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The area of a square garden is A square feet and the

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Manager
Joined: 09 Jun 2010
Posts: 78
The area of a square garden is A square feet and the [#permalink]

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26 Jul 2010, 00:56
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The area of a square garden is A square feet and the perimeter is p feet. If a=2p+9, what is the perimeter of the garden, in feet?

A. 28
B. 36
C. 40
D. 56
E. 64
[Reveal] Spoiler: OA
Manager
Joined: 16 Apr 2010
Posts: 221
Re: Gemotery complex, pleas help [#permalink]

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26 Jul 2010, 01:39
Hi,

The perimeter of the square is 4*length of one side.
The area of the square = Side*Side

Let x = side of the square
a = 2p+9
x*x = 2(x/4) +9
x*x - x/2 - 9 = 0
Solve for x and the area is x*x.

regards,
Jack
Math Expert
Joined: 02 Sep 2009
Posts: 39589
Re: Gemotery complex, pleas help [#permalink]

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26 Jul 2010, 01:49
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Expert's post
xmagedo wrote:
the area of a square garden is A square feet and the perimeter is p feet. If a=2p+9, what is the perimeter of the garden, in feet?
28
36
40
56
64

thanks
tell me about the formula !

The above solution is not right.

Let the side of garden be $$x$$ feet, then: $$area=a=x^2$$ and $$perimeter=p=4x$$. Given: $$a=2p+9$$ --> $$x^2=2*4x+9$$ --> solving for $$x$$: $$x=-1$$ (not a valid solution as $$x$$ represents the length and therefore must be positive) or $$x=9$$ --> $$perimeter=p=4x=36$$.

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Manager
Joined: 09 Jun 2010
Posts: 78
Re: Gemotery complex, pleas help [#permalink]

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26 Jul 2010, 01:54
Pleas can you explain how x=9 ? I didnt get this point
thanks
Math Expert
Joined: 02 Sep 2009
Posts: 39589
Re: Gemotery complex, pleas help [#permalink]

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26 Jul 2010, 02:30
xmagedo wrote:
Pleas can you explain how x=9 ? I didnt get this point
thanks

You'll have quadratic equation $$x^2-8x-9=0$$ and you should solve it for $$x$$ --> $$x=-1$$ or $$x=9$$.
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Senior Manager
Joined: 18 Sep 2009
Posts: 355
Re: Gemotery complex, pleas help [#permalink]

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26 Jul 2010, 18:24
If side = s
A=s*s
P= 4s
A = 2(p)+9
= 2(4s)+9
= 8s +9
s*s = 8s+9
s*s-8s-9=0
s*s-9s+s-9=0
s(s-9)+1(s-9)=0
(s-9) (s+1)=0
s=9 or -1
s=9
P=4(s) = 4(9)=36
Manager
Joined: 22 Jun 2010
Posts: 57
Re: Gemotery complex, pleas help [#permalink]

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27 Jul 2010, 03:42
xmagedo wrote:
the area of a square garden is A square feet and the perimeter is p feet. If a=2p+9, what is the perimeter of the garden, in feet?
28
36
40
56
64

thanks
tell me about the formula !

You can also solve this via using the given numbers in the answer choices!

Of course you need to be aware of the basic properties as outlined by the other posts above (a = x^2 and p = 4x)

Starting with D you will notice that x=14 is way too big for your area (14^2) and will not satisfy: a=2p+9

--> Eliminate D and E

Now pick B (its either too big, then its A, or too small then you know its C or it is B itsself)

And picking B indeed solves the problem! (36/4 --> 9; a= 9^2 = 81 and 81=2x36+9)

Cheers,
André
Director
Joined: 03 Sep 2006
Posts: 871
Re: GMAT Prep CAT PS: Area of Square Garden [#permalink]

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20 Dec 2010, 22:03
1
KUDOS
$$a^2=A$$
$$4*a=P$$
$$a^2=8*a+9$$
$$a^2-8a-9=0$$
$$(a-9)(a+1)=0$$
$$a=9$$
$$P=4*a=36$$
Re: GMAT Prep CAT PS: Area of Square Garden   [#permalink] 20 Dec 2010, 22:03
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