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GMAT Prep 1 #3_PS Triangles inscribed #10

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by kwah » Mon Mar 19, 2012 11:32 am
I have attached a question from GMAT Prep Test 1.

What is the most efficient way to achieve this result?

Answer:B

Please advise, thank you.
K
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by LalaB » Mon Mar 19, 2012 12:11 pm
since these two lines are perpendicular, the multiple of their slopes is equal to (-1)
(1-0)/(-sqrt3-0)*(t-0)/(s-0)=-1
1/(sqrt3)=s/t

s=1

p.s. please, before posting a question, use the search engine. almost all of these questions were discussed trillions times

and also I kindly ask u not to attach a file, but post a question,using copy-paste :)
it annoys to download these attachments
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by kwah » Mon Mar 19, 2012 6:00 pm
I apologize for the inconvenience LalaB.

I tried to copy and paste the questions off GMAT Prep however, the program will not allow me to do so.

Thanks,
K
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by Anurag@Gurome » Mon Mar 19, 2012 7:20 pm
kwah wrote:I have attached a question from GMAT Prep Test 1.

What is the most efficient way to achieve this result?

Answer:B

Please advise, thank you.
K

Point P and Q lies on the same circle with center at (0, 0).
Thus, (s² + t²) = (-√3)² + 1² = 3 + 1 = 4

Again line segments OP and OQ are perpendicular.
Thus (slope of OP)*(slope of OQ) = -1

Slope of OP = 1/(-√3) = -(1/√3)
=> Slope of OQ = (t - 0)/(s - 0) = t/s = (-1)/(-1/√3) = √3
=> t = √3s

Thus, (s² + (√3s)²) = 4
=> (s² + 3s²) = 4
=> s² = 1
=> s = ±1

As point Q lies in the first quadrant s = 1.
The correct answer is B.
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by [email protected] » Wed Mar 21, 2012 11:11 pm
Using the slope formula makes the question more simpler and takes less time...

You could also have done is used the hypotenuse formula and equaled the two radii...

That takes a lot of time...
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