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# Good One from GMATPrep

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SVP
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Good One from GMATPrep [#permalink]  22 Jan 2006, 23:24
Solve this
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Manager
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Since OP and OQ are radius:
s^2+t^2 = 4...(1)
From pythagoras, (t-1)^2+(s+root(3))^2 = 8 .... (2)

Solving 1 and 2, s=1, t=root(3) or s=-1, t=-root(3). Since s,t are in the first quadrant, s=1, t=root(3).

Last edited by watever on 25 Jan 2006, 02:59, edited 1 time in total.
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Re: Good One from GMATPrep [#permalink]  22 Jan 2006, 23:52
Bhai wrote:
Solve this

PA=1 OA=sqrt3 so OP=2 10x Pifagorus(not sure if this right spelling).Hence OP=OQ=2 cause they are both radius of this circle PQ=sqrt(2^2+2^2)=sqrt8.So the volume of S would be -sqrt3+sqrt8[/img]
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Manager
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Another simple way of getting one equation is to compare the slopes of OP and OQ. They should be negative reciprocals.

t/s*-1/root(3) = -1 which gives t=s*root(3). This can be used together with s^2+t^2=4 to give s=1, t=root(3).
CEO
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Length OP = SQRT(3+1) = 2
This is a semicircle so OP = OQ = 2

Angle of OP is 60 with -ve side of x axis.

So angle of OQ with +ve x axis = 180 - 90-60 = 30

So Cos(30) = SQRT(3)/2 = s/2 so

s = SQRT(3)
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watever wrote:
Another simple way of getting one equation is to compare the slopes of OP and OQ. They should be negative reciprocals.

t/s*-1/root(3) = -1 which gives t=s*root(3). This can be used together with s^2+t^2=4 to give s=1, t=root(3).

Perfecto watever. s = 1 is the answer
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S=1.

Using distance formula for distance between (0,0) and ( -sqrt(3),1) gives radius = 2.

Also, using pythagoras theorem, we get sqrt(3)s = t. Since radius = 2, distance between (s,t) and (0,0) is s^2 + t^2 = 4. Substituting, t = sqrt(3)s, we get s = 1.
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ps_dahiya wrote:
Length OP = SQRT(3+1) = 2
This is a semicircle so OP = OQ = 2

Angle of OP is 60 with -ve side of x axis.

So angle of OQ with +ve x axis = 180 - 90-60 = 30

So Cos(30) = SQRT(3)/2 = s/2 so

s = SQRT(3)

I did the wrong calculations. Cos(60) is 1/2 and Cos(30) is SQRT(3)/2. But I took totally opposite
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Quote:
Perfecto watever. s = 1 is the answer

Thanks for the encouragement Bhai. You have some relation to Mumbai?
SVP
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watever wrote:
Quote:
Perfecto watever. s = 1 is the answer

Thanks for the encouragement Bhai. You have some relation to Mumbai?

Well who is not related to Mumbai these days? But not with any Bhai. My bhaigiri is very decent.
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yups got 1 but after 5 min
Manager
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Bhai wrote:
watever wrote:
Quote:
Perfecto watever. s = 1 is the answer

Thanks for the encouragement Bhai. You have some relation to Mumbai?

Well who is not related to Mumbai these days? But not with any Bhai. My bhaigiri is very decent.

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watever wrote:
Since OP and OQ are radius:
s^2+t^2 = 4...(1)
From pythagoras, (t-1)^2+(s+root(3))^2 = 8 .... (2)

Solving 1 and 2, s=1, t=root(3) or s=-1, t=-root(3). Since s,t are in the first quadrant, s=-1, t=root(3).

watever: you meant s=1, right? (as opposed to '-1')

also you don't show the substitution of 1 and 2, did you solve for t in equation 1, t=sqrt(4-s^2) into the second equation? I did this and got the same answer but I was wondering if you omitted the work because there was an intuitive solution that you instantly knew.
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Quote:
watever: you meant s=1, right? (as opposed to '-1')

also you don't show the substitution of 1 and 2, did you solve for t in equation 1, t=sqrt(4-s^2) into the second equation? I did this and got the same answer but I was wondering if you omitted the work because there was an intuitive solution that you instantly knew.

Yeah, I meant s=1, I edited the post. Also, I did omit the working for sake of brevity. I however advise my second method be used which used the slopes as one of the equation.

So 1st eqn is s^2 + t^2 = 4 and the second is t/s*-1/root(3) = -1 which gives t=s*root(3). So substition and calculations become highly easy as compared to the first method.
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