Hello, MATNTRNG!

I solved it, but I'm certain there is a more elegant way.

Code:: x H b-2x E x : B *-------*-------*-------* C |* : *:* | | * : * : * | | * : * : * | a | * : * :a * | a | * : * : * | | * : * : * | | *:* : *| A *-------*-------*-------* D : x F b-2x G x : : - - - - - b - - - - - :

Draw .Draw

Let .[1]

In right triangle .[2]

Equate [2] and [1]: .

Square both sides: .

. . Hence: .

. .

. . . . . .

. . . . . .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Fascinating!

The answer is the ratio of the width and length

. . times the length of the diagonal.

Does this suggest a more direct solution?

. .I don't know!