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A tent is in the shape of a cylinder, surmounted by a conical top. If the height and diameter of the cylindrical part are $3.5$ m and $6$ m, and slant height of the top is $4.2$ m, find the area of canvas used for making the tent. Also, find the cost of canvas of the tent at the rate of ₹500 per m$^2$.
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Radius ($r$) = $3$m (1/2 Mark)
CSA of cylinder = $2\pi rh = 2 \times \frac{22}{7} \times 3 \times 3.5 = 66$ m$^2$ (1 Mark)
CSA of cone = $\pi rl = \frac{22}{7} \times 3 \times 4.2 = 39.6$ m$^2$ (1 Mark)
Area of canvas = $66 + 39.6 = 105.6$ m$^2$ (1/2 Mark)
Cost = $500 \times 105.6 = \text{\text{Rs} }52800$ (1 Mark)
CSA of cylinder = $2\pi rh = 2 \times \frac{22}{7} \times 3 \times 3.5 = 66$ m$^2$ (1 Mark)
CSA of cone = $\pi rl = \frac{22}{7} \times 3 \times 4.2 = 39.6$ m$^2$ (1 Mark)
Area of canvas = $66 + 39.6 = 105.6$ m$^2$ (1/2 Mark)
Cost = $500 \times 105.6 = \text{\text{Rs} }52800$ (1 Mark)