tested

Theory:

The movement of the minute hand forms a sector of a circle.

The length of the minute hand acts as the radius.

In a clock:

60 minutes → 360°
1 minute → \(6^\circ\)

Sector Area Formula:

\[ A = \frac{\theta}{360^\circ} \times \pi r^2 \]

Solution Roadmap:

Convert time into angle
Use sector area formula
Simplify step-by-step
Write final answer in proper unit

Solution:

Length of minute hand = radius, \(r = 14\ cm\)

Step 1: Angle covered in 1 minute

\[ \frac{360^\circ}{60} = 6^\circ \]

Step 2: Angle covered in 5 minutes

\[ \theta = 5 \times 6^\circ = 30^\circ \]

Step 3: Apply sector area formula

\[ A = \frac{30^\circ}{360^\circ} \times \pi \times 14^2 \]

Step 4: Simplify fraction

\[ \frac{30}{360} = \frac{1}{12} \]

\[ A = \frac{1}{12} \times \pi \times 196 \]

Step 5: Substitute \(\pi = \frac{22}{7}\)

\[ A = \frac{1}{12} \times \frac{22}{7} \times 196 \]

Step 6: Simplify step-by-step

\[ 196 \div 7 = 28 \]

\[ A = \frac{1}{12} \times 22 \times 28 \]

\[ A = \frac{616}{12} \]

\[ A = \frac{154}{3} \]

\[ A \approx 51.33\ cm^2 \]

Final Answer: \(A = \frac{154}{3}\ cm^2\)

Exam Significance:

Classic application-based question combining time and geometry
Tests angle conversion + sector formula
Frequently asked in CBSE as 3–4 mark problem
Important for NTSE, Olympiads, and aptitude-based exams

Geometric representation of chord and sector

Theory:

Area of sector: \(\displaystyle A = \frac{\theta}{360^\circ} \times \pi r^2\)
Area of segment = Area of sector − Area of triangle
For central angle \(90^\circ\), triangle formed is right-angled

Solution Roadmap:

Find area of triangle using \(\frac{1}{2}r^2\sin\theta\)
Find area of minor sector
Subtract to get minor segment
Find major sector using remaining angle

Solution:

Radius, \(r = 10\ cm\)
Central angle, \(\theta = 90^\circ\)

Step 1: Area of triangle AOB

\[ A_{\triangle} = \frac{1}{2} r^2 \sin \theta \]

\[ A_{\triangle} = \frac{1}{2} \times 10^2 \times \sin 90^\circ \]

\[ A_{\triangle} = \frac{1}{2} \times 100 \times 1 \]

\[ A_{\triangle} = 50\ cm^2 \]

Step 2: Area of minor sector

\[ A_{\text{sector}} = \frac{90}{360} \times 3.14 \times 10^2 \]

\[ A_{\text{sector}} = \frac{1}{4} \times 3.14 \times 100 \]

\[ A_{\text{sector}} = 78.5\ cm^2 \]

Step 3: Area of minor segment

\[ A_{\text{segment}} = 78.5 - 50 \]

\[ A_{\text{segment}} = 28.5\ cm^2 \]

Step 4: Area of major sector

Remaining angle:

\[ 360^\circ - 90^\circ = 270^\circ \]

\[ A_{\text{major sector}} = \frac{270}{360} \times 3.14 \times 100 \]

\[ A_{\text{major sector}} = \frac{3}{4} \times 3.14 \times 100 \]

\[ A_{\text{major sector}} = 235.5\ cm^2 \]

Area of Minor Segment = \(28.5\ cm^2\)
Area of Major Sector = \(235.5\ cm^2\)

Exam Significance:

Very important multi-concept problem (sector + triangle + segment)
Frequently appears as 4–5 mark question in CBSE boards
Tests conceptual clarity of segment vs sector
High relevance for NTSE, Olympiads, and JEE foundation

Arc, sector and corresponding segment

Theory:

