CBSE Notes, Lectures

CBSE - Mathematics - Surface Areas and Volumes

Surface Areas and Volumes

NCERT Exercise Exercise 13.2

A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see Fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

Let R be external radius and r be the internal radius h be the length of the pipe.

R  = 4.4/2 cm = 2.2 cm

r = 4/2 cm = 2 cm
h = 77 cm
(i) Inner curved surface = 2πrh cm2
= 2 × 22/7 × 2 × 77cm2
= 968 cm2

(ii) Outer curved surface = 2πRh cm2

= 2 × 22/7 × 2.2 × 77 cm2
= 1064.8 cm2

(iii) Total surface area of a pipe = Inner curved surface area + outer curved surface area + areas of two bases

= 2πrh + 2πRh + 2π(R2 – r2)
= [968 + 1064.8 + (2 × 22/7) (4.84 – 4)] cm2
= (2032.8 + 44/7 × 0.84) cm2
= (2032.8 + 5.28) cm= 2038.08 cm2

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