A shamiana is a cylinder of radius 2.1 m and height 3 m with a conical top of the same radius and height 2.8 m. Find (i) the canvas needed, with no floor, and its cost at ₹ 120 per m², (ii) the volume of air inside. (Use π = 22/7.)

1. 1 · Write what is given and what is to be found. Given r = 2.1 m, cylinder height h = 3 m, cone height H = 2.8 m, rate = ₹ 120 per m². To find: canvas area, its cost and the volume of air.
2. 2 · Draw the figure. Draw a cylinder with a cone on top. Show the joining circle dotted. Mark r, h, H and the slant height l.
3. 3 · Find the slant height. l = √(r² + H²) = √(4.41 + 7.84) = √12.25 = 3.5 m.
4. 4 · Write the formulas.
   • Canvas = 2πrh + πrl = πr(2h + l).
   • Volume = πr²h + (1/3)πr²H = πr²(h + H/3).
5. 5 · Find the canvas and its cost. πr = (22/7) × 2.1 = 6.6. Canvas = 6.6 × (6 + 3.5) = 6.6 × 9.5 = 62.7 m². Cost = 62.7 × 120 = ₹ 7524.
6. 6 · Find the volume of air. πr² = (22/7) × 2.1 × 2.1 = 13.86. Volume = 13.86 × 3 + 13.86 × 2.8 ÷ 3 = 41.58 + 12.936 = 54.516 m³, about 54.52 m³.
7. 7 · Check and box the answer. Area in m², volume in m³, cost in ₹. Box: canvas 62.7 m², cost ₹ 7524, air 54.52 m³.

Scoring tip: Canvas needs the slant height l, but volume needs the vertical height H. Mixing them up is the most common slip.