Arc length: \(\displaystyle l = \frac{\theta}{360^\circ} \times 2\pi r\)
Sector area: \(\displaystyle A = \frac{\theta}{360^\circ} \times \pi r^2\)
Segment area = Sector area − Triangle area

Solution Roadmap:

Find arc length using angle proportion
Find sector area
Find triangle area using trigonometry
Subtract to get segment area

Solution:

Radius, \(r = 21\ cm\)
Angle, \(\theta = 60^\circ\)

(i) Length of arc

\[ l = \frac{60}{360} \times 2\pi \times 21 \]

\[ l = \frac{1}{6} \times 2 \times \frac{22}{7} \times 21 \]

\[ l = \frac{1}{6} \times 2 \times 22 \times 3 \]

\[ l = 22\ cm \]

(ii) Area of sector

\[ A = \frac{60}{360} \times \pi \times 21^2 \]

\[ A = \frac{1}{6} \times \frac{22}{7} \times 441 \]

\[ A = \frac{1}{6} \times 22 \times 63 \]

\[ A = 231\ cm^2 \]

(iii) Area of segment

Step 1: Area of triangle AOB

\[ A_{\triangle} = \frac{1}{2} r^2 \sin 60^\circ \]

\[ A_{\triangle} = \frac{1}{2} \times 21^2 \times \frac{\sqrt{3}}{2} \]

\[ A_{\triangle} = \frac{441\sqrt{3}}{4} \]

Step 2: Area of segment

\[ A_{\text{segment}} = 231 - \frac{441\sqrt{3}}{4} \]

Length of arc = \(22\ cm\)

Area of sector = \(231\ cm^2\)

Area of segment = \(231 - \frac{441\sqrt{3}}{4}\ cm^2\)

Exam Significance:

Classic 3-in-1 problem (arc + sector + segment)
Frequently asked as 4–5 mark long question
Tests understanding of geometry + trigonometry integration
Highly important for Olympiads and JEE foundation

Arc, sector and corresponding segment

Theory:

Arc length: \(\displaystyle l = \frac{\theta}{360^\circ} \times 2\pi r\)
Sector area: \(\displaystyle A = \frac{\theta}{360^\circ} \times \pi r^2\)
Segment area = Sector area − Triangle area

Solution Roadmap:

Find arc length using angle proportion
Find sector area
Find triangle area using trigonometry
Subtract to get segment area

Solution:

Radius, \(r = 21\ cm\)
Angle, \(\theta = 60^\circ\)

(i) Length of arc

\[ l = \frac{60}{360} \times 2\pi \times 21 \]

\[ l = \frac{1}{6} \times 2 \times \frac{22}{7} \times 21 \]

\[ l = \frac{1}{6} \times 2 \times 22 \times 3 \]

\[ l = 22\ cm \]

(ii) Area of sector

\[ A = \frac{60}{360} \times \pi \times 21^2 \]

\[ A = \frac{1}{6} \times \frac{22}{7} \times 441 \]

\[ A = \frac{1}{6} \times 22 \times 63 \]

\[ A = 231\ cm^2 \]

(iii) Area of segment

Step 1: Area of triangle AOB

\[ A_{\triangle} = \frac{1}{2} r^2 \sin 60^\circ \]

\[ A_{\triangle} = \frac{1}{2} \times 21^2 \times \frac{\sqrt{3}}{2} \]

\[ A_{\triangle} = \frac{441\sqrt{3}}{4} \]

Step 2: Area of segment

\[ A_{\text{segment}} = 231 - \frac{441\sqrt{3}}{4} \]

Length of arc = \(22\ cm\)

Area of sector = \(231\ cm^2\)

Area of segment = \(231 - \frac{441\sqrt{3}}{4}\ cm^2\)

Exam Significance:

Classic 3-in-1 problem (arc + sector + segment)
Frequently asked as 4–5 mark long question
Tests understanding of geometry + trigonometry integration
Highly important for Olympiads and JEE foundation

