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M11 #32

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I3igDmsu
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M11 #32 [#permalink] New post   01 Jul 2009, 20:34
If a square is inscribed in a circle of radius 4, what is the area of the square?

(C) 2008 GMAT Club - m11#32

* 8
* 16
* 32
* 48
* 64

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The diameter of the circle is the diagonal of the square. The side of the square is \(2*\sqrt{8}\) , therefore the area of the square is 32. The correct answer is C.

How do you get 2*sqrt(8) for each side? I was getting 4 which is x^2+x^2=8^2, so x=4. Please explain how do solve for the sides using Pythagorean theorem.


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sdrandom1
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Re: M11 #32 [#permalink] New post   01 Jul 2009, 20:57
If a square is inscribed in a circle of radius 4, what is the area of the square?

* 8
* 16
* 32
* 48
* 64

The diameter is 8 = Diagonal of the square.
But diagonal of a square with side a \(= a\sqrt{2}\)
Hence if \(8 = a\sqrt{2}\), then we can see that side of the sqaure \(= a = \frac{8}{\sqrt{2}}\)
Therefore, area \(= a^2 = \left(\frac{8}{\sqrt{2}}\right)^2 = 32\)
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I3igDmsu
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Re: M11 #32 [#permalink] New post   02 Jul 2009, 05:09
Still not understanding. Any help?
I think its a notation problem above. what does the /\ mean?
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I3igDmsu
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Re: M11 #32 [#permalink] New post   07 Jul 2009, 05:00
sdrandom1 wrote:
If a square is inscribed in a circle of radius 4, what is the area of the square?

The diameter is 8 = Diagonal of the square.
But diagonal of a square with side \(a = a\sqrt{2}\)
Hence if \(8 = a\sqrt{2}\), then we can see that side of the sqaure = \(a = 8/\sqrt{2}\)
Therefore, area = \(a^2 = (8/\sqrt{2})^2 = 32\)


Got it now, the math formulas weren't working, its easier to understand now above. Thanks sdrandom1.
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Re: M11 #32 [#permalink] New post   07 Jul 2009, 06:25
I've edited the post above so that math expressions work.

In order for the math expressions to work fine, one needs to enclose the whole expression in the "m" tag, which is next to the "fraction" tag in the edit page. This won't work (without the "m" tag):

Code:
[square_root]x^2 -2x +1[/square_root] = [square_root](x-1)^2[/square_root] = |x-1|

\sqrt{x^2 -2x +1} = \sqrt{(x-1)^2} = |x-1|

This will work (just add the "m" tag):

Code:
[m][square_root]x^2 -2x +1[/square_root] = [square_root](x-1)^2[/square_root] = |x-1|[/m]

\(\sqrt{x^2 -2x +1} = \sqrt{(x-1)^2} = |x-1|\)

Remember to use only one "m" tag for the whole expression, no need to enclose every single unit in it.

Hope this helps.

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Re: M11 #32 [#permalink]
07 Jul 2009, 06:25
Moderator:
Bunuel
Math Expert
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