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OG Quant: Geometry

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Director
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OG Quant: Geometry [#permalink] New post 20 May 2008, 16:19
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If arc PQR above is a semicircle, what is the length of diameter PR?
(1) a = 4
(2) b = 1
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Re: OG Quant: Geometry [#permalink] New post 20 May 2008, 16:49
D.

First thing I'd like to say is DS with geometry it is rarely C or E, and in this case C should raise some red flags because it is just to damn easy (aka Trap.)

I'm not great with geometry but I'd say with similar triangles here we know they are both right triangles and we know that we have another right triangle at Q because any time you draw a triangle with the diameter as one of the sides you know it will form a right triangle.

So all we need to know is one of the sides A or B.
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Re: OG Quant: Geometry [#permalink] New post 20 May 2008, 16:53
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D

Angle PQR is a 90 degrees angle (by definition, I did not just make that up)

Let PQ = C
let QR = D

a^2 + 2^2 = C^2

20 = C^2


D^2 = 2^2 + b^2

((a+b)^2) = C^2 + D^2 -----> ((a+b)^2) = 20 + 2^2 + b^2

a^2 + 2ab + b^2 = 20 + 2^2 + b^2

16 + 8b + b^2 = 24 + b^2

8b = 8

b = 1
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Re: OG Quant: Geometry [#permalink] New post 20 May 2008, 17:53
Can't you also prove A is suff with Similar Triangle Theory??? Is this accurate?
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Re: OG Quant: Geometry [#permalink] New post 21 May 2008, 00:29
jimmyjamesdonkey wrote:
Can't you also prove A is suff with Similar Triangle Theory??? Is this accurate?


why not? a/2 =2/b -->ab=4, so D
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Director
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Re: OG Quant: Geometry [#permalink] New post 21 May 2008, 04:29
AB should be 5, not 4. But I'm saying you can solve by similar triangles, which would be a lot faster than the formula answer above.
Re: OG Quant: Geometry   [#permalink] 21 May 2008, 04:29
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