NCERT Solutions for Class 10 Maths Chapter 13

Important NCERT Questions

Surface areas and Volumes,

NCERT Books for Session 2022-2023

CBSE Board and UP Board Others state Board

EXERCISE 13.5

Page No:258

Questions No:6

# 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.

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Let ABC is a cone, which is converted into a frustum by cutting the top. let r₁ and r₂ be the radius of two ends and h be its height.

In ΔABG and ΔADF, DF ∥ BG

∴ ΔABG ∼ ΔADF

DF/BG = AF/AG = AD/AB

⇒ r₂/r₁ = (h₁ – h/h₁) = (l₁ – l/l₁) ⇒ r₂/r₁ = 1 – h/h₁ = 1 – l/l₁

⇒ r₂/r₁ = 1 – l/l₁ ⇒ l/l₁ = 1 – r₂/r₁ = r₁ – r₂/r₁

⇒ l₁/l = (r₁/r₁ – r₂) ⇒ l₁ = r₁l/r₁ – r₂

Curved surface area of frustum = CSA of cone ABC – CSA of cone ADE = πr₁l₁ – πr₂(l₁ – l)

= πr₁(r₁l/r₁ – r₂) – πr₂((r₁l/r₁ – r₂) – l) = (πr₁²l/r₁ – r₂) – πr₂ (r₁l – r₁l + r₂l/r₁ – r₂)

= (πr₁²l/r₁ -r₂) – (πr₂²l/r₁ – r₂) = πl (r₁² – r₂²/r₁ – r₂) = π(r₁ + r₂)l

total surface area of frustum = Curved surface area + Area of upper circle + Area of lower circle.

= π(r₁ + r₂)l + πr₂² + πr₁²