Question

Please help!

A water tank in the form of an inverted frustum of a cone has an altitude of 8 ft., and upper and lower radii of 6 ft. and 4 ft., respectively. Find the volume of the water tank and the wetted part of the tank if the depth of the water is 5 ft.

Answer 1

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Answer 2

When the depth is 5 ft, the tank is 5/8 full, so the top radius of the water is (4 ft) + (5/8)(6 ft - 4 ft)
= 4 ft + (5/8)(2 ft)
= 4 ft + 1.25 ft = 5.25 ft.

The slant height 5.154 ft is the square root of
[(5 ft)^2 + (5.25ft - 4ft)^2]
= sqrt(25 ft^2 + 1.5625 ft^2)
= 5.15388... feet

A formula for the surface area of the frustum
(see Wikipedia, etc) results in
A = pi(4.00 ft + 5.25 ft)*5.154 ft + pi*(4.00 ft)^2
= 200.0 ft^2

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