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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]

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.

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.

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.