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# PS: Geometry

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Senior Manager
Joined: 02 Feb 2004
Posts: 345
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Kudos [?]: 26 [0], given: 0

PS: Geometry [#permalink]  23 Aug 2005, 06:44
00:00

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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
pls explain
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Senior Manager
Joined: 29 Nov 2004
Posts: 486
Location: Chicago
Followers: 1

Kudos [?]: 10 [0], given: 0

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
_________________

Fear Mediocrity, Respect Ignorance

Last edited by ranga41 on 23 Aug 2005, 10:41, edited 1 time in total.
Manager
Joined: 28 Jun 2005
Posts: 217
Followers: 1

Kudos [?]: 3 [0], given: 0

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
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