Question

9. An engineer is designing a trench

to handle run-off from a reservoir.

To accommodate the flow of water,

it needs to have a specific cross-

sectional area. Starting with a 5-foot

wall on one side, the bottom of the

trench will slope downward 2 feet

for every 7 feet across from either

side to the center.

5 + 2/x

The cross-sectional area of each

half must be 50 square feet.

Part A

is A =

x² +

What equation in standard form

can be solved to find the distance

to the center of the trench? The

formula for the area of a trapezoid
. Th.

b₁ + b₂

2

x +

11. T

=0

Part B

How far should the center of the

trench be from either side? Round

to the nearest foot.

feet

Answer 1

A) A = x² +35x -350

B) x ≈ 8 ft

Explanation:

You want a standard form equation that can be solved for the width of a trench that has an area of 50 square feet and a bottom that slopes away from a 5 ft side wall down 2 ft for 7 ft across. You also want the solution.

Part A

The area is that of a trapezoid. One base is 5 ft. For width x, the other base is 5 +2/7x. Then the area is ...

A = 1/2(b1 +b2)h

A = 1/2(5 +(5 +2/7x))(x)

We want this area to be 50 ft², so ...

50 = 1/2(10 +2/7x)(x) = x(x/7 +5)

Subtracting 50 and multiplying by 7 gives the standard form equation ...

x² +35x -350 = 0

Part B

The solution to the equation is suggested by the graph (second attachment) to be ...

x = 17.5 +√(656.25) ≈ 8.117

The trench is about 8 feet from side wall to center.

__

Additional comment

The first attachment shows the cross section of the trench, along with its area. The dimensions shown are rounded from the values used to compute the area.

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