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M09 #07

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Senior Manager
Joined: 29 Jul 2009
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M09 #07 [#permalink]

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07 Oct 2009, 06:11
Hi everyone, I read the explanation for this question but I'm not convinced how if the angle ABC is 90, then we can conclude that $$a^2=\frac{1}{a}$$

According to me if the angle ABC is 90, then the triangle becomes isosceles but the sides are $$a^2$$ and $$\frac{2}{a}$$. Because is a triangle rectangle $$(\frac{2}{a})^2 = (a^2)^2 + (a^2)^2$$ ---> $$2 = a^6$$

Can anyone explain why my reasoning is wrong?
Thank you
CIO
Joined: 02 Oct 2007
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Kudos [?]: 874 [0], given: 334

Re: M09 #07 [#permalink]

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07 Oct 2009, 06:43
Here's the discussion you need to have a look at:
ds-triangle-m09q07-72173.html

You can search for the questions you need to ask a question of in this thread:
gmat-club-tests-master-threads-all-tests-78599.html

We try to avoid posting the duplicate threads. Thank you for cooperation .
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Senior Manager
Joined: 29 Jul 2009
Posts: 314
Followers: 4

Kudos [?]: 286 [0], given: 9

Re: M09 #07 [#permalink]

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07 Oct 2009, 07:45
I'm sorry I tried to look for that thread but couldn't find it. The normal search does not work properly, and I cannot access to the other thread you post. I'll avoid duplicating threads next time. sorry for the inconvenience.
CIO
Joined: 02 Oct 2007
Posts: 1218
Followers: 95

Kudos [?]: 874 [0], given: 334

Re: M09 #07 [#permalink]

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07 Oct 2009, 08:03
Sorry, there was a typo. You must be able to access both threads now.
mikeCoolBoy wrote:
I'm sorry I tried to look for that thread but couldn't find it. The normal search does not work properly, and I cannot access to the other thread you post. I'll avoid duplicating threads next time. sorry for the inconvenience.

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Re: M09 #07   [#permalink] 07 Oct 2009, 08:03
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M09 #07

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