Question

A square is inscribed in another square such that each vertex of inner square divides the side of the outside square into intervals of length `x` and `y` so that `x > y`. Find the ratio of inscribed square to outer square?

Answer 1

Please see below.

Explanation:

As a square is inscribed in another square such that each vertex of inner square divides the side of the outside square into intervals of length `x` and `y` so that `x > y`, it appears as follows:

It is apparent that each side of outer square is `x+y` and its area is `(x+y)^2` or `x^2+y^2+2xy`.

Further using Pythagoras theorem, each side of inner square is `sqrt(x^2+y^2)` and its area is `x^2+y^2`.

Hence, area of outer square exceeds that of inner square by `2xy`.

and ratio of inscribed square to outer square is `(x^2+y^2)/(x^2+y^2+2xy)`