Segment formed by 120° chord

Theory:

Sector area: \(\displaystyle A = \frac{\theta}{360^\circ} \times \pi r^2\)
Triangle area: \(\displaystyle A = \frac{1}{2} r^2 \sin\theta\)
Segment area = Sector area − Triangle area

Solution Roadmap:

Find sector area using \(\pi = 3.14\)
Find triangle area using \(\sin 120^\circ = \sin 60^\circ\)
Subtract to obtain segment area

Solution:

Radius, \(r = 12\ cm\)
Angle, \(\theta = 120^\circ\)

Step 1: Area of sector

\[ A_{\text{sector}} = \frac{120}{360} \times 3.14 \times 12^2 \]

\[ = \frac{1}{3} \times 3.14 \times 144 \]

\[ = \frac{452.16}{3} \]

\[ A_{\text{sector}} = 150.72\ cm^2 \]

Step 2: Area of triangle AOB

\[ A_{\triangle} = \frac{1}{2} r^2 \sin 120^\circ \]

\[ = \frac{1}{2} \times 144 \times \frac{\sqrt{3}}{2} \]

\[ = \frac{144\sqrt{3}}{4} \]

\[ = 36\sqrt{3} \]

\[ = 36 \times 1.73 \]

\[ A_{\triangle} = 62.28\ cm^2 \]

Step 3: Area of segment

\[ A_{\text{segment}} = 150.72 - 62.28 \]

\[ A_{\text{segment}} = 88.44\ cm^2 \]

Area of segment = \(88.44\ cm^2\)

Exam Significance:

Very common segment-based numerical problem
Tests correct use of \(\sin(120^\circ)\) identity
Frequent in CBSE as 3–4 mark question
Important for precision-based exams (NTSE, Olympiad)

Horse grazing as quadrant inside square

Theory:

Horse tied at corner → grazing region is a quadrant
Area of quadrant: \(\displaystyle A = \frac{1}{4}\pi r^2\)
If radius ≤ side of square, full quadrant lies inside field

Solution Roadmap:

Compute quadrant area for \(r = 5\)
Compute quadrant area for \(r = 10\)
Find increase by subtraction

Solution:

Side of square = 15 m

(i) When rope length = 5 m

Radius, \(r = 5\ m\)

\[ A_1 = \frac{1}{4} \times 3.14 \times 5^2 \]

\[ = \frac{1}{4} \times 3.14 \times 25 \]

\[ = \frac{78.5}{4} \]

\[ A_1 = 19.625\ m^2 \]

(ii) When rope length = 10 m

Radius, \(r = 10\ m\)

\[ A_2 = \frac{1}{4} \times 3.14 \times 10^2 \]

\[ = \frac{1}{4} \times 3.14 \times 100 \]

\[ = \frac{314}{4} \]

\[ A_2 = 78.5\ m^2 \]

Step 3: Increase in grazing area

\[ \text{Increase} = A_2 - A_1 \]

\[ = 78.5 - 19.625 \]

\[ = 58.875\ m^2 \]

(i) Grazing area = \(19.625\ m^2\)
(ii) Increase in grazing area = \(58.875\ m^2\)

Exam Significance:

Classic real-life application of sectors
Tests understanding of quadrant geometry in constraints
Frequently appears as case-based CBSE question
Important for visual reasoning in competitive exams

Circle divided into 10 equal sectors

Theory:

Circumference: \(\displaystyle C = 2\pi r\)
Each diameter = full straight length across circle
Sector angle = \(\displaystyle \frac{360^\circ}{10} = 36^\circ\)
Sector area: \(\displaystyle A = \frac{\theta}{360^\circ}\pi r^2\)

Solution Roadmap:

Find circumference
Add length of 5 diameters
Find area of one sector

Solution:

Diameter = 35 mm
Radius, \(r = \frac{35}{2} = 17.5\ mm\)

