Case Study Questions on Chapter 6 ‘Measuring Space - Perimeter and Area’ for Class 9 Maths

This set of case-study questions for Chapter 6 ‘Measuring Space: Perimeter and Area’ offers short, focused problem scenarios designed to build exam-ready skills in geometry topics covered in Class 9. Instead of asking ‘find the area of this triangle,’ case study questions give you a real-life scenario like a running track, a school garden, a tiled floor, a clock and then ask you several questions based on it. Step-by-step solutions show clear reasoning, highlight key formula use, and point out typical traps students face during tests. A free PDF with answer keys and timed practice sets is included for offline revision, classroom worksheets, and homework ideal for quick revision and targeted skill building before tests.

Case Study Questions Class 9 Chapter 6: Measuring Space: Perimeter and Area With Answers

CBSE's case-based questions sometimes called source-based or passage-based questions, usually give you a short passage describing a real situation, sometimes with a small figure, followed by 4 - 5 questions of increasing difficulty, a couple of MCQs to check basic understanding, a short-answer question that needs a calculation, and a longer question that asks you to interpret or justify your answer.

Case Study 1: The History of π

Read the passage and answer the questions below.

Maths teacher Ms. Kavitha is teaching her class about the history of the constant π. She explains that in ancient Mesopotamia (around 1900 BCE), mathematicians set π = 3 + 1/8 = 3.125 after observing that a circle's perimeter was larger than the perimeter of a hexagon inscribed within it. Archimedes of Syracuse (around 250 BCE) used polygons with up to 96 sides to show that 3(10/71) < π < 3(1/7). Indian mathematician Āryabhaṭa (499 CE) gave the value 62832/20000 = 3.1416, describing it as asanna (approximate). Zu Chongzhi (480 CE) discovered the fraction 355/113 ≈ 3.1415929. Finally, Mādhava of Sangamagrāma discovered the first exact formula: π/4 = 1 − 1/3 + 1/5 − 1/7 + ...

Q.1 What does the ratio C/D represent for any circle?

(a) Area divided by radius

(b) Circumference divided by diameter

(c) Circumference divided by radius

(d) Area divided by circumference

Q.2 State whether the following is True or False: π = 22/7.

Q.3 Who introduced the symbol π, and what does the letter π stand for?

Q.4 Why is π called an irrational number? How is this different from rational numbers like 1/3 or 7/11?

Q.5 Āryabhaṭa described his value of π as asanna (approximate). Why was this a profound insight for the time?

Solutions: 

Q.1: (b) Circumference divided by diameter

Q.2: False.

π ≈ 22/7, but π ≠ 22/7. π is irrational and cannot be expressed exactly as any fraction.

Q.3: Welsh mathematician William Jones introduced the symbol π in 1706. It is the first letter of the Greek word perimetros, meaning perimeter. The symbol was later popularised by Leonhard Euler.

Q.4: A rational number can be written as a fraction a/b where a and b are integers and b ≠ 0. When expressed as a decimal, rational numbers either terminate (like 1.4) or repeat in a pattern (like 1/3 = 0.333... or 1/7 = 0.142857142857...).

π cannot be written as any such fraction. Its decimal expansion goes on forever with no repeating pattern (3.14159265358...). Such numbers are called irrational. Lambert proved this in 1761.

Q.5: By calling his value asanna (approaching/approximate), Āryabhaṭa was suggesting that π could not be expressed exactly as a simple fraction recognising its irrational nature centuries before Lambert formally proved it in 1761. This shows remarkable mathematical intuition and distinguishes Āryabhaṭa's understanding from earlier civilisations that simply used approximations without questioning whether an exact fractional value could even exist.

 

Case Study 2: The Triangular Land Survey 

Read the passage and answer the questions below.

A civil engineer is surveying three triangular plots of land for a housing project. The side lengths are:

Plot A: 13 m, 14 m, and 15 m

Plot B: Sides in the ratio 5 : 12 : 13 with perimeter 60 m

Plot C: An equilateral triangle with each side 10 m

The land will be levelled at a cost of ₹80 per m².

