Chapter 11 : Class 10 : Areas Related To Circles : Case Study Question Bank

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Chapter 11: Areas Related to Circles

Class 10 Standard Mathematics (Case Study Bank)

Curated By Akash Srivastva

✨ Interactive TLMs, Worksheets & Digital Study Notes: www.akashmaths.online

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Case Study 1 Clean Energy 4 Marks

Wind Turbine Sweep Zone

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A coastal wind turbine blade of length \(r = 42\text{ m}\) rotates about center hub O, sweeping a circular safety sector zone (use \(\pi = \frac{22}{7}\)).

Q1. Arc Length Calculation 1 Mark

Calculate the exact length of the arc swept by the outer tip of the blade when it rotates through an angle of 60°.

(A) 22 m

(B) 44 m

(C) 88 m

(D) 66 m

Q2. Sector Area 1 Mark

Calculate the exact area of the wind sweep sector when the turbine blade rotates through an angle of 90°.

(A) 1150 m²

(B) 1280 m²

(C) 1386 m²

(D) 1540 m²

Q3. Minor Segment Area 2 Marks

Find the area of the minor segment bounded by the chord connecting the endpoints of the 90° arc.

(A) 504 m²

(B) 480 m²

(C) 520 m²

(D) 462 m²

OR (Alternative Q3)

Calculate the exact area of the segment swept by the blade during a 60° rotation (take √3 ≈ 1.732).

(A) 140.50 m²

(B) (924 − 441√3) m² (≈ 160.19 m²)

(C) 175.25 m²

(D) 190.00 m²

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Arc length l = (θ/360°) × 2πr = (60°/360°) × 2 × (22/7) × 42 = (1/6) × 2 × 22 × 6 = 44 m.	[0.5 Mark] for formula. [0.5 Mark] for 44 m.
Q2	Area = (90°/360°) × (22/7) × 42 × 42 = (1/4) × 22 × 6 × 42 = 1386 m².	[1 Mark] for 1386 m².
Q3	Area of △ = (1/2) × 42 × 42 = 882 m². Area of Segment = 1386 − 882 = 504 m².	[1 Mark] for triangle area. [1 Mark] for segment area 504 m².
Q3 (OR)	Sector(60°) = (60°/360°) × (22/7) × 42² = 924 m². Equilateral △ = (√3/4) × 42² = 441√3 ≈ 763.81 m². Segment = 924 − 441√3 ≈ 160.19 m².	[1 Mark] for sector and triangle areas. [1 Mark] for (924 − 441√3) m².

Case Study 2 Aerospace Telecom 4 Marks

LEO Satellite Beam Footprint

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A telecommunications satellite S projects a conical radio beam forming a circular sector footprint on Earth with slant radius \(r = 210\text{ km}\) (use \(\pi = \frac{22}{7}\)).

Q1. Sector Perimeter 1 Mark

If the signal beam covers a sector with central angle θ = 60°, calculate the total perimeter of this sector.

(A) 640 km

(B) 580 km

(C) 620 km

(D) 700 km

Q2. Coverage Area 1 Mark

Calculate the exact coverage area (in km²) of the signal footprint for a central angle of 60°.

(A) 21,500 km²

(B) 23,100 km²

(C) 24,200 km²

(D) 25,000 km²

Q3. Geometric Segment & Expansion 2 Marks

If central angle is 90°, calculate the area of the circular segment bounded by the chord of the footprint.

(A) 11,200 km²

(B) 13,500 km²

(C) 12,600 km²

(D) 10,800 km²

OR (Alternative Q3)

If engineers increase central angle from 60° to 120°, calculate the exact increase in coverage area.

(A) 23,100 km²

(B) 21,000 km²

(C) 25,200 km²

(D) 26,400 km²

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Arc length l = (60°/360°) × 2 × (22/7) × 210 = 220 km. Perimeter = l + 2r = 220 + 2(210) = 640 km.	[0.5 Mark] for arc length 220 km. [0.5 Mark] for perimeter 640 km.
Q2	Area = (60°/360°) × (22/7) × 210 × 210 = 23,100 km².	[1 Mark] for 23,100 km².
Q3	Sector(90°) = (1/4) × (22/7) × 210² = 34,650 km². Triangle = (1/2) × 210² = 22,050 km². Segment = 34,650 − 22,050 = 12,600 km².	[1 Mark] for sector and triangle areas. [1 Mark] for 12,600 km².
Q3 (OR)	Area(120°) = (120°/360°) × (22/7) × 210² = 46,200 km². Increase = 46,200 − 23,100 = 23,100 km².	[1 Mark] for 120° sector area. [1 Mark] for increase = 23,100 km².

