A square piece of tin, 10 inches on a side, is to have four equal squares cut from its corners, as shown. If the edges are then to be folded

Let the sides of the cut squares  = x....and when we fold these up, the dimensions of the floor area of the box are just (10 - 2x) (10-2x)...and this area = 36 sq in

So we have

(10 - 2x)^2 = 36    take the square root of both sides

10 - 2x = ±√36.......so we have

2x = 10 ±√36

x = (10 ±√36) / 2

So x = (10 + 6) / 2    or   x = (10 - 6) / 2

So, x = 8 or x  = 2

But, we must reject 8 because that would mean the sides of the floor area would be (10 - 2(8)) = (10 -16) = -6 inches  ...and this is impossible.....

So the sides of the floor area are (10 - 2(2)) = (10 - 4) = 6 in.   And the height is just x = 2 in. .....I'm assuming that, since I can't see a picture, that "depth" equates to "height."

The square floor must have sides of 6 inches.  The depth of the folded sides of the box must be (10 - 6)/2 = 2 inches.

Beaten to it by Chris this time!!

That's OK, Alan......I still gave you some points !!!

And I will give you both some points