CBSE & State Board Question Bank | 9th - 12th

Deepa Maths Academy • Global Examination Portal

CBSE Class 10 Maths Chapter 11 Areas Related to Circles Model Questions - 4 Marks

Secure University-Grade Repository for Model Assessments, Board Examinations, and Step-by-Step Solutions.

Portal ID: DMA-EXAM-2026 • Security: Encrypted • Status: Active Examination

SECTION D — Case-Based/Source-Based Integrated Questions [4 Marks Each]

• 4 Marks Q1. Case Study: The Circular Park and Fountain
  A municipal corporation designs a circular park with a central fountain. The park has a radius of $14\text{ metres}$. Two straight walking paths make a sector of angle $60^\circ$ at the center $O$, enclosing a flower bed.
  1 Marks Sub-question (a): Find the area of the flower bed sector formed by the $60^\circ$ angle.

  Answer:
  1. Area of sector = $\frac{\theta}{360^\circ} \times \pi r^2$
  2. Calculation = $\frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{6} \times 616 = \frac{308}{3}\text{ m}^2$
  
  Result: $\frac{308}{3}\text{ m}^2$ (or $102.67\text{ m}^2$)

  1 Marks Sub-question (b): Calculate the length of the boundary arc of this sector.

  Answer:
  1. Arc length = $\frac{\theta}{360^\circ} \times 2\pi r$
  2. Calculation = $\frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 14 = \frac{1}{6} \times 88 = \frac{44}{3}\text{ m}$
  
  Result: $\frac{44}{3}\text{ m}$ (or $14.67\text{ m}$)

  2 Marks Sub-question (c): If a chord is drawn joining the ends of the arc, find the area of the minor segment formed by the chord and arc.

  Answer:
  1. Area of segment = Area of sector - Area of triangle
  2. Area of triangle = $\frac{1}{2} r^2 \sin(60^\circ) = \frac{1}{2} \times 196 \times \frac{\sqrt{3}}{2} = 49\sqrt{3}\text{ m}^2$
  3. Segment Area = $\frac{308}{3} - 49\sqrt{3}\text{ m}^2$
  
  Result: $\left(\frac{308}{3} - 49\sqrt{3}\right)\text{ m}^2$
• 4 Marks Q2. Case Study: The Ferris Wheel Illumination
  During a local festival, a giant circular Ferris wheel of radius $21\text{ metres}$ rotates slowly around its center $O$. A spectator watches illuminated passenger capsules moving along the rim. An arc $AB$ subtends an angle of $90^\circ$ at the center $O$.
  1 Marks Sub-question (a): Find the length of the arc $AB$ traced by the capsule.

  Answer:
  1. Arc length = $\frac{90^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21$
  2. Calculation = $\frac{1}{4} \times 132 = 33\text{ m}$
  
  Result: $33\text{ m}$

  1 Marks Sub-question (b): What is the area swept by the radius vector as the capsule moves from $A$ to $B$?

  Answer:
  1. Sector Area = $\frac{90^\circ}{360^\circ} \times \frac{22}{7} \times 21^2$
  2. Calculation = $\frac{1}{4} \times 1386 = 346.5\text{ m}^2$
  
  Result: $346.5\text{ m}^2$

  2 Marks Sub-question (c): If the wheel completes $\frac{1}{4}$ of a full rotation in 2 minutes, find the linear speed of the capsule along the rim.

  Answer:
  1. Distance for $\frac{1}{4}$ rotation = Arc length = $33\text{ m}$
  2. Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{33}{2} = 16.5\text{ m/min}$
  
  Result: $16.5\text{ m/min}$
• 4 Marks Q3. Case Study: The Arch Bridge Design
  An engineer designs an arch bridge supported by a large steel circular arc. The radius of the circular arch is $10\text{ metres}$, and a chord $AB$ subtends a right angle ($90^\circ$) at the center $O$.
  1 Marks Sub-question (a): State the formula for the area of a sector of angle $\theta$ and radius $r$.

