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  1. Chapter 13 Class 10 Surface Areas and Volumes
  2. Serial order wise
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Ex 13.5, 7 Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained. There are two cones OCD & OAB We are given Height of frustum = h Slant height of frustum = l Radius PB = r1 Radius QD = r2 We need to find Curved Surface Area & Total Surface Area Here, We need to write h1, l1, h2, l2 in terms of h and l Volume of frustum = Volume of cone OAB – Volume of cone OCD = 1/3 𝜋〖𝑟_1〗^2 ℎ_1−1/3 𝜋〖𝑟_2〗^2 ℎ_2 In Δ OPB & Δ OQD ∠ BOP = ∠ DOQ ∠ OPB = ∠ OQD So, Δ OPB ∼ Δ OQD ∴ 𝑃𝐵/𝑄𝐷 = 𝑂𝐵/𝑂𝐷 𝑟_1/𝑟_2 = ℎ_1/ℎ_2 Putting h1 = h + h2 𝑟_1/𝑟_2 = (ℎ + ℎ_2)/ℎ_2 𝑟_1/𝑟_2 = ℎ/ℎ_2 + ℎ_2/ℎ_2 𝑟_1/𝑟_2 = ℎ/ℎ_2 + 1 𝑟_1/𝑟_2 −1 = ℎ/ℎ_2 (𝑟_1 − 𝑟_2)/𝑟_2 = ℎ/ℎ_2 ℎ_2 ((𝑟_1 − 𝑟_2)/𝑟_2 ) = ℎ ℎ_2 = ℎ(𝑟_2/(𝑟_1 − 𝑟_2 )) From (1) Volume of frustum = 1/3 𝜋〖𝑟_1〗^2 ℎ_1−1/3 𝜋〖𝑟_2〗^2 ℎ_2 Putting h1 = h + h2 = 1/3 𝜋〖𝑟_1〗^2 (ℎ+ℎ_2)−1/3 𝜋〖𝑟_2〗^2 ℎ_2 From (2): Putting ℎ_2 = ℎ(𝑟_2/(𝑟_1 − 𝑟_2 )) = 1/3 𝜋〖𝑟_1〗^2 (ℎ+ℎ(𝑟_2/(𝑟_1 − 𝑟_2 )))−1/3 𝜋〖𝑟_2〗^2 ℎ(𝑟_2/(𝑟_1 − 𝑟_2 )) = 1/3 𝜋〖𝑟_1〗^2 ℎ(1+(𝑟_2/(𝑟_1 − 𝑟_2 )))−1/3 𝜋〖𝑟_2〗^2 ℎ(𝑟_2/(𝑟_1 − 𝑟_2 )) = 1/3 𝜋〖𝑟_1〗^2 ℎ ((𝑟_1 − 𝑟_2 + 𝑟_2)/(𝑟_1 − 𝑟_2 )) – 1/3 𝜋〖𝑟_2〗^2 ℎ(𝑟_2/(𝑟_1 − 𝑟_2 )) = 1/3 𝜋〖𝑟_1〗^2 ℎ(𝑟_1/(𝑟_1 − 𝑟_2 )) – 1/3 𝜋〖𝑟_2〗^2 ℎ(𝑟_2/(𝑟_1 − 𝑟_2 )) = 1/3 𝜋ℎ(〖𝑟_1〗^3/(𝑟_1 − 𝑟_2 )) – 1/3 𝜋ℎ(〖𝑟_2〗^3/(𝑟_1 − 𝑟_2 )) = 1/3 𝜋ℎ ((〖𝑟_1〗^3 − 〖𝑟_2〗^3)/(𝑟_1 − 𝑟_2 )) Using a3 – b3 = (a – b) (a2 + b2 + ab) = 1/3 𝜋ℎ ((𝑟_1 − 𝑟_2 )(〖𝑟_1〗^2+ 〖𝑟_2〗^2 + 𝑟_1 𝑟_2 )/(𝑟_1 − 𝑟_2 )) = 1/3 𝜋ℎ(〖𝑟_1〗^2+ 〖𝑟_2〗^2 + 𝑟_1 𝑟_2 ) ∴ Volume of frustum = 1/3 𝜋ℎ(〖𝑟_1〗^2+ 〖𝑟_2〗^2 + 𝑟_1 𝑟_2 )

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