More details with Trapazoidal lagoon.?

Question

My "trapazoidal lagoon" is a holding pond shaped like a trapazoid. The bottom measures 614ft x 415ft. If the depth of the water increases one foot the surface dimensions increase by 6ft. How can I calculate the cf in the pond for a given depth.

This is a real life problem. It is not a trick question!

Answer 1

The water in the pond would look like an upside down truncated rectangular pyramid or a 'pyramidal frustum'.

The volume of such a figure is given by:
V = 1/3 h (A1 + A2 + sqrt(A1*A2))

A1 is the area of the bottom of the pond.
A2 is the area at a depth of h feet.

For h = 1:
A1 = 614 * 415 = 254,810 sq. ft.
A2 = 620 * 421 = 261,020 sq. ft.
sqrt(A1*A2) = 257,896
V(1) = 1/3 * 1 (254810 + 261020 + 257896)
V(1) = 773726 / 3
V(1) = 257,908.667 cu. ft.

For h = 2:
A1 = 614 x 415
A2 = 626 x 427

For h = 3:
A1 = same as before
A2 = (626+6) x (427+6)
etc.

You should be able to plug in the numbers rather easily to the formula to get any depth.

Answer 2

Is this a trick question?