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 13.4
Page No:257
Questions No:2
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
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Circumference of upper part of frustum = 18 cm
⇒ 2πr₁ = 18 ⇒r₁ = 9/π
Circumference of lower part of frustum = 6 cm
⇒ 2πr₂ = 6 ⇒r₂ = 3/π
Height of frustum = 4 cm
Curved surface area of the frustum = π (r₁ + r₂)l
= π (9/π+3/π)4 = 12 × 4 = 48 cm²
Hence, the curved area of the frustum is 48 cm².
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