Question 311546
The ancient Greeks thought that the
most pleasing shape for a rectangle was one for which the
ratio of the length to the width was approximately 8 to 5,
the golden ratio. If the length of a rectangular painting is
2 ft longer than its width, then for what dimensions would
the length and width have the golden ratio?
:
Here's one way
:
Let x = the multiplier
then
8x = the length
and
5x = the width
:
" length of a rectangular painting is 2 ft longer than its width,"
8x = 5x + 2
8x - 5x = 2
3x = 2
x = {{{2/3}}} is the multiplier
then
8*{{{2/3}}} = {{{16/3}}} = 5{{{1/3}}} ft is the length
and
5*{{{2/3}}} = {{{10/3}}} = 3{{{1/3}}} ft is the width
:
We can see there is a 2' difference