One corner of a square is folded to its centre to form an irregular pentagon.the areas of the pentagon and of the square are consecutive integers.what is the area of the square?

Answer 1

Make a sketch.
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