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If arc PQR is a semicircle, what is the diameter PR? 1) a=4

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Director
Joined: 17 Sep 2005
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If arc PQR is a semicircle, what is the diameter PR? 1) a=4 [#permalink]  16 Jun 2006, 08:39
If arc PQR is a semicircle, what is the diameter PR?
1) a=4
2) b=1

Regards,
Brajesh
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GMAT Club Legend
Joined: 07 Jul 2004
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Location: Singapore
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Angle PQR = 90 degrees

St1:
Let b be x, can use pythagora's therem to solve for x and hence hte dimeter.

St2:
Let a be x, and using the same method to solve for x.
Manager
Joined: 15 Jun 2006
Posts: 70
Location: Seoul
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Re: DC - Semicircle [#permalink]  16 Jun 2006, 09:23
because if either part of the triangle (left or right) is set,
the remaining part will be determined.

However, I feel hard to prove it.
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You go, we go

Intern
Joined: 01 Jun 2006
Posts: 22
Location: Toronto
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hm.....until you guys gave me the answer, I didn't know what to do.
But I think I know how to prove.
Let's make the point, where line a and b meet, W.
If you look carefully, triangle PQW is similiar to triangle PQR.
Then, you can use the ratio to solve this problem by using statement 1 and statement 2 seperately (Triangle PQR is also similar to triangle QWR)
I used the pythagora's theory to get line PQ, not for the entire problem.
But...when can I solve this kind of problem without getting an answer? Bad...
Director
Joined: 17 Sep 2005
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fankor wrote:
hm.....until you guys gave me the answer, I didn't know what to do.

This was nice fankor

fankor wrote:

But...when can I solve this kind of problem without getting an answer? Bad...

This is again

Well I can only say Practice, Practice and more Practice.

Regards,
Brajesh
CEO
Joined: 20 Nov 2005
Posts: 2910
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 19

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This is D.

Lets call point which is between P and R on the diameter as X.

Angle PQR = 90, QX = 2

St1: a = 4 then PQX can be found. Then we can find XQR and in turn can find b. SUFF

St2: b = 1 then RQX can be found. Then we can find XQP and in turn can find a. SUFF
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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