Case Study for Class 10 Maths Chapter 11 Areas Related to Circles

The Case Study Questions for Class 10 Maths Chapter 11 "Areas Related to Circles" include short, real life problem situations that have clear answers and step by step solutions to help students gain confidence for exams. It covers important topics including reviewing the circumference and area of a circle (C = 2πr, A = πr²), understanding sectors and calculating their areas, finding the area of major and minor sectors, calculating the length of arc of a sector, understanding segments and finding their areas, distinguishing between major and minor segments, calculating the area of combinations of plane figures involving circles, solving problems involving wheels and their revolutions, finding areas of circular paths and rings, calculating areas of circular fields with pathways, and applying circle area concepts to real-world scenarios like designing roundabouts, calculating fabric for circular tables, and finding garden areas. These practice questions help the students in better understanding of the concepts, handling circle area problems smoothly and to be faster and accurate for their board exams. A free PDF is included for offline timed practice.

Introduction to Case Study Questions on Areas Related to Circles

Radius and Diameter

The radius of a circle is the distance from the centre of the circle to any point on its boundary. The diameter is the distance across the circle, passing through the centre, connecting two opposite points on the boundary.

Relationship: Diameter = 2 × Radius, or d = 2r

Circumference of a Circle

The circumference is the total distance around the boundary of a circle essentially the "perimeter" of a circle.

Circumference = 2πr or equivalently Circumference = πd

Here, π (pi) is a constant approximately equal to 3.14159, commonly rounded to 3.14 or used as the fraction 22/7 in calculations.

Area of a Circle

The area of a circle is the total flat surface enclosed within its boundary.

Area = πr²

This formula uses only the radius. If you are given the diameter instead, remember to divide it by 2 first to get the radius before applying the formula.

Sector and Segment Basics

A sector of a circle is a "pie-slice" shaped region bounded by two radii and an arc.

Area of a Sector = (θ/360°) × πr², where θ is the angle of the sector at the centre.

Length of Arc = (θ/360°) × 2πr

A segment is the region between a chord and the arc it cuts off — it is different from a sector because it does not include the two straight radius lines.

Case Study 1: Designing a Circular Garden

A residential society wants to develop a circular garden in the centre of their colony. The garden will have a radius of 14 metres. The society also plans to build a circular walking path of width 3.5 metres around the garden, just outside its boundary. The society secretary asks an engineering student living in the colony to calculate the area of the garden, the area of the walking path, and the total cost of materials using circle formulas. (Use π = 22/7)

Questions

(i) What is the area of the circular garden?

(ii) What is the radius of the garden including the walking path?

(iii) What is the total area of the garden plus the walking path?

(iv) What is the area of the walking path alone?

(v) If the cost of laying the path is ₹150 per square metre, what is the total cost?

Solution:

Answer 1: Radius of garden, r = 14 m

Area = πr² = (22/7) × 14 × 14 = (22/7) × 196 = 22 × 28 = 616 m²

Answer 2: The walking path is 3.5 m wide, added outside the garden boundary.

New radius (garden + path) = 14 + 3.5 = 17.5 m

Answer 3: Total area (garden + path) = πR² = (22/7) × 17.5 × 17.5

= (22/7) × 306.25 = 22 × 43.75 = 962.5 m²

Answer 4: Area of path alone = Total area − Area of garden

= 962.5 − 616 = 346.5 m²

Answer 5: Cost = Area of path × rate per m²

= 346.5 × 150 = ₹51,975

Case Study 2: Running Track Around a Park

A sports authority is designing a circular running track around a central circular park. The park has a radius of 35 metres. The running track surrounds the park and has a uniform width of 7 metres. Athletes will run along the inner edge of the track for their races. The sports coach asks the trainees to calculate the relevant measurements before marking out lanes. (Use π = 22/7)

Questions:

(i) What is the circumference of the inner edge of the track (which touches the park boundary)?

(ii)  What is the radius of the outer edge of the track?

(iii) What is the circumference of the outer edge of the track?

(iv) What is the area of the running track (the track region only, not the park)?

(v) If an athlete runs one full lap along the inner edge, how much distance do they cover?

Solution:

Answer 1: Radius of park (inner edge of track) = 35 m

Circumference = 2πr = 2 × (22/7) × 35 = 2 × 22 × 5 = 220 m

Answer 2:Track width = 7 m, added outside the park.

