Answer:
It is easy to see that the height of the triangle that is folded over is 1/4 of the diagonal of the square, so ....
let the side of the square be x
then its diagonal is √2x
and the height of the triangle bent over is √2x/4
and the fold at which the bending takes place would be √2x/2 (we have 45° angles)
area of triangle bent over = (1/2)(√2x/2)(√2x/4)
= 2x^2/16 = x^2 /8
area of pentagon = x^2 - x^2/8 = 7x^2/8
x^2 - 7x^2/8 = 1 (difference between consecutive integers is 1)
times 8
8x^2 - 7x^2 = 8
x^2 = 8
the area of the triangle is x^2/8 = 8/8 = 1 square unit