Case Study 3 Smart Agriculture 4 Marks

Automated Pivot Irrigation Sprinkler

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A center-pivot irrigation sprinkler arm of length \(r = 35\text{ m}\) rotates about fixed pivot point O, watering circular sectors of agricultural crops (use \(\pi = \frac{22}{7}\)).

Q1. Outer Arc Length 1 Mark

If the sprinkler arm rotates through an angle of 36°, calculate the exact length of the curved outer arc.

(A) 18 m

(B) 22 m

(C) 24 m

(D) 28 m

Q2. Watered Sector Area 1 Mark

Find the exact area of the field watered when the sprinkler arm sweeps through an angle of 72°.

(A) 770 m²

(B) 720 m²

(C) 800 m²

(D) 690 m²

Q3. Minor Segment & Field Expansion 2 Marks

For a sector of angle 120°, find the area of the minor segment between the straight path and arc (take √3 ≈ 1.732).

(A) 720.50 m²

(B) 780.15 m²

(C) 752.91 m²

(D) 742.80 m²

OR (Alternative Q3)

If setting expands from 60° to 120°, calculate the exact increase in watered field area.

(A) 616.00 m²

(B) 1925/3 m² (≈ 641.67 m²)

(C) 700.00 m²

(D) 550.50 m²

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	l = (36°/360°) × 2 × (22/7) × 35 = (1/10) × 220 = 22 m.	[1 Mark] for l = 22 m.
Q2	Area = (72°/360°) × (22/7) × 35 × 35 = (1/5) × 22 × 5 × 35 = 770 m².	[1 Mark] for 770 m².
Q3	Sector(120°) = (1/3) × (22/7) × 35² = 3850/3 ≈ 1283.33 m². Triangle = (1/2) × 35² × sin 120° = (1225√3)/4 ≈ 530.42 m². Segment = 1283.33 − 530.42 = 752.91 m².	[1 Mark] for sector and triangle areas. [1 Mark] for 752.91 m².
Q3 (OR)	Increase = Area(120°) − Area(60°) = 3850/3 − 1925/3 = 1925/3 ≈ 641.67 m².	[1 Mark] for difference setup. [1 Mark] for 1925/3 m².

Case Study 4 Smart UI/UX 4 Marks

Digital Fitness Band Screen

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A circular smartwatch display has active screen radius \(r = 28\text{ mm}\). Health metrics highlight progress sectors with central angle θ (use \(\pi = \frac{22}{7}\)).

Q1. Progress Arc Length 1 Mark

If the progress arc subtends a central angle of 45°, find the exact length of this arc in mm.

(A) 22 mm

(B) 24 mm

(C) 28 mm

(D) 18 mm

Q2. Highlighted Screen Area 1 Mark

Calculate the exact area of the circular sector highlighted when central angle reaches 120°.

(A) 780.50 mm²

(B) 2464/3 mm² (≈ 821.33 mm²)

(C) 850.00 mm²

(D) 2156/3 mm²

Q3. Goal Segment & Widgets 2 Marks

When central angle is 90° (25% goal), calculate the area of the minor segment formed by the chord.

(A) 210 mm²

(B) 240 mm²

(C) 224 mm²

(D) 256 mm²

OR (Alternative Q3)

If the circular screen is divided into 8 equal widget sectors, calculate the area of each sector.

(A) 308 mm²

(B) 320 mm²

(C) 296 mm²

(D) 315 mm²

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	l = (45°/360°) × 2 × (22/7) × 28 = (1/8) × 2 × 22 × 4 = 22 mm.	[1 Mark] for l = 22 mm.
Q2	Area = (120°/360°) × (22/7) × 28 × 28 = (1/3) × 22 × 4 × 28 = 2464/3 ≈ 821.33 mm².	[1 Mark] for 2464/3 mm².
Q3	Sector(90°) = (1/4) × (22/7) × 28² = 616 mm². Triangle = (1/2) × 28 × 28 = 392 mm². Segment = 616 − 392 = 224 mm².	[1 Mark] for sector and triangle areas. [1 Mark] for 224 mm².
Q3 (OR)	θ = 360°/8 = 45°. Area = (1/8) × (22/7) × 28² = 308 mm².	[1 Mark] for angle 45°. [1 Mark] for 308 mm².

