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Ex 13.7, 9 - A heap of wheat is in form of cone whose - Ex 13.7

  1. Chapter 13 Class 9 Surface Areas and Volumes
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Ex 13.7, 9 A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required. Radius of cone = r = ๐ท๐‘–๐‘Ž๐‘š๐‘’๐‘ก๐‘’๐‘Ÿ/2 = (10.5/2) m = 5.25 m Height of the cone = h = 3 m Volume of the cone (heap) = 1/3 ฯ€r2h = (1/3ร—22/7ร—5.25ร—5.25ร—3) m3 = 86.625 m3 We have to find area of canvas required, i.e., curved surface area of cone Curved Surface Area of cone = ฯ€r๐‘™ We know that ๐‘™2 = h2 + r2 ๐‘™2 = 32 + (5.25)2 ๐‘™2 = 9 + 27.5625 ๐‘™2 = 36.5625 ๐‘™ = โˆš36.5625 ๐‘™ = 6.046 Curved surface area of cone = ฯ€r๐‘™ = (22/7ร—5.25ร—6.046) m2 = 99.756 m2

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