the golden ratio- the ancient greeks thought that the most pleasing shape for a rectangle was one for which the ratio of the length to the was 8 to 5, the golden ratio. if the length of a rectangular painting is 2 ft.longer than its width, then for what dimension would the length and width have the golden ratio?
2007-10-19 14:48:20 · 4 answers · asked by mormar 1 in Science & Mathematics ➔ Mathematics
w = width in feet
then w+2 = length in feet
So 8/5 = (w+2)/w
Find the cross products, then solve for w:
8w = 5(w+2)
8w = 5w + 10
3w = 10
w = 3 1/3 ft
So the length must be 2 + 3 1/3 = 5 1/3 ft
I hope this helps!
2007-10-19 14:54:45 · answer #1 · answered by math guy 6 · 0⤊ 0⤋
If we denote the width of the painting with "x" then the length of the painting is x+2. Next we can produce an equation to find out what x is:
(x+2)/x = 8/5
5x + 10 = 8x
5x + 10 - 8x = 0
10 - 3x = 0
10 = 3x
thus x = 10/3 (approx 3.3 ft)
The painting dimensions would have to be 3.3 width and 5.3 length to produce the golden ratio.
2007-10-19 21:56:01 · answer #2 · answered by sammichbabe 1 · 0⤊ 0⤋
Let the width be x
Length = x + 2
Length/Width = 8/5
(x + 2)/x = 8/5
5(x + 2) = 8x
5x + 10 = 8x
3x = 10
x = 10/3
x + 2 = 16/3
The dimensions would be 16/3 ft by 10/3 ft.
2007-10-19 22:02:51 · answer #3 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
you can reduce ratios so you can also multiply them
2007-10-19 21:53:36 · answer #4 · answered by Anonymous · 0⤊ 0⤋