Algebra 1: Structure and MEthod problem help?

Question

A rectangular garden 30 meters by 40 meters has two paths of equal width crossing through each other. Find the width of each path is the total area covered by the paths is 325m^2

The paths intersect at a right angle like a cross.

Answer 1

If we assume the paths are at right angles (one parallel straight across the width, and one straight across the length) you can make the following equation:

W * 30m + W * 40m - W**2 = 325 sq.m.
Then solve for W.

Notes:
1.) W = the path width
2.) W**2 (that's "W squared" in FORTRAN notation) takes into account that both paths cover the area where they cross and you don't want to count that square of land two times, just one time.

Answer 2

I think we need more info. Are the paths running parallel with the sides?

I am assuming it does and answering based on that.

they are equal and they overlap so the total area (325) + the width squared (the overlap) equals the width times seventy (30+40)

325+W^2 = W*70

If you cant solve it you can use it to check your answer. It is an even number (that is why I assume the paths are parallel) and you can get a rough estimate by dividing the total area by the lengths of the paths.

Answer 3

Assuming the paths form 4 smaller rectangles, their area using w for their width is 30w + 40w - w^2 (subtract the part where they cross so as not to count it twice)

So 70w - w^2 = 325, or w^2 - 70w + 325 = 0

Factor this by FOIL to get w.