ICSE Board Question Based on Mensuration Chapter of M.L Aggarwal for class10

In this question a figure is given which is in the form of metal pipe of a certain height given.

The inner diameter of a cross-section and the outer one is given

Find its (i) inner curved surface area (ii) outer curved surface area (iii) total surface area.

This is the Question Number 19, Exercise 17.1 of M.L Aggarwal.

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# The given figure shows a metal pipe 77 cm long. The inner diameter of a cross-section is 4 cm and the outer one is 4.4 cm. Find its (i) inner curved surface area (ii) outer curved surface area (iii) total surface area.

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Given height of the metal pipe = 77 cmInner diameter = 4 cmInner radius, r = d/2 = 4/2 = 2 cmOuter diameter = 4.4 cmOuter radius, R = d/2 = 4.4/2 = 2.2 cm(i)Inner curved surface area = 2rh= 2×(22/7)×2×77= 968 cm^{2}Hence the inner surface area is 968 cm^{2}.(ii)Outer curved surface area = 2Rh= 2×(22/7)×2.2×77= 1064.8 cm^{2}Hence the outer curved surface area is 968 cm^{2}.(iii)Area of ring = (R^{2}-r^{2})= (22/7)×(2.2^{2}-2^{2})= (22/7)×(4.84-4)= (22/7)×0.84= 2.64Total surface area = inner surface area + outer surface area + area of two rings= 968+ 1064.8+ 2×2.64= 2038.08 cm^{2}Hence the total surface area of the metal pipe is 2038.08 cm^{2}.