Canvas needed for a tent (cylinder + cone, r = 10 m)\nA tent is formed by a cylinder of radius 10 m and height 5 m, surmounted by a cone of the same radius with slant height 15 m. How much canvas is required if 20% extra is allowed for folding/stitching? (Use π = 22/7. Ignore floor.)

Introduction / Context:
Canvas required equals the lateral areas (outer surface only) since tents typically exclude the base/floor unless specified. Here a cylinder is topped by a cone. We compute each lateral area and then add 20% wastage for folds and stitching.

Given Data / Assumptions:

• Cylinder: radius r = 10 m, height h = 5 m
• Cone: same radius r = 10 m, slant height l = 15 m
• π = 22/7; base is not included (canvas excludes floor).
• Extra allowance = 20% of computed lateral area.

Concept / Approach:
Lateral area (cylinder) = 2 * π * r * h. Lateral area (cone) = π * r * l. Sum, then multiply by 1.20 to account for extra canvas.

Step-by-Step Solution:

Verification / Alternative check:
If a floor were included, add πr^2 = (22/7) * 100 = 2200/7 ≈ 314.286, yielding ≈ 1320 m^2 even then. None of the given options match ≈ 943 m^2 or ≈ 1320 m^2.

Why Other Options Are Wrong:
All numeric choices are far larger than either interpretation (with or without base). Hence the correct selection within this set is “None of these.”

Common Pitfalls:
Including the base by default or using circumference * slant height for the cone incorrectly. Maintain unit consistency and apply the 20% only after summing areas.

Final Answer:
None of these