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Re: GMATPREP- semicircle [#permalink]
13 Nov 2008, 11:06

1

This post received KUDOS

prasun84 wrote:

good one really... at 1st sight, s looks as sqrt3

radius=2 now, distance between P and Q=2sqrt2

now (s+sqrt3)^2 + (t-1)^2 =8 to solve this equation i need sqrt3 value for T, hence t=sqrt3

Hence s=1. (if u look at the equation closely, it'll be clear....)

it s an interesting method we will have to rely on hit and trial to get the answer My approach

Starting from the -ve X axix the order of angles formed will be 30,60,30,60 hence applying tan60 in the triangle created in the first quadrant will have sides hypotenuse= 2 base=1 (which is s) perpendicular =root 3(which is t)

Re: GMATPREP- semicircle [#permalink]
13 Nov 2008, 20:42

Thats a very intelligent ans. but shldnt slope of OP be -1/sqrt(3)? (slope=difference in y/difference in x)

alpha_plus_gamma wrote:

calculate radius of the circle radius^2= (sqrt(3))^2 + 1^2 radius = 2 (cant be -ve)

the equation of line passing through point P is,

y = -sqrt(3)x [ slope = (-sqrt(3) - 0) / (1-0) and y intercept of the line is 0) The line passing through point Q is perpendicular to the above line hence its equation is y = sqrt(3)x i.r t = sqrt(3)*s

Now OQ being radius can be expressed as OQ^2 = s^2 + t^2 2^2 = s^2 + (sqrt(3)*s)^2 4 = S^2 * 4 S^2 = 1

Re: GMATPREP- semicircle [#permalink]
13 Nov 2008, 20:48

Wow ! wht a shortcut. Applying angle formula never occured to me. +1 from me

hibloom wrote:

prasun84 wrote:

good one really... at 1st sight, s looks as sqrt3

radius=2 now, distance between P and Q=2sqrt2

now (s+sqrt3)^2 + (t-1)^2 =8 to solve this equation i need sqrt3 value for T, hence t=sqrt3

Hence s=1. (if u look at the equation closely, it'll be clear....)

it s an interesting method we will have to rely on hit and trial to get the answer My approach

Starting from the -ve X axix the order of angles formed will be 30,60,30,60 hence applying tan60 in the triangle created in the first quadrant will have sides hypotenuse= 2 base=1 (which is s) perpendicular =root 3(which is t)

Re: GMATPREP- semicircle [#permalink]
14 Nov 2008, 11:28

There's a much faster way to do this problem. To find the coordinates of q, all you have to do is switch the x- and y- values of the coordinates of P, then multiply the new y-value (t) by -1.

To understand why, imagine the full circle. Draw any two perpendicular lines through the center. They will intersect the circle at (a, b), (b, -a), (-a, -b), and (-b, a).