the length of a cylindrical rod is twice its radius if its radius is 42 cm find the area of a road leveled by it in one revolution
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Answered by
21
Radius = 42 cm
Length = 42 x 2 = 84 cm
Find 1 revolution:
1 revolution = 2πrh
1 revolution = 2π(42)(84) = 12,672 cm²
Answer: The area covered by 1 revolution is 12,672 cm²
TheLostMonk:
i think you did mistake in multiplication .
Answered by
20
given radius = 42 cm
length of cylindrical roller = twice of its radius
= 2 × 42 = 84 cm
height of cylinderical roller will be=length of roller=84 cm
so now
area leveled by the roller in one revolution= curved surface area of cylinder=2πrh
= 2 × 22/7 × 42 × 84
= 2× 22 × 6 × 84 = 22176 cm ^2
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Your Answer : 22176 cm^2
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length of cylindrical roller = twice of its radius
= 2 × 42 = 84 cm
height of cylinderical roller will be=length of roller=84 cm
so now
area leveled by the roller in one revolution= curved surface area of cylinder=2πrh
= 2 × 22/7 × 42 × 84
= 2× 22 × 6 × 84 = 22176 cm ^2
______________________________
Your Answer : 22176 cm^2
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