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# In the figure above, line segment AD is the diameter of circle O, line

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Re: In the figure above, line segment AD is the diameter of circle O, line [#permalink]
Bunuel chetan2u ScottTargetTestPrep -
Can you let me know where I'm going wrong?

BC = 4
BE = EC = 2+x = say Y
So BC^2 = Y^2+Y^2
16 = 2Y^2
8 = Y^2
Y = 2Root2
since 2+x= 2Root2
x = 2root2-2
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Re: In the figure above, line segment AD is the diameter of circle O, line [#permalink]
TarPhi wrote:
Bunuel chetan2u ScottTargetTestPrep -
Can you let me know where I'm going wrong?

BC = 4
BE = EC = 2+x = say Y
So BC^2 = Y^2+Y^2
16 = 2Y^2
8 = Y^2
Y = 2Root2
since 2+x= 2Root2
x = 2root2-2

Response:

Your solution assumes that BC^2 should equal BE^2 + EC^2, which is not necessarily true. That is the Pythagorean theorem, and it holds only for right triangles. If the angle BEC were a right angle, your approach would be correct; however, we have no such information.
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Re: In the figure above, line segment AD is the diameter of circle O, line [#permalink]
D it is..

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Re: In the figure above, line segment AD is the diameter of circle O, line [#permalink]
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