  Answer:
  Result: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$

  1 Marks Sub-question (b): Calculate the area of the minor sector corresponding to the $90^\circ$ angle.

  Answer:
  1. Area = $\frac{90^\circ}{360^\circ} \times \pi \times 10^2 = \frac{1}{4} \times 100\pi = 25\pi\text{ m}^2$
  
  Result: $25\pi\text{ m}^2$ (or $78.54\text{ m}^2$)

  2 Marks Sub-question (c): Find the exact area of the minor segment enclosed by the bridge chord and the circular arch.

  Answer:
  1. Segment Area = Sector Area - Triangle Area
  2. Triangle Area = $\frac{1}{2} \times 10 \times 10 = 50\text{ m}^2$
  3. Segment Area = $(25\pi - 50)\text{ m}^2$
  
  Result: $(25\pi - 50)\text{ m}^2$
• 4 Marks Q4. Case Study: The Circular Traffic Roundabout
  A city traffic planner designs a circular roundabout of inner radius $7\text{ metres}$ surrounded by a concrete walkway of uniform width $3.5\text{ metres}$.
  1 Marks Sub-question (a): Find the outer radius of the roundabout including the walkway.

  Answer:
  1. Outer Radius $R = \text{Inner radius} + \text{width} = 7 + 3.5 = 10.5\text{ m}$
  
  Result: $10.5\text{ m}$

  1 Marks Sub-question (b): Calculate the area of the inner circular roundabout.

  Answer:
  1. Area = $\frac{22}{7} \times 7^2 = 154\text{ m}^2$
  
  Result: $154\text{ m}^2$

  2 Marks Sub-question (c): Find the total area of the concrete walkway (annulus) surrounding the roundabout.

  Answer:
  1. Walkway Area = $\pi(R^2 - r^2) = \frac{22}{7}(10.5^2 - 7^2)$
  2. Calculation = $\frac{22}{7}(110.25 - 49) = \frac{22}{7}(61.25) = 192.5\text{ m}^2$
  
  Result: $192.5\text{ m}^2$
• 4 Marks Q5. Case Study: The Stained Glass Window
  An artist creates a decorative circular stained-glass window of radius $14\text{ cm}$ containing an inscribed equilateral triangle $ABC$ in the middle, leaving the remaining region filled with colored glass.
  1 Marks Sub-question (a): State the formula for the area of an equilateral triangle with side length $a$.

  Answer:
  Result: $\text{Area} = \frac{\sqrt{3}}{4} a^2$

  1 Marks Sub-question (b): Calculate the area of the circular window.

  Answer:
  1. Area = $\frac{22}{7} \times 14^2 = 616\text{ cm}^2$
  
  Result: $616\text{ cm}^2$

  2 Marks Sub-question (c): If the side length of the equilateral triangle is $14\sqrt{3}\text{ cm}$, find the area of the shaded regions (colored glass) outside the triangle but inside the circle.

  Answer:
  1. Triangle Area = $\frac{\sqrt{3}}{4} (14\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 588 = 147\sqrt{3}\text{ cm}^2$
  2. Shaded Area = Circle Area - Triangle Area = $616 - 147\sqrt{3}\text{ cm}^2$
  
  Result: $(616 - 147\sqrt{3})\text{ cm}^2$
• 4 Marks Q6. Case Study: The Circular Running Track and Coach
  A sports academy features two concentric circular running tracks with radii $70\text{ m}$ and $77\text{ m}$ respectively.
  1 Marks Sub-question (a): Find the circumference of the inner running track.

  Answer:
  1. Inner Circumference = $2 \times \frac{22}{7} \times 70 = 440\text{ m}$
  
  Result: $440\text{ m}$

  1 Marks Sub-question (b): Find the circumference of the outer running track.

  Answer:
  1. Outer Circumference = $2 \times \frac{22}{7} \times 77 = 484\text{ m}$
  
  Result: $484\text{ m}$

  2 Marks Sub-question (c): Calculate the total area of the running track and the extra distance run by an athlete on the outer boundary compared to the inner boundary in one full lap.

