A square sheet of paper has area 15 cm^2. The front side is white and the back side is black. A corner of the sheet is lifted and placed so that the crease is at a 45 degrees angle. If the fold is such that the visible black area is equal to the visible white area, how many centimeters long is the crease? Express your answer in simplest radical form.
+129840 Let the equal sides of the black triangle = x = length of the fold
And the area of this triangle = (1/2)x^2 = x^2/2
If the area of the square sheet of paper = 15cm^2, rhen one side = √15 cm
And the "white" area = x ( √15 -x) + √15 ( √15 - x) = (x + √15) (√15 - x) =
-x^2 + 15
So....since the two areas are equal, we have
x^2/2 = -x^2 + 15 add x^2 to both sides
3x^2 / 2 = 15 multiply both sides by 2/3
x^2 = 10 take the positive root of both sides
x = √10 in = the length of the fold
