Coordinate Geometry Question - Equilateral triangle. - The Beat The GMAT Forum

Coordinate Geometry Question - Equilateral triangle.

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vineeshp: Legendary Member; Posts: 965; Joined: Thu Jan 28, 2010 12:52 amFollowed by: 34 members; Thanked: 156 times; GMAT Score:720

by vineeshp » Thu Mar 31, 2011 8:09 am

If points A and B are on the y-axis in the figure, what is the area of equilateral triangle ABC?
(1) Coordinates of point B are (0, 5âˆš3). (2) Coordinates of point C are (6, 3âˆš3).

Figure attached.

Please answer and explain.

Attachments

Vineesh,
Just telling you what I know and think. I am not the expert.

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Source: Beat The GMAT — Data Sufficiency | Thu Mar 31, 2011 8:09 am

tpr-becky: GMAT Instructor; Posts: 509; Joined: Wed Apr 21, 2010 1:08 pmFollowed by: 85 members; Location: Irvine, CA; Thanked: 199 times; GMAT Score:750

by tpr-becky » Thu Mar 31, 2011 8:31 am

Equilateral triangle on the GMAT is code for 30:60:90 right triangle so that once we get one side we can get the height. Also, we know that in an equilateral triangle all sides are equal.

Therefore all we need it to find one side of the triangle and we can answer this question.

From your figure and statements:

1) - knowing one set of coordinates will not give you a side of the triangle - therefore you can't solve and are left with BCE as answer choices

2) from this we can tell that the height of the triangle is 6 (this distance from the y-axis to C) and because we can get the height we can use the 30:60:90 ratio to find each of the other sides and thus the Base. Therefore this answer is sufficient and the answer is B.

Last edited by tpr-becky on Thu Mar 31, 2011 9:28 am, edited 1 time in total.

Becky
Master GMAT Instructor
The Princeton Review
Irvine, CA

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MAAJ: Master | Next Rank: 500 Posts; Posts: 243; Joined: Sun Jul 12, 2009 7:12 amFollowed by: 2 members; Location: Dominican Republic; Thanked: 31 times; GMAT Score:480

by MAAJ » Thu Mar 31, 2011 9:14 am

Uhmmmm... IMHO it should be (B)

Statement 2 provide us with "X sqrt(3)" which is 6 [the height] and from that we can determine that X = 6/sqrt(3)[the base].

So we have the base and the height of a 30:60:90 triangle, we can determine its Area = b*h/2 -> (6/sqrt(3) * 6)/2 = 18/sqrt(3)

If we multiple the area of the 30:60:90 triangle by 2 then we get the full equilateral triangle 2 * 18/sqrt(3) = 36/sqrt(3)

What's the OA?

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tpr-becky: GMAT Instructor; Posts: 509; Joined: Wed Apr 21, 2010 1:08 pmFollowed by: 85 members; Location: Irvine, CA; Thanked: 199 times; GMAT Score:750

by tpr-becky » Thu Mar 31, 2011 9:27 am

I appologize, you are exactly correct. if you can find the height of at 30:60:90 you can solve for all of the sides of the equilateral. I will edit my response and thanks for pointing out my flaw.

Becky
Master GMAT Instructor
The Princeton Review
Irvine, CA

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vineeshp: Legendary Member; Posts: 965; Joined: Thu Jan 28, 2010 12:52 amFollowed by: 34 members; Thanked: 156 times; GMAT Score:720

by vineeshp » Thu Mar 31, 2011 6:02 pm

Thank you both of you..

Vineesh,
Just telling you what I know and think. I am not the expert.

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