Ground rolled by a cylindrical roller – area covered: A roller has diameter 84 cm and length 120 cm. It makes 500 complete revolutions to level a playground. Find the total area rolled (in m^2).

Introduction / Context:
Each revolution of a cylindrical roller covers a rectangular strip whose width equals the roller length and whose length equals the circumference of the circular cross-section. The total area equals (area per revolution) × (number of revolutions).

Given Data / Assumptions:

• Diameter d = 84 cm = 0.84 m → circumference C = πd
• Length L = 120 cm = 1.2 m
• Revolutions n = 500
• Area per revolution = C * L

Concept / Approach:
Compute area = n * (πd * L). Using π = 22/7 simplifies the arithmetic here to an exact multiple, a common exam design.

Step-by-Step Solution:
Per revolution area = π * 0.84 * 1.2 = π * 1.008 m^2Using π = 22/7 → per rev = (22/7) * 1.008 = (22 * 1.008) / 7 ≈ 3.168 m^2Total area = 500 * 3.168 = 1584 m^2

Verification / Alternative check:
Work in centimeters: C = π * 84; strip width = 120; per rev area = 84π * 120 cm^2 = 10080π cm^2; times 500 → 5,040,000π cm^2. Divide by 10,000 to convert to m^2 → 504π ≈ 1584 (since π ≈ 22/7).

Why Other Options Are Wrong:
1632 and 1532 are nearby but not equal to 504π; 1817 is too large; 1600 is a rounded guess, not the exact value.

Common Pitfalls:
Using radius instead of diameter in circumference; forgetting unit conversion cm ↔ m; overlooking that area per revolution is linear in both circumference and roller length.

Final Answer:
1584