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One 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 a Golden Rectangle, the sides of the remaining rectangle are proportional to the original rectangle. So the length and width of the original rectangle satisfy L/W=W/L-W. 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?

Submitted: 228 days and 11 hours ago.

Category: Math

Value: $15

Status: CLOSED

Accepted Answer

Let the width be W inches.

36/W = W/(36 - W)

36(36 - W) = W^2

W^2 + 36W - 1296 = 0

W = [-36 ± √(36^2 - 4 * 1 * -1296)]/2(1)

W = [-36 ± √6480]/2

W = 22.25 inches.

Accepts:
Expert:	Chirag
Pos. Feedback:	99.3 %
Answered:	4/8/2009

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Accepted Answer

Let the width be W inches. 36/W = W/(36 - W) 36(36 - W) = W^2 W^2 + 36W - 1296 = 0 W = [-36 ± √(36^2 - 4 * 1 * -1296)]/2(1) W = [-36 ± √6480]/2 W = 22.25 inches.

Related Math Questions

Let the width be W inches.

36/W = W/(36 - W)

36(36 - W) = W^2

W^2 + 36W - 1296 = 0

W = [-36 ± √(36^2 - 4 * 1 * -1296)]/2(1)

W = [-36 ± √6480]/2

W = 22.25 inches.