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In the figure, each side of square ABCD has length 1, the

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Manager
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In the figure, each side of square ABCD has length 1, the [#permalink] New post 27 Nov 2007, 13:42
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A
B
C
D
E

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In the figure, each side of square ABCD has length 1, the length of line
Segment CE is 1, and the length of line segment BE is equal to the length
Of line segment DE. What is the area of the triangular region BCE?
a. 1/3
b. (2^-2 )/4
c. 1/2
d. (2^-2)/2
e. 3/4

See attached file for figure
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 [#permalink] New post 27 Nov 2007, 14:26
Expert's post
area(BCE)=area(BGE)-area(BGC)

area(BCE)=1/2*1/√2*(1+h)-1/2*1/√2*h=1/2*1/√2=2^(-3/2)

but 2^(-3/2) is not among possible solutions. What is wrong?
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Manager
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 [#permalink] New post 27 Nov 2007, 14:37
walker wrote:
area(BCE)=area(BGE)-area(BGC)

area(BCE)=1/2*1/√2*(1+h)-1/2*1/√2*h=1/2*1/√2=2^(-3/2)

but 2^(-3/2) is not among possible solutions. What is wrong?



Your figure is right. I was confused with the figure. The answer is D.

You are on right track. BG=GC = root(2)/2 (half of diagonal of square)

Once we know BG and GC,rest is as you said.
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  [#permalink] 27 Nov 2007, 14:37
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In the figure, each side of square ABCD has length 1, the

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