  Answer:
  1. Track Area = $\frac{22}{7}(77^2 - 70^2) = \frac{22}{7}(5929 - 4900) = 3234\text{ m}^2$
  2. Extra Distance = $484 - 440 = 44\text{ m}$
  
  Result: Area = $3234\text{ m}^2$, Extra Distance = $44\text{ m}$
• 4 Marks Q7. Case Study: The Horse Grazing Field
  A horse is tied to a peg at one corner of a square grassy field of side $20\text{ m}$ by means of a $7\text{ m}$ long rope.
  1 Marks Sub-question (a): What is the central angle formed by the corner of a square field?

  Answer:
  Result: $90^\circ$

  1 Marks Sub-question (b): Calculate the area of the field that the horse can graze with the $7\text{ m}$ rope.

  Answer:
  1. Grazing Area = $\frac{90^\circ}{360^\circ} \times \frac{22}{7} \times 7^2 = \frac{1}{4} \times 154 = 38.5\text{ m}^2$
  
  Result: $38.5\text{ m}^2$

  2 Marks Sub-question (c): If the rope length is increased to $14\text{ m}$, find the increase in the grazing area.

  Answer:
  1. New Area = $\frac{90^\circ}{360^\circ} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times 616 = 154\text{ m}^2$
  2. Increase in Area = $154 - 38.5 = 115.5\text{ m}^2$
  
  Result: $115.5\text{ m}^2$
• 4 Marks Q8. Case Study: The Clock Face Geometry
  An antique clock face has a minute hand of length $14\text{ cm}$ and an hour hand of length $7\text{ cm}$.
  1 Marks Sub-question (a): Find the angle swept by the minute hand in $15\text{ minutes}$.

  Answer:
  1. Minute hand sweeps $360^\circ$ in $60\text{ minutes}$ ($6^\circ\text{ per minute}$).
  2. Angle in 15 minutes = $15 \times 6^\circ = 90^\circ$
  
  Result: $90^\circ$

  1 Marks Sub-question (b): Calculate the area swept by the minute hand in $15\text{ minutes}$.

  Answer:
  1. Area = $\frac{90^\circ}{360^\circ} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times 616 = 154\text{ cm}^2$
  
  Result: $154\text{ cm}^2$

  2 Marks Sub-question (c): Calculate the area swept by the hour hand in $2\text{ hours}$.

  Answer:
  1. Hour hand sweeps $360^\circ$ in $12\text{ hours}$ ($30^\circ\text{ per hour}$). In 2 hours, angle = $60^\circ$.
  2. Area = $\frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 7^2 = \frac{1}{6} \times 154 = \frac{77}{3}\text{ cm}^2$
  
  Result: $\frac{77}{3}\text{ cm}^2$ (or $25.67\text{ cm}^2$)
• 4 Marks Q9. Case Study: The Triangular Land Plot
  A real estate developer plans a triangular park $\triangle ABC$ which is equilateral with side length $12\text{ cm}$. Circular flower beds are designed at each vertex with a radius equal to half the side length ($6\text{ cm}$).
  1 Marks Sub-question (a): What is the measure of each interior angle of an equilateral triangle?

  Answer:
  Result: $60^\circ$

  1 Marks Sub-question (b): Calculate the total area of the three circular sectors at the vertices.

  Answer:
  1. Total Area = $3 \times \left(\frac{60^\circ}{360^\circ} \times \pi \times 6^2\right) = 3 \times (6\pi) = 18\pi\text{ cm}^2$
  
  Result: $18\pi\text{ cm}^2$ (or $56.55\text{ cm}^2$)

  2 Marks Sub-question (c): Find the area of the triangular park lying outside the three circular sectors.

  Answer:
  1. Triangle Area = $\frac{\sqrt{3}}{4} \times 12^2 = 36\sqrt{3}\text{ cm}^2$
  2. Remaining Area = Triangle Area - Sectors Area = $(36\sqrt{3} - 18\pi)\text{ cm}^2$
  
  Result: $(36\sqrt{3} - 18\pi)\text{ cm}^2$
• 4 Marks Q10. Case Study: The Satellite Dish
  A telecommunications engineer models a satellite tracking wheel of diameter $84\text{ cm}$.
  1 Marks Sub-question (a): Calculate the circumference of the tracking wheel.

