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Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained
Consider a frustum of a cone with h as height, l as the slant height, r₁ and r₂ as the radii of the ends where r₁ > r₂
CSA of the frustum of the cone = 1/3πh (r₁² + r₂² + r₁r₂)
where r₁, r₂, and h are the radii and height of the frustum of the cone respectively.
Extend side BC and AD of the frustum of cone to meet at O.
The frustum of a cone can be viewed as a difference of two right circular cones OAB and OCD.
Let h₁ and l₁ be the height and slant height of cone OAB and h₂ and l₂ be the height and slant height of cone OCD respectively.
In ΔAPO and ΔDQO
∠APO = ∠DQO = 90° (Since both cones are right circular cones)
∠AOP = ∠DOQ (Common)
Therefore, ΔAPO ∼ ΔDQO ( A.A criterion of similarity)
AP/DQ = AO/DO = OP/OQ (Corresponding sides of similar triangles are proportional)
⇒ r₁/r₂ = l₁/l₂ = h₁/h₂
⇒ r₁/r₂ = h₁/h₂ or ⇒ r₂/r₁ = h₂/h₁
Subtracting 1 from both sides
r₁/r₂ - 1 = h₁/h₂ - 1
(r₁ - r₂)/r₂ = (h₁ - h₂)/h₂
(r₁ - r₂)/r₂ = h/h₂ ....(i) [From figure h₁ - h₂ = h]
h₂ = hr₂/(r₁ - r₂) ...... (i)
Now, considering r₂/r₁ = h₂/h₁
Subtracting 1 from both sides we get
r₂/r₁ - 1 = h₂/h₁ - 1
(r₂ - r₁)/r₁ = (h₂ - h₁)/h₁
(r₁ - r₂)/r₁ = (h₁ - h₂)/h₁
(r₁ - r₂)/r₁ = h/h₁
h₁ = hr₁/(r₁ - r₂) ....(ii)
Volume of frustum of cone = Volume of cone OAB - Volume of cone OCD
= 1/3 πr₁²h₁ - 1/3 πr₂²h₂
= 1/3π [r₁²h₁ - r₂²h₂]
= 1/3π [r₁² × hr₁/(r₁ - r₂) - r₂² × hr₂(r₁ - r₂)] [Using (i) and (ii)]
= 1/3π [hr₁³/(r₁ - r₂) - hr₂³/(r₁ - r₂)]
= 1/3πh [(r₁³ - r₂³)/(r₁ - r₂)]
= 1/3πh [(r₁ - r₂)(r₁² + r₁r₂ + r₂²)/(r₁ - r₂)] [Since, (a³ - b³) = (a - b)(a² + ab + b²)]
= 1/3πh (r₁² + r₂² + r₁r₂)
Hence proved, volume of the frustum = 1/3πh (r₁² + r₂² + r₁r₂)
Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.5 Question 7
The formula for the volume of the frustum of a cone is derived (1/3)πh (r₁2 + r₂2 + r₁r₂).
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