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semicircle

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VP
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semicircle  [#permalink]

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New post 29 Jan 2009, 04:26
Pls explain ur answers

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Re: semicircle  [#permalink]

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New post 29 Jan 2009, 08:48
Drop a line from P to x axis and name it m.
Also Drop a line from Q to x axis and name it n.

using 30:60:90 ratio in triangle pom:

angle pom = 30,
angle pmo = 90,
angle mpo = 60,

similarly in triangle qon,
angle qon = 60,
angle qno = 90,
angle oqn = 30,

since po = qo = 2

again using 30:60:90 ratio in triangle, on = s = 1.

on = s is opposit to angle oqn = 30.
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Re: semicircle  [#permalink]

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New post 29 Jan 2009, 09:05
Here is my approach.

OP ^2= (-sqrt(3))^2 + 1^2
-- OP =2 and OQ=2

SLOPE OF OP = Y2-Y1/X2-X1 = (1-0)/(-SQRT(3) -1) = -1/SQRT(3)
SLOPE OF OQ = SQRT(3)
becaue OP and OQ are perpendicular.

t-0/s-0 = sqrt(3) ---> t=s *sqrt(3)
s*s + t*t = 2*2
s*s (3+1)=4 --> s=1
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Re: semicircle  [#permalink]

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New post 29 Jan 2009, 15:26
I agree with S = 1. Here is my approach.

OP and OQ are the radii of the circle. From OP, we can find the radius = 2 (aplying distance between two points formula)
OP = OQ ie applying distance forumal between OQ will give s^2 + t^2 = 4 ---eq 1.

In triangle POQ, <O = 90 and two sides are equal. By 45:45:90 rules, sides will be 1;1:sqrt 2. So the distance between PQ = 2 sqrt 2.

Now apply distance forumal for PQ equating it to 2 (sqrt 2) we get sqrt(3)*s = t. ----Eq 2

Substituting eq 2 in eq 1, we get s = 1.

I doubt this problem can be solved with in 2 min when we look at the problem for the firs time. I almost took 6 min and trying since I know I can solve but definitely not with in 2 min.

Good to know some other alternate approaches as well. :)
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Re: semicircle  [#permalink]

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New post 29 Jan 2009, 20:12
GMAT TIGER wrote:
Drop a line from P to x axis and name it m.
Also Drop a line from Q to x axis and name it n.

using 30:60:90 ratio in triangle pom:

angle pom = 30,
angle pmo = 90,
angle mpo = 60,

similarly in triangle qon,
angle qon = 60,
angle qno = 90,
angle oqn = 30,

since po = qo = 2

again using 30:60:90 ratio in triangle, on = s = 1.

on = s is opposit to angle oqn = 30.



Nice answer!

Just a quick question... how do you know that it is a 30-60-90 triangle?

--== Message from GMAT Club Team ==--

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If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: semicircle &nbs [#permalink] 29 Jan 2009, 20:12
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