Question

A room is shaped like a trapezoid. There are windows on the walls that are parallel. The distance between the windows is x. The length of one windowed wall is 3 less than the x, and the length of the other windowed wall is 3 more than x. The area of the room is 354 ft². Find the distance between the windows to the nearest tenth of a foot.

Answer 1

while your language is most ambiguous, it appears that

x((x-3)+(x+3))/2 = 354
making x = 18.8

Answer 2

What did you divide 2 by? (x-3)+(x+3)?

Answer 3

if a trapezoid has parallel bases a and b, and height h,

area = h(a+b)/2

Answer 4

I don't understand how you found x.

Answer 5

You describe x as the distance between the windows. Reading the language of the problem, That made me think that x was the distance between the parallel walls.

So, one wall was x-3 and the other was x+3 in length

Their average length is ((x-3)+(x+3))/2 = x

so, since the area of a trapezoid is height (x, the distance between the walls) times the average of the two bases, 2x, we have

x(x) = 354
x^2 = 354
x = √354 = 18.8

If this is not what was intended, clean up the language some.

Answer 6

Okay, I understand now. Sorry if I was bothering you. Thank you so much for your help.