Outer radius = 35 + 7 = 42 m

Answer 3: Circumference (outer edge) = 2πR = 2 × (22/7) × 42 = 2 × 22 × 6 = 264 m

Answer 4:Area of track = Area of outer circle − Area of inner circle

Area of outer circle = πR² = (22/7) × 42 × 42 = (22/7) × 1764 = 22 × 252 = 5544 m²

Area of inner circle = πr² = (22/7) × 35 × 35 = (22/7) × 1225 = 22 × 175 = 3850 m²

Area of track = 5544 − 3850 = 1694 m²

Answer 5: One lap along the inner edge equals the inner circumference.

Distance covered = 220 metres

Case Study 3: Pizza and Food Packaging Design

A pizza company sells two sizes of pizza a Regular pizza with a diameter of 28 cm and a Large pizza with a diameter of 35 cm. The company wants to compare how much more pizza area customers get when they upgrade from Regular to Large. The packaging team also wants to calculate the area of a single triangular slice when the Large pizza is cut into 8 equal slices. (Use π = 22/7)

Questions:

(i) What is the radius of the Regular pizza and the Large pizza?

(ii) What is the area of the Regular pizza?

(iii) What is the area of the Large pizza?

(iv) How much extra area does a customer get by choosing Large over Regular?

(v) If the Large pizza is cut into 8 equal slices, what is the area of one slice (sector)?

Solution:

Answer 1: Radius of Regular pizza = 28 ÷ 2 = 14 cm

Radius of Large pizza = 35 ÷ 2 = 17.5 cm

Answer 2: Area of Regular pizza = πr² = (22/7) × 14 × 14 = (22/7) × 196 = 22 × 28 = 616 cm²

Answer 3: Area of Large pizza = πR² = (22/7) × 17.5 × 17.5 = (22/7) × 306.25 = 962.5 cm²

Answer 4: Extra area = Area of Large − Area of Regular

= 962.5 − 616 = 346.5 cm²

A customer gets 346.5 cm² more pizza by choosing Large.

Answer 5: Since the pizza is cut into 8 equal slices, each slice has a central angle of 360° ÷ 8 = 45°.

Area of one slice (sector) = (θ/360°) × πR²

= (45/360) × 962.5

= (1/8) × 962.5

= 120.31 cm²

Case Study 4: Water Fountain Construction

A city park authority is constructing a circular water fountain. The fountain's water basin has a radius of 6 metres. Around the basin, a circular pathway of width 2 metres is being built for visitors to walk around. The authority also plans to install a decorative circular hedge ring at the outer edge of the pathway. The lead architect asks an assistant to compute the area calculations needed for budgeting tiles and water. (Use π = 3.14)

Questions:

(i) What is the area of the water basin?

(ii) What is the radius of the outer boundary including the pathway?

(iii) What is the area of the pathway alone?

(iv) What is the circumference of the outer boundary of the pathway?

(v) If tiles cost ₹180 per square metre, find the total cost of tiling the pathway.

Solution:

Answer 1: Radius of basin, r = 6 m

Area = πr² = 3.14 × 6 × 6 = 3.14 × 36 = 113.04 m²

Answer 2: Outer radius = 6 + 2 = 8 m

Answer 3: Area of outer circle = πR² = 3.14 × 8 × 8 = 3.14 × 64 = 200.96 m²

Area of pathway = Area of outer circle − Area of basin

= 200.96 − 113.04 = 87.92 m²

Answer 4: Circumference of outer boundary = 2πR = 2 × 3.14 × 8 = 50.24 m

Answer 5: Cost = Area of pathway × rate

= 87.92 × 180 = ₹15,825.60

Important Topics from Areas Related to Circles for Case Studies

Area of a Circle

The area of a circle is calculated using Area = πr², where r is the radius. This is the most frequently used formula across nearly every case study in this chapter, whether the question involves a garden, a pizza, a fountain, or a stadium field.

Circumference of a Circle

The circumference is calculated using Circumference = 2πr, and it represents the total boundary length of a circle. This formula is essential for problems involving fencing, running tracks, wheel rotations, and border materials.

Area of a Sector

The area of a sector is calculated using Sector Area = (θ/360°) × πr², where θ is the central angle of the sector in degrees. This formula is used whenever a problem describes a "slice" or "portion" of a circle, such as a pizza slice or a marked performance area in a stadium.

Area of Composite Circular Figures

Composite circular figures combine a circle with another shape, or combine two circles together (such as a path around a garden). The general strategy is to calculate the area of the larger combined shape and the area of the smaller inner shape separately, and then either add or subtract them depending on what the question requires. This method is used consistently across pathway, fountain, and running track case studies, and mastering it is essential for scoring well in CBSE Class 10 Maths Areas Related to Circles case study questions.

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