Gmat Prep Geometry

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by lion147 » Sun Jul 06, 2008 6:20 pm
onesome wrote:Can you please explain how you are identifying that this triangle is a special triangle = 30-60-90 triangle?
I can see 90 but Im not sure how you conclude 30-60 for other 2.
The first triangle on the left has the X axis as it's base with a length of sroot(3), the height is 1 and there's a right angle in between; this must be a 30-60-90 triangle.

The same triangle is on the right, but rotated; the radius (2) is the longest side, we also have two angles - 30 and 90, so this also must be a 30-60-90 triangle.

Also, you don't need to use the 30-60-90 rule, you can use pythag once you know one angle is 90 degrees, but it would be better to know the shortcut as a time saver.

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by GMATDavid » Mon Jul 07, 2008 6:45 am
Yes, the 30-60-90 rule is very convenient to know as the GMAT test writers like to use it.

All 30-60-90 triangles have sides with the same ratios. x, x-root3 and 2x. So you could have 1, root3, 2 as here, or 3, 3-root3, 6 in a bigger triangle.

Every diagonal line in the coordinate plane is the hyptenuse of a right triangle. Here, using the x and y values (the two sides) you can recognize the 30-60-90 ratio.

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by Ian Stewart » Mon Jul 07, 2008 9:25 am
I could suggest a different approach here. Notice that OP and OQ are perpendicular lines, and are the same length. We can use the simple fact that they are perpendicular to get the answer directly here, with no calculations.

To review the theory (and I'll use concrete numbers to make things easier): if 2/3 is the slope of a line L, we know that -3/2 is the slope of a perpendicular line M. Now, what does the slope '2/3' mean? It means you need to go right 3 units in order that L will rise by 2 units. That's the definition of the slope. On the perpendicular line M, with slope -3/2, you would need to go right 2 units in order for the line to 'rise' -3 units (i.e. fall by 3 units). The vertical and horizontal changes get 'reversed' on the perpendicular line, and one of the changes becomes negative. Thus, if (0,0) is a point on both L and M, then we know that (3, 2) will be a point on L, and (2, -3) will be a point on the perpendicular line M. Notice that each of these points is the same distance from (0,0).

The relationship between perpendicular lines becomes much clearer if you draw a few examples on the co-ordinate plane. Anyway, if you understand this relationship well, the question can be solved almost immediately:

On OP, if we go right by sqrt(3) units, the line falls by 1 unit. On the perpendicular line OQ, if we go right by 1 unit, the line must rise by sqrt(3) units, since OQ is perpendicular to OP. Thus, point Q must be (1, sqrt(3)).
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by Ian Stewart » Mon Jul 07, 2008 9:30 am
Aha, I just followed the link Stuart posted above, and a while ago he suggested the same solution as I just did, with a very clear explanation as well- nice one!