(i) Total length of wire

Step 1: Circumference

\[ C = 2\pi r \]

\[ = 2 \times \frac{22}{7} \times 17.5 \]

\[ = 2 \times \frac{22}{7} \times \frac{35}{2} \]

\[ = 22 \times 5 \]

\[ C = 110\ mm \]

Step 2: Length of 5 diameters

\[ = 5 \times 35 = 175\ mm \]

Step 3: Total length

\[ \text{Total} = 110 + 175 \]

\[ = 285\ mm \]

(ii) Area of each sector

Angle of each sector:

\[ \theta = \frac{360^\circ}{10} = 36^\circ \]

Step 1: Apply sector formula

\[ A = \frac{36}{360} \times \pi \times (17.5)^2 \]

\[ = \frac{1}{10} \times \frac{22}{7} \times 306.25 \]

\[ = \frac{1}{10} \times 22 \times 43.75 \]

\[ = \frac{962.5}{10} \]

\[ A = 96.25\ mm^2 \]

Total length of silver wire = \(285\ mm\)
Area of each sector = \(96.25\ mm^2\)

Exam Significance:

Combination of perimeter + sector geometry
Tests clarity in multi-step construction problems
Common CBSE question with 2–4 marks weightage
Important for visual geometry reasoning

Area between two consecutive ribs

Theory:

Circle divided into equal sectors → each sector angle = \(\frac{360^\circ}{n}\)
Area of sector: \(\displaystyle A = \frac{\theta}{360^\circ} \times \pi r^2\)

Solution Roadmap:

Find central angle between two ribs
Apply sector area formula
Simplify step-by-step

Solution:

Radius, \(r = 45\ cm\)
Number of ribs = 8

Step 1: Angle between two ribs

\[ \theta = \frac{360^\circ}{8} = 45^\circ \]

Step 2: Apply sector area formula

\[ A = \frac{45}{360} \times \pi \times 45^2 \]

\[ = \frac{1}{8} \times \pi \times 2025 \]

Step 3: Substitute \(\pi = \frac{22}{7}\)

\[ A = \frac{1}{8} \times \frac{22}{7} \times 2025 \]

Step 4: Simplify

\[ A = \frac{22 \times 2025}{56} \]

\[ A = \frac{44550}{56} \]

\[ A \approx 795.54\ cm^2 \]

Area between two consecutive ribs = \(795.54\ cm^2\)

Exam Significance:

Direct sector-based application problem
Tests ability to convert equal division into angle
Common CBSE 2–3 mark question
Important for speed-based competitive exams

Theory:

Each wiper sweeps a sector of a circle
Area of sector: \(\displaystyle A = \frac{\theta}{360^\circ} \times \pi r^2\)
Total area = sum of both sectors (since no overlap)

Solution Roadmap:

Find area swept by one wiper
Multiply by 2 (no overlap)
Simplify step-by-step

Solution:

Radius, \(r = 25\ cm\)
Angle, \(\theta = 115^\circ\)

Step 1: Area swept by one wiper

\[ A_1 = \frac{115}{360} \times \pi \times 25^2 \]

\[ = \frac{115}{360} \times \pi \times 625 \]

Step 2: Substitute \(\pi = \frac{22}{7}\)

\[ A_1 = \frac{115}{360} \times \frac{22}{7} \times 625 \]

Step 3: Multiply numerator

\[ 115 \times 22 \times 625 = 1581250 \]

\[ A_1 = \frac{1581250}{2520} \]

Step 4: Simplify

\[ A_1 \approx 627.48\ cm^2 \]

Step 5: Total area for two wipers

\[ A = 2 \times 627.48 \]

\[ A \approx 1254.96\ cm^2 \]

Total area cleaned = \(1254.96\ cm^2\)

Exam Significance:

Tests understanding of multiple sector addition
Important case: non-overlapping regions
Common CBSE 3–4 mark application question
Useful in real-life geometry problems

Theory:

Lighthouse coverage forms a sector of a circle
Area of sector: \(\displaystyle A = \frac{\theta}{360^\circ} \times \pi r^2\)

