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Senior Manager
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PS FRAME [#permalink] New post 07 Jun 2005, 07:53
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A
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D
E

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:) hello
Does anyone plz can help me with this one THANKS
Plz look at the attachment for the frame graph

15. 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 length and width of the frame, what is the length of the picture, in inches?
(A) 9 SQUARE ROOT OF 2
(B) 3/2
(C) 9/SQUARE ROOT OF 2
(D) 15(1-1/SQROOT OF 2
(E) 9/2
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 [#permalink] New post 09 Jun 2005, 12:29
it's weird. should not it be

xy=18*15-xy xy=9*15

x/y=18/15 9*15/y^2=18/15 y=15/(2^1/2)
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 [#permalink] New post 09 Jun 2005, 13:31
l/w=18/15=6/5=>w=5/6l. 18*15-(w*l)=135=>270-(5l/6*l)=135=>135-5l^2/6=0=>l=sqrt162=9sqrt2=A)

i tried a 2nd way. (15-2x)*(18-2x)=135. its 4x^2-66x+135=0. how could i solve this eq ?
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 [#permalink] New post 09 Jun 2005, 21:37
b/a=18/15
ab=18*15/2

1) 15b=18a
=>15b^2=18ab=18*18*15/2
b^2=18*18*/2
b=18/2*sqrt(2)=9sqrt(2)
A
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  [#permalink] 09 Jun 2005, 21:37
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