Case Study 5 Automotive Engineering 4 Marks

Windshield Wiper Blades

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An EV rear windshield wiper of arm length \(R = 35\text{ cm}\) sweeps a sector through central angle \(\theta = 120^\circ\) about pivot O (use \(\pi = \frac{22}{7}\)).

Q1. Outer Tip Arc Length 1 Mark

Calculate the distance (arc length) traversed by the outermost tip in a 120° sweep.

(A) 66.67 cm

(B) 220/3 cm (≈ 73.33 cm)

(C) 80.00 cm

(D) 250/3 cm

Q2. Maximum Swept Area 1 Mark

Find the maximum area swept if the full length of 35 cm is fitted with cleaning rubber.

(A) 3850/3 cm² (≈ 1283.33 cm²)

(B) 1150.00 cm²

(C) 1350.50 cm²

(D) 3500/3 cm²

Q3. Actual Cleaned Area 2 Marks

If innermost 14 cm is non-cleaning support and only outer 21 cm cleans, find actual area cleaned.

(A) 980 cm²

(B) 1024 cm²

(C) 1078 cm²

(D) 1120 cm²

OR (Alternative Q3)

If sweep angle is 90° and active blade length extends to R = 42 cm from O, find area cleaned.

(A) 1250 cm²

(B) 1420 cm²

(C) 1300 cm²

(D) 1386 cm²

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	l = (120°/360°) × 2 × (22/7) × 35 = (1/3) × 220 = 220/3 ≈ 73.33 cm.	[1 Mark] for 220/3 cm.
Q2	Area = (120°/360°) × (22/7) × 35² = 3850/3 ≈ 1283.33 cm².	[1 Mark] for 3850/3 cm².
Q3	Area = (120°/360°) × (22/7) × (35² − 14²) = (1/3) × (22/7) × 21 × 49 = 22 × 49 = 1078 cm².	[1 Mark] for concentric difference. [1 Mark] for 1078 cm².
Q3 (OR)	Area = (90°/360°) × (22/7) × 42² = (1/4) × 22 × 6 × 42 = 1386 cm².	[1 Mark] for formula setup. [1 Mark] for 1386 cm².

Case Study 6 Eco-Architecture 4 Marks

Eco-Dome Glass Segment Design

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A circular skylight of radius \(R = 14\text{ m}\) (center O) has an equilateral triangle framework ABC inscribed inside it, creating three circular segment glass panels (use \(\pi = \frac{22}{7}\), \(\sqrt{3} \approx 1.732\)).

Q1. Total Skylight Area 1 Mark

Find the total surface area of the central circular skylight (R = 14 m).

(A) 580 m²

(B) 616 m²

(C) 640 m²

(D) 528 m²

Q2. Subtended Angle & Arc Length 1 Mark

Determine the angle subtended by chord AB at O, and calculate its arc length.

(A) 120°; 88/3 m (≈ 29.33 m)

(B) 90°; 22 m

(C) 60°; 44/3 m

(D) 120°; 44 m

Q3. Segment & Triangle Area 2 Marks

Calculate the surface area of one of the three circular segment glass panels.

(A) 105.20 m²

(B) 115.80 m²

(C) 120.46 m²

(D) 128.50 m²

OR (Alternative Q3)

Find the total surface area of the inscribed equilateral triangular frame ABC.

(A) 147√3 m² (≈ 254.60 m²)

(B) 196√3 m²

(C) 120√3 m²

(D) 240.50 m²

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Area = πR² = (22/7) × 14 × 14 = 616 m².	[1 Mark] for 616 m².
Q2	Angle θ = 360°/3 = 120°. Arc length = (120°/360°) × 2 × (22/7) × 14 = 88/3 ≈ 29.33 m.	[0.5 Mark] for angle 120°. [0.5 Mark] for 88/3 m.
Q3	Sector(120°) = 616/3 ≈ 205.33 m². △OAB = (1/2) × 14² × sin 120° = 49√3 ≈ 84.87 m². Segment = 205.33 − 84.87 = 120.46 m².	[1 Mark] for sector and triangle areas. [1 Mark] for 120.46 m².
Q3 (OR)	Triangle ABC area = 3 × Area(△OAB) = 3 × 49√3 = 147√3 ≈ 254.60 m².	[1 Mark] for 3 × Area(△OAB). [1 Mark] for 147√3 m².

Case Study 7 Sustainable Jewelry 4 Marks

Recycled Gold Brooch

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A circular gold brooch of radius \(r = 35\text{ mm}\) has 8 radial gold spokes dividing it into 8 equal sectors centered at O (use \(\pi = \frac{22}{7}\)).