Solution Roadmap:

Substitute given angle and radius
Compute \(r^2\)
Multiply step-by-step carefully

Solution:

Angle, \(\theta = 80^\circ\)
Radius, \(r = 16.5\ km\)

Step 1: Write formula

\[ A = \frac{80}{360} \times 3.14 \times (16.5)^2 \]

Step 2: Simplify fraction

\[ \frac{80}{360} = \frac{2}{9} \]

\[ A = \frac{2}{9} \times 3.14 \times (16.5)^2 \]

Step 3: Square the radius

\[ (16.5)^2 = 272.25 \]

\[ A = \frac{2}{9} \times 3.14 \times 272.25 \]

Step 4: Multiply

\[ 3.14 \times 272.25 = 854.865 \]

\[ A = \frac{2}{9} \times 854.865 \]

\[ A = \frac{1709.73}{9} \]

\[ A \approx 189.97\ km^2 \]

Area of sea warned = \(189.97\ km^2\)

Exam Significance:

Real-life sector application problem
Tests accuracy with decimal calculations
Common CBSE 2–3 mark question
Important for precision-based competitive exams

One design = sector − triangle

Theory:

Each design = sector − triangle
Sector area: \(\displaystyle \frac{\theta}{360^\circ}\pi r^2\)
Triangle area: \(\displaystyle \frac{1}{2} r^2 \sin\theta\)

Solution Roadmap:

Find area of one sector
Find area of triangle
Subtract to get one design
Multiply by 6 and then by cost rate

Solution:

Radius, \(r = 28\ cm\)
Number of designs = 6
Angle of each sector:

\[ \theta = \frac{360^\circ}{6} = 60^\circ \]

Step 1: Area of one sector

\[ A_{\text{sector}} = \frac{60}{360} \times \frac{22}{7} \times 28^2 \]

\[ = \frac{1}{6} \times \frac{22}{7} \times 784 \]

\[ = \frac{1}{6} \times 22 \times 112 \]

\[ A_{\text{sector}} = 410.67\ cm^2 \]

Step 2: Area of triangle

\[ A_{\triangle} = \frac{1}{2} r^2 \sin 60^\circ \]

\[ = \frac{1}{2} \times 28^2 \times \frac{\sqrt{3}}{2} \]

\[ = \frac{784\sqrt{3}}{4} \]

\[ = 196\sqrt{3} \]

\[ = 196 \times 1.7 = 333.2\ cm^2 \]

Step 3: Area of one design

\[ A_{\text{design}} = 410.67 - 333.2 \]

\[ = 77.47\ cm^2 \]

Step 4: Area of 6 designs

\[ A_{\text{total}} = 77.47 \times 6 = 464.82\ cm^2 \]

Step 5: Cost

\[ \text{Cost} = 464.82 \times 0.35 \]

\[ = 162.69 \]

Cost of designs = ₹162.69

Exam Significance:

Classic sector − triangle composite problem
Tests correct use of \(\sin 60^\circ\) and approximation
Frequent CBSE 4–5 mark question
Important for multi-step numerical accuracy

\(\dfrac{p}{180}2\pi R\)
\(\dfrac{p}{180}2\pi R^2\)
\(\dfrac{p}{360}2\pi R\)
\(\dfrac{p}{720}2\pi R^2\)

Theory:

The standard formula for the area of a sector is:

\[ A = \frac{\theta}{360^\circ} \times \pi R^2 \]

Solution Roadmap:

Recall standard sector area formula
Compare each option with correct expression
Simplify options where needed

Solution:

Correct formula:

\[ A = \frac{p}{360} \times \pi R^2 \]

Check Option D:

\[ \frac{p}{720} \times 2\pi R^2 \]

\[ = \frac{p}{720} \times 2 \times \pi R^2 \]

\[ = \frac{p}{360} \times \pi R^2 \]

This matches the standard formula.

