Friday, March 1, 2024

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 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 2


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




No comments: