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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]

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27 Nov 2007, 14:42
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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
Attachments

Geometry.doc [24 KiB]

Geometry.doc [24 KiB]

_________________

--gregspirited

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CEO
Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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27 Nov 2007, 15:26
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?
Attachments

GC56301.gif [ 11.57 KiB | Viewed 1088 times ]

Kudos [?]: 4596 [0], given: 360

Manager
Joined: 07 Oct 2005
Posts: 141

Kudos [?]: 145 [0], given: 0

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27 Nov 2007, 15:37
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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--gregspirited

Kudos [?]: 145 [0], given: 0

27 Nov 2007, 15:37
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