Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained
Solution:
Consider a frustum of a cone with h as height, l as the slant height, r₁ and r₂ as radii of the ends where r₁ > r₂
To Prove:
(i) CSA of the frustum of the cone = πl (r₁ + r₂)
(ii) TSA of the frustum of the cone = πl (r₁ + r₂) + πr₁² + πr₂²
where r₁, r₂, h and l are the radii height and slant height of the frustum of the cone respectively.
Construction:
Extended side BC and AD of the frustum of cone to meet at O
Proof:
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 (AA criterion of similarity)
AP/DQ = AO/DO = OP/OQ (Corresponding sides of similar triangles are proportional)
⇒ r₁/r₂ = l₁/l₂ = h₁/h₂
⇒ r₁/r₂ = l₁/l₂ or ⇒ r₂/r₁ = l₂/l₁
Subtracting 1 from both sides we get
r₁ / r₂ - 1 = l₁ / l₂ - 1
(r₁ - r₂)/r₂ = (l₁ - l₂) / l₂
(r₁ - r₂)/r₂ = l/l₂ [From diagram, l₁ - l₂ = l]
l₂ = lr₂/(r₁ - r₂).... (i)
or r₂ / r₁ = l₂ / l₁
Subtracting 1 from both sides we get
r₂ / r₁ - 1 = l₂ / l₁ - 1
(r₂ - r₁) / r₁ = (l₂ - l₁) / l₁
(r₁ - r₂) / r₁ = (l₁ - l₂) / l₁
(r₁ - r₂) / r₁ = l / l₁
l₁ = lr₁/(r₁ - r₂) .... (ii)
(i) CSA of frustum of cone = CSA of cone OAB - CSA of cone OCD
= πr₁l₁ - πr₂l₂
= π (r₁l₁ - r₂l₂)
= π [(r₁ × lr₁/(r₁ - r₂) - r₂ × lr₂/(r₁ - r₂)] [Using (i) and (ii)]
= π [(lr₁² - lr₂²)/(r₁ - r₂)]
= π [l(r₁² - r₂²)/(r₁ - r₂)]
= π [l(r₁ - r₂)(r₁ + r₂)/(r₁ - r₂)] [Since, a² - b² = (a - b)(a + b)]
= π l(r₁ + r₂)
TSA of frustum of cone = CSA of frustum + Area of lower circular end + Area of top circular end
= πl (r₁ + r₂) + πr₁² + πr₁²
Therefore, CSA of the frustum of the cone = πl (r₁ + r₂)
TSA of the frustum of the cone = πl (r₁+ r₂) + πr₁² + πr₂²
Hence Proved.
☛ Check: Class 10 Maths NCERT Solutions Chapter 13
Video Solution:
Derive the formula for the curved surface area and total surface area 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 6
Summary:
The formula for the curved surface area and total surface area of the frustum of a cone are πl(r₁+r₂) and πl(r₁+r₂)+πr₁² +πr₂² respectively where r₁,r₂, h and l are the radii,height and slant height of the frustum of the cone respectively has been derived.
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