Therefore, correct answer is:

D. \(\dfrac{p}{720} \times 2\pi R^2\)

Exam Insight (Fast Trick):

Area must contain \(R^2\) → eliminate A and C
Correct denominator must reduce to 360
Only option D simplifies correctly

θ	Sector Area	Arc Length	Triangle Area	Segment Area
30°	πr²/12	πr/6	r²/4	πr²/12 − r²/4
45°	πr²/8	πr/4	r²√2/4	πr²/8 − r²√2/4
60°	πr²/6	πr/3	√3r²/4	πr²/6 − √3r²/4
90°	πr²/4	πr/2	r²/2	(π/4 − 1/2)r²
120°	πr²/3	2πr/3	√3r²/4	πr²/3 − √3r²/4
180°	πr²/2	πr	r²	πr²/2 − r²

Chapter 11 — Areas Related to Circles

Theory:

Solution Roadmap:

Solution:

Final Answer: \(A \approx 18.857\ cm^2\)

Exam Significance:

Theory:

Solution Roadmap:

Solution:

Final Answer: \(A = 9.625\ cm^2\)

Exam Significance:

Theory:

Solution Roadmap:

Solution:

Final Answer: \(A = \frac{154}{3}\ cm^2\)

Exam Significance:

Theory:

Solution Roadmap:

Solution:

Area of Minor Segment = \(28.5\ cm^2\) Area of Major Sector = \(235.5\ cm^2\)

Exam Significance:

Theory:

Solution Roadmap:

Solution:

Length of arc = \(22\ cm\) Area of sector = \(231\ cm^2\) Area of segment = \(231 - \frac{441\sqrt{3}}{4}\ cm^2\)

Exam Significance:

Theory:

Solution Roadmap:

Solution:

Length of arc = \(22\ cm\) Area of sector = \(231\ cm^2\) Area of segment = \(231 - \frac{441\sqrt{3}}{4}\ cm^2\)

Exam Significance:

Theory:

Solution Roadmap:

Solution:

Area of segment = \(88.44\ cm^2\)

Exam Significance:

Theory:

Solution Roadmap:

Solution:

(i) Grazing area = \(19.625\ m^2\) (ii) Increase in grazing area = \(58.875\ m^2\)

Exam Significance:

Theory:

Solution Roadmap:

Solution:

Total length of silver wire = \(285\ mm\) Area of each sector = \(96.25\ mm^2\)

Exam Significance:

Theory:

Solution Roadmap:

Solution:

Area between two consecutive ribs = \(795.54\ cm^2\)

Exam Significance:

Theory:

Solution Roadmap:

Solution:

Total area cleaned = \(1254.96\ cm^2\)

Exam Significance:

Theory:

Solution Roadmap:

Solution:

Area of sea warned = \(189.97\ km^2\)

Exam Significance:

Theory:

Solution Roadmap:

Solution:

Cost of designs = ₹162.69

Exam Significance:

Theory:

Solution Roadmap:

Solution:

D. \(\dfrac{p}{720} \times 2\pi R^2\)

Exam Insight (Fast Trick):

Chapter Complete!

🗺️ Concept Map — Chapter Overview

📚 Core Definitions

What is a Sector?

What is a Segment?

Important Angle Cases

📐 All Formulas at a Glance

Quick Reference — Special Angles

Area of Minor Segment = \(28.5\ cm^2\)
Area of Major Sector = \(235.5\ cm^2\)

Length of arc = \(22\ cm\)

Area of sector = \(231\ cm^2\)

Area of segment = \(231 - \frac{441\sqrt{3}}{4}\ cm^2\)

Length of arc = \(22\ cm\)

Area of sector = \(231\ cm^2\)

Area of segment = \(231 - \frac{441\sqrt{3}}{4}\ cm^2\)

(i) Grazing area = \(19.625\ m^2\)
(ii) Increase in grazing area = \(58.875\ m^2\)

Total length of silver wire = \(285\ mm\)
Area of each sector = \(96.25\ mm^2\)