![]() #GK#, in the middle, is equal to #DC# because #DE# and #CF# are drawn perpendicular to #GK# and #AB# which makes #CDGK # a rectangle. The large base is #HJ# which consists of three segments: Since we have to find an expression for #V#, the volume of the water in the trough, that would be valid for any depth of water #d#, first we need to find an expression for the large base of trapezoid #CDHJ# in terms of #d# and use it to calculate the area of the trapezoid. The volume of water is calculated by multiplying the area of trapezoid #CDHJ# by the length of the trough. This change affects the length of the large base of the trapezoids at both ends. The water in the trough forms a smaller trapezoidal prism whose length is the same as the length of the trough.īut the trapezoids in the front and the back of the water prism are smaller than those of the trough itself because the depth of the water #d# is smaller than the depth of the trough.Īs the water level varies in the trough, #d# changes. The water level in the trough is shown by blue lines. The trapezoidal prism has a maximum volume of 8.171 L (498.631 in 3) for x 11.076 cm (4.361 in) Using equations (15) through (18), determine the dimensions of the prism: Note: Like the vertical configuration, the graph in Figure 7 could make students think that any x -value between 0 and 12.7, may be used as a length for the trapezoid’s leg. The volume of prism is calculated by multiplying the area of the trapezoid #ABCD# by the length of the trough.īut we are asked to figure out the volume of the water in the trough, and the trough is not full. The trough itself is a trapezoidal prism. The front and back of the trough are isosceles trapezoids. ![]() The figure above shows the trough described in the problem. ![]()
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