Question 312152: Once principle used by the ancient Greeks to get shapes that are pleasing to the eye in art and architecture was the Golden Rectangle. If a square is removed from one end of the Golden Rectangle, the sides of the remaining rectangle are proportional to the original rectangle. So the lenght and width of the original rectangle satisfy L/W = W/L-W. If the length of a Golden Rectangle is 10 meters, then what is its width? Also, an artist wants her painting to be in the shape of a golden rectangle. If the length of the painting is 36 inches, then what should be the width?
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! Once principle used by the ancient Greeks to get shapes that are pleasing to the eye in art and architecture was the Golden Rectangle. If a square is removed from one end of the Golden Rectangle, the sides of the remaining rectangle are proportional to the original rectangle. So the lenght and width of the original rectangle satisfy L/W = W/L-W. If the length of a Golden Rectangle is 10 meters, then what is its width? Also, an artist wants her painting to be in the shape of a golden rectangle. If the length of the painting is 36 inches, then what should be the width?
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If the length of a Golden Rectangle is 10 meters, then what is its width?
L/W = W/(L-W)
10/W = W/(10-W)
10(10-W) = W^2
100-10W = W^2
0 = W^2+10W-100
Solve by applying the quadratic formula. Doin so yields:
W={6.18, -16.18}
Toss out the negative solution leaving:
W = 6.18 meters
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If the length of the painting is 36 inches, then what should be the width?
L/W = W/(L-W)
36/W = W/(36-W)
36(36-W) = W^2
1296-36W = W^2
0 = W^2+36W-1296
Solve by applying the quadratic formula. Doin so yields:
W={22.25, -58.25}
Toss out the negative solution leaving:
W = 22.25 inches
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