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The figure above shows a rectangular garden that is

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tarek99
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The figure above shows a rectangular garden that is [#permalink] New post   28 Nov 2007, 09:34
The figure above shows a rectangular garden that is comprised of a central plot and a border surrounding the central plot; the border and the central plot have the same area. The garden has length 25 feet and width 20 feet. If the length and width of the central plot have the same ratio as the length and width of the garden, what is the length of the central plot, in feet?

a) 25 sqrt(2)
b) 20 sqrt(2)
c) 20(1-sqrt(2))
d) ( 25 sqrt(2) ) / 2
e) (25/2) ( 25/2)


Please show your steps



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Re: The figure above shows a rectangular garden that is [#permalink] New post   28 Nov 2007, 10:29
tarek99 wrote:
The figure above shows a rectangular garden that is comprised of a central plot and a border surrounding the central plot; the border and the central plot have the same area. The garden has length 25 feet and width 20 feet. If the length and width of the central plot have the same ratio as the length and width of the garden, what is the length of the central plot, in feet?

a) 25 sqrt(2)
b) 20 sqrt(2)
c) 20(1-sqrt(2))
d) ( 25 sqrt(2) ) / 2
e) (25/2) ( 25/2)


Please show your steps


The area of the global rectangle is 25*20, while the area of the central plot is l*w. Our aim is to calculate l. From the argument we know that the area of the border is equal to the area of the central plot, so 25*20 - lw=lw, thus lw=250. let's have a system with the other hipothesis l/w=25/20 and substitute w=20/25 l in the first equation. then we have l powered 2 =250*5/4 and it follows that the answer is d. doubts?
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Re: The figure above shows a rectangular garden that is [#permalink] New post   28 Nov 2007, 10:34
I've been struggling with hard sums (quadratic equations and all that), but I think the answer might simply be process of elimination.

The answer has to be in the region of something less than 25, maybe 4 or 5 or 6, say 19 -21.

a) 25 sqrt(2)
Impossible = this comes to more than 25
b) 20 sqrt(2)
Impossible = this comes to more than 25
c) 20(1-sqrt(2))
Impossible - this is negative!
d) ( 25 sqrt(2) ) / 2
This is roughly 18
e) (25/2) ( 25/2)
This is more than 12^2, > 144

So there's only one possible answer - D
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Re: The figure above shows a rectangular garden that is [#permalink] New post   28 Nov 2007, 10:52
the OA is D. and I think marcodonzelli showed a very nice explanation to reach to an answer.
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Re: The figure above shows a rectangular garden that is [#permalink] New post   28 Nov 2007, 11:22
tarek99 wrote:
The figure above shows a rectangular garden that is comprised of a central plot and a border surrounding the central plot; the border and the central plot have the same area. The garden has length 25 feet and width 20 feet. If the length and width of the central plot have the same ratio as the length and width of the garden, what is the length of the central plot, in feet?

a) 25 sqrt(2)
b) 20 sqrt(2)
c) 20(1-sqrt(2))
d) ( 25 sqrt(2) ) / 2
e) (25/2) ( 25/2)


Please show your steps


z is the width of the frame. X is the length of the picture, Y is the width of the picture.

x+2z=25 y+2z=20 x/y=25/20 --> 5w/4w w is a multiplier.

25*20= total area so 500 -(x*y) = x*y so x*y=250.

x-y=5 y=x-5 --> x(x-5)=250 --> x^2-5x-250=0

From here I dont really know how to solve this quadratic.

I did what Raffie did. Only D works
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Re: The figure above shows a rectangular garden that is [#permalink] New post   28 Nov 2007, 13:12
Expert Reply
shortcut:

S1/S0=1/2 ==> L1/L0=1/√2 ==> 25/√2 :-D



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Re: The figure above shows a rectangular garden that is [#permalink]
28 Nov 2007, 13:12
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