  Answer:
  1. Circumference = $\pi \times d = \frac{22}{7} \times 84 = 264\text{ cm}$
  
  Result: $264\text{ cm}$

  1 Marks Sub-question (b): Find the distance covered by the wheel in $1\text{ complete revolution}$.

  Answer:
  1. Distance per revolution = Circumference = $264\text{ cm}$ ($2.64\text{ m}$)
  
  Result: $264\text{ cm}$ (or $2.64\text{ m}$)

  2 Marks Sub-question (c): How many complete revolutions must the wheel make to cover a distance of $7.92\text{ km}$?

  Answer:
  1. Total Distance = $7.92\text{ km} = 792,000\text{ cm}$
  2. Revolutions = $\frac{792,000}{264} = 3000$
  
  Result: $3000\text{ revolutions}$
• 4 Marks Q11. Case Study: The Planetary Orbit Model
  In an astronomy lab simulation, a planetary orbit is modeled as a circle of radius $21\text{ cm}$. An arc of the orbit sweeps out a sector whose area is $\frac{5}{18}$ of the total area of the circular orbit.
  1 Marks Sub-question (a): State the fraction relation given for the sector area.

  Answer:
  Result: $\text{Sector Area} = \frac{5}{18} \times \text{Total Area}$

  1 Marks Sub-question (b): Find the central angle of this planetary sector.

  Answer:
  1. Central Angle = $\frac{5}{18} \times 360^\circ = 100^\circ$
  
  Result: $100^\circ$

  2 Marks Sub-question (c): Calculate the length of the arc corresponding to this sector.

  Answer:
  1. Arc Length = $\frac{100^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21 = \frac{5}{18} \times 132 = \frac{110}{3}\text{ cm}$
  
  Result: $\frac{110}{3}\text{ cm}$ (or $36.67\text{ cm}$)
• 4 Marks Q12. Case Study: The Merry-Go-Round Safety Barrier
  A children's playground features a square safety barrier of side $14\text{ metres}$ built around a circular merry-go-round base that touches all four sides of the square.
  1 Marks Sub-question (a): What is the diameter of the largest circular merry-go-round that can fit inside the square?

  Answer:
  1. Diameter = Side length of square = $14\text{ m}$
  
  Result: $14\text{ m}$

  1 Marks Sub-question (b): Calculate the area of the circular merry-go-round base.

  Answer:
  1. Area = $\frac{22}{7} \times 7^2 = 154\text{ m}^2$
  
  Result: $154\text{ m}^2$

  2 Marks Sub-question (c): Find the area of the square region lying outside the circular merry-go-round base.

  Answer:
  1. Square Area = $14^2 = 196\text{ m}^2$
  2. Outside Area = $196 - 154 = 42\text{ m}^2$
  
  Result: $42\text{ m}^2$
• 4 Marks Q13. Case Study: The Archery Target
  An archery target consists of concentric circular scoring rings. The innermost bullseye has a radius of $3.5\text{ cm}$ and the next outer ring expands the radius to $7\text{ cm}$.
  1 Marks Sub-question (a): Calculate the area of the bullseye circle.

  Answer:
  1. Area = $\frac{22}{7} \times (3.5)^2 = 38.5\text{ cm}^2$
  
  Result: $38.5\text{ cm}^2$

  1 Marks Sub-question (b): Calculate the total area enclosed up to the outer ring limit.

  Answer:
  1. Area = $\frac{22}{7} \times 7^2 = 154\text{ cm}^2$
  
  Result: $154\text{ cm}^2$

  2 Marks Sub-question (c): Find the area of the ring band (annulus) between the bullseye and the outer ring limit.

  Answer:
  1. Ring Area = $154 - 38.5 = 115.5\text{ cm}^2$
  
  Result: $115.5\text{ cm}^2$
• 4 Marks Q14. Case Study: The Radar Tracking Station
  A coastal radar station tracks ships across a circular surveillance zone of radius $21\text{ km}$. The radar beam sweeps a sector with a central angle of $120^\circ$.
  1 Marks Sub-question (a): Write the formula for the area of a sector of angle $120^\circ$.

