A silo is constructed using a cylinder with a hemisphere on top. The circumference of the hemisphere and the circumference of the cylinder are equal. The diameter of the circular base of the cylinder is 10 feet. The cylinder is 40 feet tall. One of the circular bases on the cylinder is in contact with the ground. What is the area of the surface of the silo that will be exposed to rain, wind, and sun?

Question

Mathematics

Rene Ward

13 November, 23:41

A silo is constructed using a cylinder with a hemisphere on top. The circumference of the hemisphere and the circumference of the cylinder are equal. The diameter of the circular base of the cylinder is 10 feet. The cylinder is 40 feet tall. One of the circular bases on the cylinder is in contact with the ground. What is the area of the surface of the silo that will be exposed to rain, wind, and sun?

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Answers (2)

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Reilly Palmer · Answer 1

The surface area for this case is given by:

A = 2 * pi * r * h + (1/2) * 4 * pi * r ^ 2

Where,

r: cylinder radius (same sphere radius)

h: cylinder height

Substituting values we have:

A = 2 * 3.14 * 5 * 40 + (1/2) * 4 * 3.14 * 5 ^ 2

A = 1413 feet ^ 2

Answer:

The area of the surface of the silo that will be exposed to rain, wind, and sun is:

A = 1413 feet ^ 2

Glenn · Answer 2

Surface area will be = hemisphere + cylinder - 1 base of cylinder

hemisphere surface area = 2 π r^ 2

hemisphere surface area - area of circle = 2 π r^2 - π r^2 = π r^2

hemisphere surface area - area of circle = π (5) ^2 = 78.5 sq. ft

cylinder - area of circle = 2 π r h + π r^2 = 1256 + 78.5 = 1334.5 sq. ft

surface area = 1413 sq. ft