Solution To Math Squares Problem

 

The  solution is straightforward once one constructs an additional square around D with one side on the horizontal line.  Then the 4 right triangles surrounding square D are identical.  We then let the sides (lengths) of those triangles be x and y.

The area of square D is then equal to the area of the square around square D minus the area of the 4 right triangles surrounding square D.   Working:

A = (x + y)²   - 4 (½ xy ) =  x² +  2xy + y² - 2xy = x² + y²

^{The  length of a side of square D is then:} Ö x² + y ²

We next construct a square around square B so the four right triangles surrounding square B are identical - and the lengths of the sides can then be derived.  This would be from the already derived lengths of the triangles and rectangles written in terms of x and y.   Thus, the length of the side of square surrounding square B is:

 L _{(b )}  =  2 (x + y).   The area of square B is equal to the area of the square around square B minus the area of the four right triangles surrounding square B.  Working:

 A _b   = 2(x + y)²  - 4 [½ (2x) (2y)] = 

4x² +  8xy + 4y²  - 8xy  =   4x² +  4y² 

The length of the side of square B itself is then:

L _b   =  Ö 4(x² +  4y²)   =   2Ö x² + y ²

which is twice the length of a side of square D.