Ex 13.1, 7 - Chapter 13 Class 10 Surface Areas and Volumes (Term 2)

Transcript

Ex 13.1, 7 A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.) Area of the canvas = Curved area of cone + Curved surface area of the cylinder Curved surface area of cone Diameter of cone = Diameter of cylinder = 4 cm. So, Radius = Diameter/2 = 4/2 = 2 m & slant height = 𝑙 = 2.8 m. Curved surface area of cone = 𝜋𝑟𝑙 = 22/7×2×2.8 = 22 ×2×0.4 = 17.6 m2 Curved surface area of cylinder Diameter of cylinder = 4 cm. So, Radius = Diameter/2 = 4/2 = 2 m & Height = h = 2.1 m Curved surface area of cylinder = 2𝜋rh = 2 × 22/7×2×2.1 = (2 × 22 × 2 × 21)/(7 × 10) = 132/5 = 26.4 m2 Area of the canvas = Curved area of cone + Curved surface area of the cylinder = 17.6 + 26.4 = 44 m2 Now, Cost of the canvas of the tent for 1m2 = Rs 500 Cost of the canvas of the tent for 44 m2 = 500 × 44 = Rs 22000 Hence, Total cost = 22,000 rupees