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PS: Geometry

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Senior Manager
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PS: Geometry [#permalink] New post 23 Aug 2005, 06:44
00:00
A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
pls explain
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Senior Manager
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 [#permalink] New post 23 Aug 2005, 08:27
By SIN relation BO = 1/(sqrt 2)

By TAN relation A0 = 0C = 1/(sqrt 2) so AC = 2/(sqrt 2)

area of triangle = 1/2*AC*BO = 1/2
So area of ABO = 1/4
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Last edited by ranga41 on 23 Aug 2005, 10:41, edited 1 time in total.
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 [#permalink] New post 23 Aug 2005, 08:58
the triangle is a right isosceles triangle with base and height = 1

Area of ABO = half of area of ABC = 1/2*1/2*1*1=1/4
  [#permalink] 23 Aug 2005, 08:58
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