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Balvinder
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geometry 01 Jul 2007, 07:57
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please explain..



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Himalayan
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geometry 01 Jul 2007, 18:03
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Balvinder
please explain..
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I DID NOT GET IT? any idea?
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UMB
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geometry 02 Jul 2007, 12:50
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I have seen this problem before. I know the source (S25-31). Something was wrong here. I don’t still change my mind. Although the RA (or OA) is D. I don’t think it is the only possible answer. Moreover it seems to me it is difficult , if not impossible, to come to this solution. Area of BCE, under the same conditions, could be ½ (if CE is perpendicular to ABCD square) or (sqrt2)/4 (if CE lies on the same plane with ABCD square.).
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Juaz
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geometry 03 Jul 2007, 08:32
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UMB
I have seen this problem before. I know the source (S25-31). Something was wrong here. I don’t still change my mind. Although the RA (or OA) is D. I don’t think it is the only possible answer. Moreover it seems to me it is difficult , if not impossible, to come to this solution. Area of BCE, under the same conditions, could be ½ (if CE is perpendicular to ABCD square) or (sqrt2)/4 (if CE lies on the same plane with ABCD square.).
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Thats what I thought earlier. Area = 1/2* 1*1*sin135 = 1/2*(1/sqrt2) = sqrt2/4. :?
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Andrey2010
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geometry 03 Jul 2007, 08:32
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sorry for stupid question :oops:
I see the answer :oops:
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[email protected]
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geometry 04 Jul 2007, 06:24
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I have read this question somewhere but don't know where.


Draw the diagonal BD.
BD = sqrt(2)
Area of triangle BDE = 1/2 *sqrt(2) * [1 +(1/sqrt(2)]

Area of BCE = (Area of BDE - Area of BCD)/2 = 1/(2*sqrt(2))

But I don't see this answer choice.



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