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!