Canvas and floor capacity of a shamiana
An NGO is erecting a canvas shamiana for a village health camp in Barmer. The shamiana is a cylinder of radius 5 m and height 4 m, with a cone of the same radius and height 12 m fixed on top; the circular floor of radius 5 m is not covered with canvas. The supplier's quotation mentions 350 m² of canvas, inclusive of all cutting wastage, which is the quantity the NGO has budgeted for. The camp doctor requires at least 3.5 m² of floor space per patient and expects 25 patients at a time; the organiser says the tent is comfortably large. Waterproofing of the curved surface is quoted at ₹45 per square metre, and the NGO must decide whether ₹15,000 sanctioned for it is enough. Take π = 3.14 unless an exact value simplifies the work.
1. Calculate the canvas area needed and judge whether the supplier's 350 m² is sufficient.
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Slant height of the cone: l = √(r² + h²) = √(5² + 12²) = √169 = 13 m.
Curved surface of cylinder = 2πrh = 2 × 3.14 × 5 × 4 = 125.6 m² (i.e. 40π).
Curved surface of cone = πrl = 3.14 × 5 × 13 = 204.1 m² (i.e. 65π).
Total canvas = 40π + 65π = 105π = 329.7 m². The floor is not canvassed, so it is excluded.
Judgement: 350 − 329.7 = 20.3 m², which is only 20.3/329.7 ≈ 6.2% of the canvas. So 350 m² allows under about 6% wastage. If cutting wastage is the usual 10%, the requirement becomes 329.7 × 1.10 = 362.7 m² > 350 m², and the quotation falls short. The decision therefore depends on the wastage rate, not on the surface area alone - the NGO must either confirm a wastage below 6.2% or ask for about 363 m².
2. Check the organiser's claim that the tent is comfortable for 25 patients at a time.
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Floor area = πr² = 3.14 × 25 = 78.5 m².
Space per patient at 25 patients = 78.5/25 = 3.14 m², which is less than the required 3.5 m².
Maximum patients = 78.5/3.5 = 22.43, and patients cannot be counted in fractions, so the safe capacity is 22 patients, not 23 (23 would need 80.5 m²).
For 25 patients the doctor's norm needs 25 × 3.5 = 87.5 m², which is 9 m² more floor than the tent provides. The organiser is wrong; the tent can hold at most 22 patients at a time under the stated norm. This is a typical rounding trap - rounding 22.43 up would breach the safety requirement.
3. Is ₹15,000 enough for waterproofing the curved surface? What changes if the floor is also waterproofed?
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Waterproofing applies to the curved surface only: 105π = 329.7 m².
Cost = 329.7 × 45 = ₹14,836.50, which is within ₹15,000, leaving about ₹163.50 spare.
If the floor is included as well: additional area = πr² = 78.5 m², additional cost = 78.5 × 45 = ₹3,532.50, giving a total of 14,836.50 + 3,532.50 = ₹18,369, which exceeds the sanction by ₹3,369.
So the answer is conditional: ₹15,000 suffices for the canvas surface as quoted, but not if the specification is widened to the floor. The NGO should confirm the scope of waterproofing in writing before accepting the quotation, and should note that a 1 m² error changes the cost by ₹45.