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

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Manager
Joined: 07 Oct 2005
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In the figure, each side of square ABCD has length 1, the [#permalink]  27 Nov 2007, 13:42
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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

CEO
Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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[#permalink]  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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GC56301.gif [ 11.57 KiB | Viewed 728 times ]

Manager
Joined: 07 Oct 2005
Posts: 139
Location: Boston,MA
Followers: 1

Kudos [?]: 33 [0], given: 0

[#permalink]  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.
_________________

--gregspirited

[#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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