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# M09 #07

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Senior Manager
Joined: 29 Jul 2009
Posts: 314
Followers: 4

Kudos [?]: 266 [0], given: 9

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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
Posts: 1218
Followers: 94

Kudos [?]: 838 [0], given: 334

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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:

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 [?]: 266 [0], given: 9

### Show Tags

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: 94

Kudos [?]: 838 [0], given: 334

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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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14 M09 Q11 18 19 Nov 2008, 07:03
m09 #34 6 10 Nov 2008, 04:10
18 M09 #20 19 10 Nov 2008, 03:24
24 DS : TRIANGLE (m09q07) 41 29 Oct 2008, 00:53
10 m09 q22 41 07 Sep 2008, 13:18
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# M09 #07

Moderator: Bunuel

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