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Q31

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by kajcha » Mon Sep 17, 2007 3:31 pm
It's not clear what the choice B is?

But, if this is (2^(1/2))/4.. then this is the answer...

What's the OA?
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by magical cook » Mon Sep 17, 2007 3:35 pm
Yes you're right as to how you see B...but answer seems to be D.... hmm
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by kajcha » Mon Sep 17, 2007 3:38 pm
Are you sure B is (2^(1/2))/4

Or

Is it (2^(-1/2))/4
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by magical cook » Mon Sep 17, 2007 3:39 pm
I think (2^(1/2))/4 ...
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by kajcha » Mon Sep 17, 2007 3:50 pm
Would you please tell what is D?
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by magical cook » Mon Sep 17, 2007 3:53 pm
I think (2^1/2)/2
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by kajcha » Mon Sep 17, 2007 4:01 pm
If it is NOT 2^(-1/2) then this cannot be answer..

As per my calculation answer will be root(2)/4 = 2^(1/2)/4. This is how I calculated.. I will try my best to explain w/o the pic.. Will try to upload a pic later

Connect BD and extend EC to BD. Say EC connect BD at F.

Now BD = root(2) (Diagonal of a sq)

CF = BF = BD/2 = root(2)/2

EF = 1+CF = 1 + (root(2)/2)

Area of BFE = (1/2)*(EF)*(BF) = (root(2)+1)/4 ----- (1)

Area of BFC = (1/2)*(CF)*(BF) = 1/4 ----- (2)

Area of BCE = (1) - (2) = root(2)/4
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Query on my approach????

by niks_01.27 » Mon Sep 17, 2007 4:45 pm
When I solved using 'kajcha' approach, I got the area of BCE = 1/[2 root(2)]; which is answer choice B, but initially I thought the only case when BE=DE and CE=1 ( equal to side of a square) is when CE is perpendicular to both BC & CD and/or all are parts of a cube. So the area of BCE would be equal to 1/2 (BC)(CE) = 1/2.

I still couldn't counter point my first approach. Any one here can help or give light on this??
regards
niks...
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by kajcha » Mon Sep 17, 2007 4:59 pm
One thing that I assumed in my approach was extention of EC will be perpendicular to diagonal BD... that might be wrong... still thinking... :roll:
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by kajcha » Mon Sep 17, 2007 6:00 pm
Since BE = ED, BDE forms a isoc traingle.

extension of CE will cut BD at right angle...
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by niks_01.27 » Mon Sep 17, 2007 7:03 pm
kajcha wrote:Since BE = ED, BDE forms a isoc traingle.

extension of CE will cut BD at right angle...
Every line from C will not form a right angle at BD. We have to assume here that CE is the median of the isoc triangle BDE. In that case;

Area BDE = 1/2 x root(2) x (1 + 1/root(2)) = 1/root(2) + 1/2
Arear BEF = 1/2 Area BDE = 1/2 root(2) + 1/4
Area BDC = 1/2 x root(2) x 1/root(2) = 1/2

Now Area BCE = Area BEF - Area BDC
= 1/2 root(2) + 1/4 - 1/2
= 1/2 root(2) - 1/4
= [root(2) - 1]/4

Am I missing anything??

My approach will be right if we consider a 3D figure (which again is an assumption :wink: )

I don't know what to do here.

Also would be great if you can look into https://www.beatthegmat.com/viewtopic.php?t=5035 and give your inputs.
regards
niks...
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