  Answer:
  Result: $\text{Area} = \frac{120^\circ}{360^\circ} \times \pi r^2 = \frac{1}{3} \pi r^2$

  1 Marks Sub-question (b): Calculate the surveillance area covered by the $120^\circ$ radar sweep.

  Answer:
  1. Area = $\frac{1}{3} \times \frac{22}{7} \times 21^2 = \frac{1}{3} \times 1386 = 462\text{ km}^2$
  
  Result: $462\text{ km}^2$

  2 Marks Sub-question (c): Find the perimeter of this radar sector boundary.

  Answer:
  1. Perimeter = $\text{Arc length} + 2r$
  2. Arc Length = $\frac{120^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21 = 44\text{ km}$
  3. Perimeter = $44 + 2(21) = 86\text{ km}$
  
  Result: $86\text{ km}$
• 4 Marks Q15. Case Study: The Suspension Bridge Cable
  A suspension bridge uses a copper wire of a certain length. When bent in the form of a square, it encloses an area of $484\text{ cm}^2$.
  1 Marks Sub-question (a): Find the side length of the square formed by the wire.

  Answer:
  1. Side = $\sqrt{484} = 22\text{ cm}$
  
  Result: $22\text{ cm}$

  1 Marks Sub-question (b): Determine the total length (perimeter) of the copper wire.

  Answer:
  1. Perimeter = $4 \times 22 = 88\text{ cm}$
  
  Result: $88\text{ cm}$

  2 Marks Sub-question (c): If the exact same wire is rebent into the form of a circle, calculate the area of the circle.

  Answer:
  1. Circumference = $88 \implies 2\pi r = 88 \implies r = 14\text{ cm}$
  2. Circle Area = $\frac{22}{7} \times 14^2 = 616\text{ cm}^2$
  
  Result: $616\text{ cm}^2$
• 4 Marks Q16. Case Study: The Circular Fountain Basin
  A landscape architect designs a circular fountain basin of radius $21\text{ m}$ surrounded by a uniform outer stone path $7\text{ m}$ wide.
  1 Marks Sub-question (a): Find the total radius including the fountain and path.

  Answer:
  1. Total Radius $R = 21 + 7 = 28\text{ m}$
  
  Result: $28\text{ m}$

  1 Marks Sub-question (b): Calculate the area of the stone path.

  Answer:
  1. Path Area = $\frac{22}{7}(28^2 - 21^2) = \frac{22}{7}(784 - 441) = 1078\text{ m}^2$
  
  Result: $1078\text{ m}^2$

  2 Marks Sub-question (c): Find the total cost of paving the path at the rate of $\text{₹}10$ per $\text{m}^2$.

  Answer:
  1. Total Cost = $1078 \times 10 = \text{₹}10,780$
  
  Result: $\text{₹}10,780$
• 4 Marks Q17. Case Study: The Theatre Stage Design
  A theatre stage includes a decorative semicircular wooden flooring section whose radius is $7\text{ metres}$.
  1 Marks Sub-question (a): State the formula for the area of a semicircle of radius $r$.

  Answer:
  Result: $\text{Area} = \frac{1}{2} \pi r^2$

  1 Marks Sub-question (b): Calculate the area of the semicircular stage flooring.

  Answer:
  1. Area = $\frac{1}{2} \times \frac{22}{7} \times 7^2 = 77\text{ m}^2$
  
  Result: $77\text{ m}^2$

  2 Marks Sub-question (c): Find the total perimeter (including the straight baseline diameter) of the semicircular stage.

  Answer:
  1. Perimeter = $\pi r + 2r = 22 + 14 = 36\text{ m}$
  
  Result: $36\text{ m}$
• 4 Marks Q18. Case Study: The Umbrella Ribs
  An umbrella has $8$ ribs which are equally spaced. Assuming the umbrella to be a flat circle of radius $28\text{ cm}$.
  1 Marks Sub-question (a): What is the central angle between two consecutive ribs of the umbrella?