Q.1 What is the semi-perimeter of Plot A?

(a) 18 m   (b) 21 m   (c) 24 m   (d) 42 m

Q.2 Using Heron's formula, find the area of Plot A.

Q.3 For Plot B, the sides are in the ratio 5 : 12 : 13 and the perimeter is 60 m. What are the actual side lengths?

Q.4: Show that Plot B is a right-angled triangle. Then find its area without using Heron's formula.

Q.5 Find the area of Plot C (equilateral triangle, side = 10 m) using Heron's formula, and hence find the total cost of levelling all three plots at ₹80 per m²

Solutions: 

Q.1: (b) 21 m

s = (13 + 14 + 15)/2 = 42/2 = 21 m

Q.2: s = 21 m, sides a = 13, b = 14, c = 15

Area = √[s(s−a)(s−b)(s−c)]

= √[21 × 8 × 7 × 6]

= √7056

= 84 m²

Q.3: Let sides be 5k, 12k, 13k.

5k + 12k + 13k = 60 

⇒ 30k = 60 ⇒ k = 2

Sides are 10 m, 24 m, 26 m.

Q.4:

Check: 10² + 24² = 100 + 576 = 676 = 26²  

By the converse of the Baudhāyana–Pythagoras theorem, Plot B is right-angled (right angle between the 10 m and 24 m sides).

Area = ½ × base × height = ½ × 10 × 24 = 120 m²

Q.5: Plot C: Heron's formula:

s = (10 + 10 + 10)/2 = 15 m

Area = √[15 × 5 × 5 × 5] = √1875 = 25√3 m² ≈ 43.30 m²

Total area = Plot A + Plot B + Plot C

= 84 + 120 + 25√3

= 204 + 25√3 m²

≈ 204 + 43.30 = 247.30 m²

Total cost = 247.30 × ₹80 ≈ ₹19,784

Case Study 3: The Clock

Read the passage and answer the questions below.

In a maths class, the teacher uses a wall clock to explain the concepts of arc length and sector area. The minute hand of the clock is 7 cm long. In 60 minutes, it completes one full revolution, with its tip tracing the complete circumference of a circle. The teacher asks the class to calculate how much distance the tip covers and how much area it sweeps for different time intervals. (Use π = 22/7)

 

Q.1 In 60 minutes, the minute hand sweeps through an angle of:

(a) 90°   (b) 180°   (c) 270°   (d) 360°

Q.2 What angle does the minute hand sweep in 5 minutes?

Q.3 Find the length of the arc traced by the tip of the minute hand in 15 minutes.

Q.4 Find the area swept by the minute hand in 10 minutes.

Q.5 The hour hand of the same clock is 4 cm long. In 3 hours, how much area does the hour hand sweep? (Use π = 22/7)

Solutions: 

Q.1 (d) 360°

Q.2 

In 60 minutes → 360°

In 5 minutes → 360° × 5/60 = 30°

Q.3: Angle in 15 minutes = 360° × 15/60 = 90°

Arc length = 2πr × θ/360

= 2 × (22/7) × 7 × 90/360

= 44 × 1/4

= 11 cm

Q.4: Angle in 10 minutes = 360° × 10/60 = 60°

Area of sector = πr² × θ/360

= (22/7) × 49 × 60/360

= 154 × 1/6

= 77/3 cm² ≈ 25.67 cm²

Q.5: The hour hand completes one full revolution in 12 hours.

In 3 hours:  angle = 360° × 3/12 = 90°

Area = πr² × θ/360

= (22/7) × 16 × 90/360

= (22/7) × 16 × 1/4

= (22 × 4)/7

= 88/7 cm² ≈ 12.57 cm²

 

Click below to download your free Case Study Questions PDF with worked-out examples for Class 9 Chapter 6: Measuring Space: Perimeter and Area, perfect for last-minute CBSE exam revision.

Class 9 Chapter 6: Measuring Space: Perimeter and Area Case Study PDF