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The shaded region in the figure above represents a rectangular frame [#permalink]

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18 Mar 2010, 14:34

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The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the lenght and width of the frame, what is the length of the picture, in inches?

A. \(9\sqrt2\) B. \(\frac {3}{2}\) C. \(\frac {9}{\sqrt2}\) D.\(15 ( 1 - \frac {1}{\sqrt2})\) E. \(\frac {9}{2}\)

Additional Info on the Problem Source: Paper Test Test Code 28 Section 5 # 15

The was I was approaching this question was that I found the ratio of 18:15, that is 6:5 and then trying to find the length with the given answer options to see if i can get the same area as 18*15=270.

I get [9(2)^1/2 *5/6] *9(2)^1/2 which is equal to 135. I dont see how this will be equal to 270. Maybe I dont understand the question properly.

Re: The shaded region in the figure above represents a rectangular frame [#permalink]

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28 Sep 2015, 11:08

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Re: The shaded region in the figure above represents a rectangular frame [#permalink]

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28 Sep 2015, 23:11

Expert's post

The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the lenght and width of the frame, what is the length of the picture, in inches?

A. \(9\sqrt2\) B. \(\frac {3}{2}\) C. \(\frac {9}{\sqrt2}\) D. \(15 ( 1 - \frac {1}{\sqrt2})\) E. \(\frac {9}{2}\)

Say the length and the width of the picture are \(x\) and \(y\) respectively. Since they have the same ratio as the lenght and width of the frame, then \(\frac{x}{y}=\frac{18}{15}\) --> \(y=\frac{5}{6}x\).

Next, since the frame encloses a rectangular picture that has the same area as the frame itself and the whole area is \(18*15\), then the areas of the frame (shaded region) and the picture (inner region) are \(\frac{18*15}{2}=9*15\) each.

The area of the picture is \(xy=9*15\) --> \(x*(\frac{5}{6}x)=9*15\) --> \(x^2=2*81\) --> \(x=9\sqrt{2}\).

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