Question

# An open metal bucket is in the shape of frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameter of the two circular end of buckets are $$45$$cm and $$25$$ cm, the total vertical height of the bucket is $$40$$ cm and that of cylindrical base is$$6$$ cm. Find the area of the metalic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also find the volume of water the bucket can hold. [Take $$\pi=\dfrac{22}{7}$$]

Solution

## Curved bucket surface area $$BB'C'C$$ $$=$$curved surface area of cone $$ABC$$$$-$$curved surface area of$$AB'C'$$slant height of cone $$ABC$$$$=$$$$\sqrt { { 90 }^{ 2 }+{ 22.5 }^{ 2 } }$$$$=$$$$92.77cm$$slant height of cone $$AB'C'=$$$$\sqrt { { 50 }^{ 2 }+{ 12.5 }^{ 2 } }$$$$=$$$$50.12$$$$cm$$so curved surface area of bucket $$BB'C'C=$$$$\pi \times 22.5\times 92.77-\pi \times 12.5\times 50.12$$                                                                         $$=4587{ cm }^{ 2 }$$surface area of cylindrical base$$=2\times 3.14\times { 12.5 }^{ 2 }+2\times 3.14\times 12.5\times 6\\ =1452.25{ cm }^{ 2 }$$total surface area$$=1452.25+4587=6039.25{ cm }^{ 2 }$$volume of water$$=\frac { \pi { \times 22.5 }^{ 2 }\times 90 }{ 3 } -\frac { \pi \times { 12.5 }^{ 2 }\times 50 }{ 3 } \\ =39511.67{ cm }^{ 3 }\\ =\frac { 39511.67 }{ 1000 } ℓ\\ =39.5ℓ$$             Mathematics

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