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07 Oct 2009, 06:11
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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
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
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07 Oct 2009, 06:43
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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
.
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
. 07 Oct 2009, 07:45
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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.