  Answer:
  1. Angle = $\frac{360^\circ}{8} = 45^\circ$
  
  Result: $45^\circ$

  1 Marks Sub-question (b): Calculate the total area of the flat circular umbrella.

  Answer:
  1. Area = $\frac{22}{7} \times 28^2 = 2464\text{ cm}^2$
  
  Result: $2464\text{ cm}^2$

  2 Marks Sub-question (c): Find the area of the fabric material between two consecutive ribs.

  Answer:
  1. Fabric Area = $\frac{45^\circ}{360^\circ} \times \frac{22}{7} \times 28^2 = \frac{1}{8} \times 2464 = 308\text{ cm}^2$
  
  Result: $308\text{ cm}^2$
• 4 Marks Q19. Case Study: The Circular Drainage Tunnel
  An engineering team inspects a circular drainage tunnel of radius $14\text{ cm}$. A chord subtends an angle of $90^\circ$ at the center.
  1 Marks Sub-question (a): Calculate the area of the sector formed by the $90^\circ$ angle.

  Answer:
  1. Area = $\frac{90^\circ}{360^\circ} \times \frac{22}{7} \times 14^2 = 154\text{ cm}^2$
  
  Result: $154\text{ cm}^2$

  1 Marks Sub-question (b): Calculate the area of the right-angled triangle formed by the two radii and the chord.

  Answer:
  1. Triangle Area = $\frac{1}{2} \times 14 \times 14 = 98\text{ cm}^2$
  
  Result: $98\text{ cm}^2$

  2 Marks Sub-question (c): Find the area of the minor segment enclosed by the chord and the arc.

  Answer:
  1. Segment Area = $154 - 98 = 56\text{ cm}^2$
  
  Result: $56\text{ cm}^2$
• 4 Marks Q20. Case Study: The Roundabout Landscaping
  A city planner notes that a circular decorative garden has a special property where its numerical circumference is equal to its numerical area.
  1 Marks Sub-question (a): Set up the equation equating the circumference and area formulas of a circle.

  Answer:
  Result: $2\pi r = \pi r^2$

  1 Marks Sub-question (b): Solve for the unique radius $r$ of this circular garden.

  Answer:
  1. $2\pi r = \pi r^2 \implies r = 2$
  
  Result: $r = 2\text{ units}$

  2 Marks Sub-question (c): Calculate the exact area and perimeter of a square inscribed inside this circle.

  Answer:
  1. Square diagonal = Circle diameter = $4$
  2. Square Area = $\frac{1}{2} \times d^2 = \frac{1}{2} \times 16 = 8\text{ sq. units}$
  3. Square Side = $\frac{4}{\sqrt{2}} = 2\sqrt{2}$, Perimeter = $4 \times 2\sqrt{2} = 8\sqrt{2}\text{ units}$
  
  Result: $\text{Area} = 8$, $\text{Perimeter} = 8\sqrt{2}$

Academic Repository Disclaimer & எச்சரிக்கை அறிவிப்பு :

English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.

தமிழ்: இந்த மாதிரி கணிதத் தீர்வுகள் மாணவர்களின் சுய கற்றல் மற்றும் பயிற்சித் திறனுக்காக உலகளாவிய கல்வித் தரத்தில் உருவாக்கப்பட்டுள்ளன. தேர்வுகளுக்குத் தயாராகும் போது உங்கள் அதிகாரப்பூர்வ பாடப்புத்தகத்துடன் சரிபார்க்கவும்.

மேலதிக விவரங்களுக்கு எங்களைத் தொடர்பு கொள்ளவும் → For further details, contact us →

Copyright © 2026 Deepa Maths Academy | Global Examination & Academic Portal. All Rights Reserved.

மாணவர்களுடன் பகிரவும்: