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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 Jul 2008 15:06:01 IST
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ABCD is a square .F is the mid point of AB.BE is one third of BC .If the area of triangle FBE=108cm^2,then find the length of AC.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 Jul 2008 21:19:54 IST
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AB=BF/2=AF/2
BE=BC/3
ar(FBE)=1/2(BF)(BE)
108=1/2(AB/2)(BC/3)
(36)(36)=a^2
a=36
Therefore,AC=36(ROOT2)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jul 2008 21:04:31 IST
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how is AB=BF/2=AF/2 AB/2=BF=AF
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SHREYA |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jul 2008 21:08:00 IST
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@shinee,in the question itself it is mentioned that F is midpoint of AB.So AB=BF/2=AF/2 and BF=AF=2AB
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Jul 2008 23:21:12 IST
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I think my answer is correct.Please check it and tell me if i am wrong.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Jul 2008 12:29:47 IST
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area of =




 
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God does not care about our mathematical difficulties. He integrates empirically. ~~~Albert Einstein (1879-1955)~~~~
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.~~~Pierre de Fermat (1601-1665)~~~
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Jul 2008 21:13:43 IST
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If F is the mid point of AB, then AB/2=BF=AF, AB=2BF=2AF Ab should be twice of BF and AF
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SHREYA |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Jul 2008 21:26:09 IST
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" If F is the mid point of AB, then AB/2=BF=AF, AB=2BF=2AF Ab should be twice of BF and AF"
Yes what is the problem??
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God does not care about our mathematical difficulties. He integrates empirically. ~~~Albert Einstein (1879-1955)~~~~
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.~~~Pierre de Fermat (1601-1665)~~~
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Jul 2008 20:46:05 IST
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@rudra, once see wat krish1097 has written
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SHREYA |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Jul 2008 23:37:03 IST
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What do you want to say?????
AB/2=BF=AF, AB=2BF=2AF this is correct. What is the problem?
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God does not care about our mathematical difficulties. He integrates empirically. ~~~Albert Einstein (1879-1955)~~~~
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.~~~Pierre de Fermat (1601-1665)~~~
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