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# Geometry

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Manager
Joined: 21 Feb 2009
Posts: 99

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26 Mar 2009, 16:17
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DS in Geometry

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Manager
Joined: 05 Jan 2009
Posts: 73

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26 Mar 2009, 18:19
ANS:D
recently discussed in forum.
Manager
Joined: 21 Feb 2009
Posts: 99

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27 Mar 2009, 14:47
Can you please post the link which has the explanation,Probably last time i missed it.
Senior Manager
Joined: 30 Nov 2008
Posts: 479
Schools: Fuqua

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27 Mar 2009, 15:42
IMO D.

I tired solving this problem using one of the basic rules of the triable.

Exterior angle of a triable = sum of the opposite interior angles.

It is given AB = OC.

Following can also be derived. OB = AB.
Let <BAO = <BOA = x (Because of the Isoceles triable proerty)
Let <BCO = <CBO = y. (Becuase of the isoceles triangle property)

From stmt1 : <COD = 60 Then x + y = 60 (By the rule, stated above)
Again <BAO + <BOA = <CBO ==> y + y = x ==> 2y = x(By the rule stated above)

Solving these two, we can get the value of y. Hence sufficient.

From Stmt 2: Given <BCO = 40. ==> < CBO = 40.

Again, <BAO + <BOA = <CBO ==> y + y = x ==> 2y = x(By the rule stated above)
Here we know x = 40.

Solving it, we can get the value of Y. Hence sufficient.

Other solutions can be found in the following link.

gmatprep-2-triangle-semicircle-76801.html
Manager
Joined: 28 Jul 2004
Posts: 135
Location: Melbourne
Schools: Yale SOM, Tuck, Ross, IESE, HEC, Johnson, Booth

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27 Mar 2009, 15:43
nitindas wrote:
Can you please post the link which has the explanation,Probably last time i missed it.

Don't know the post but, here is the solution:

from the stem , AB = OC = OB (Since OC and OB are radii of the same circle) --- (a)

from (1) , we have 4 2 <BAO = 20. Hence (2) is sufficient.

I have not explained here but if you understood (1) above, I think this is easy to stand.

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kris

Re: Geometry &nbs [#permalink] 27 Mar 2009, 15:43
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# Geometry

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