Q1. Total Wire Length 1 Mark

Find the total length of gold wire required for outer circular rim and 8 internal radial spokes.

(A) 480 mm

(B) 500 mm (or 50 cm)

(C) 520 mm

(D) 450 mm

Q2. Individual Sector Area 1 Mark

Determine the central angle and calculate the exact area of one sector of the brooch.

(A) 60°; 520 mm²

(B) 45°; 450 mm²

(C) 45°; 481.25 mm²

(D) 40°; 460 mm²

Q3. Quadrant Segment & Rescaling 2 Marks

Find the area of the segment formed across 2 adjacent sectors with central angle 90°.

(A) 350 mm² (or 3.5 cm²)

(B) 325 mm²

(C) 380 mm²

(D) 365 mm²

OR (Alternative Q3)

If radius is increased to r = 42 mm for the 8-sector brooch, calculate increase in area of each sector.

(A) 195.50 mm²

(B) 211.75 mm²

(C) 225.00 mm²

(D) 205.25 mm²

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Rim circumference C = 2 × (22/7) × 35 = 220 mm. 8 spokes = 8 × 35 = 280 mm. Total wire = 220 + 280 = 500 mm (50 cm).	[0.5 Mark] for circumference. [0.5 Mark] for 500 mm.
Q2	θ = 360°/8 = 45°. Area = (45°/360°) × (22/7) × 35² = (1/8) × 3850 = 481.25 mm².	[0.5 Mark] for angle 45°. [0.5 Mark] for 481.25 mm².
Q3	Sector(90°) = (1/4) × (22/7) × 35² = 962.5 mm². △ = (1/2) × 35² = 612.5 mm². Segment = 962.5 − 612.5 = 350 mm² (3.5 cm²).	[1 Mark] for sector and triangle areas. [1 Mark] for 350 mm².
Q3 (OR)	New sector area = (1/8) × (22/7) × 42² = 693 mm². Increase = 693 − 481.25 = 211.75 mm².	[1 Mark] for new sector area 693 mm². [1 Mark] for increase 211.75 mm².

Case Study 8 Urban Drainage 4 Marks

Rainwater Drainage Grates

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A circular cast-iron drainage grate of radius \(r = 28\text{ cm}\) has an inscribed regular hexagon filtering mesh, leaving 6 outer circular segments open for runoff (use \(\pi = \frac{22}{7}\), \(\sqrt{3} \approx 1.732\)).

Q1. Total Grate Area 1 Mark

Find the total surface area of the entire circular drainage grate (r = 28 cm).

(A) 2250 cm²

(B) 2464 cm²

(C) 2600 cm²

(D) 2380 cm²

Q2. Hexagon Perimeter 1 Mark

Determine the perimeter of the inscribed regular hexagon.

(A) 168 cm

(B) 140 cm

(C) 180 cm

(D) 154 cm

Q3. Open Segment & Mesh Area 2 Marks

Calculate the area of one of the 6 circular segments outside the hexagon (take √3 ≈ 1.732).

(A) 64.50 cm²

(B) 82.30 cm²

(C) 71.20 cm²

(D) 78.45 cm²

OR (Alternative Q3)

Calculate the total surface area of the inscribed regular hexagon where mesh is installed.

(A) 1850.40 cm²

(B) 1980.25 cm²

(C) 2100.00 cm²

(D) 1176√3 cm² (≈ 2036.83 cm²)

📋 View Step-by-Step Marking Scheme & Solution▼

Question	Expected Answer / Step-by-Step Solution	CBSE Marks Allocation
Q1	Area = πr² = (22/7) × 28 × 28 = 2464 cm².	[1 Mark] for 2464 cm².
Q2	Hexagon side s = r = 28 cm. Perimeter = 6 × 28 = 168 cm.	[1 Mark] for 168 cm.
Q3	Sector(60°) = 2464/6 ≈ 410.67 cm². Equilateral △ = (√3/4) × 28² = 196√3 ≈ 339.47 cm². Segment = 410.67 − 339.47 = 71.20 cm².	[1 Mark] for sector and triangle areas. [1 Mark] for 71.20 cm².
Q3 (OR)	Hexagon Area = 6 × 196√3 = 1176√3 ≈ 2036.83 cm².	[1 Mark] for 6 × Area of triangle. [1 Mark] for 1176√3 cm² (2036.83 cm²).

Live Practice: Chapter 11 Areas Related to Circles

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Case Study 1

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