An open channel is a half-pipe: a semicircular trough of internal radius 21 cm and length 10 m, flat side upwards. Its inner curved surface is to be given a waterproof coating. Find the area to be coated in m². Take π = 22/7.

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Accepted Answer

6.6. The inner curved surface is half of a cylinder's curved surface: πr × length = (22/7) × 21 × 1000 = 66000 cm². Since 1 m² = 10000 cm², that is 66000/10000 = 6.6 m². (Using the whole cylinder would wrongly give 13.2 m².)

An open channel is a half-pipe: a semicircular trough of internal radius 21 cm and length 10 $m$ , flat side upwards. Its inner curved surface is to be given a waterproof coating. Find the area to be coated in $m^{2}$ . Take π = 22/7.

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An open channel is a half-pipe: a semicircular trough of internal radius 21 cm and length 10 m, flat side upwards. Its inner curved surface is to be given a waterproof coating. Find the area to be coated in m2. Take π = 22/7.

Step-by-Step Solution

Master this topic in 30 minutes a day.

More Areas & Volumes practice

An open channel is a half-pipe: a semicircular trough of internal radius 21 cm and length 10 $m$ , flat side upwards. Its inner curved surface is to be given a waterproof coating. Find the area to be coated in $m^{2}$ . Take π = 22/7.