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Points and Hieght

Expert replies
by GMATPR » Tue Apr 22, 2008 7:03 am
Please explain this for me. Thanks


The (x, y) coordinates of points P and Q are (-2, 9) and (-7, -3), respectively. The height of equilateral triangle XYZ is the same as the length of line segment PQ. What is the area of triangle XYZ?
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Source: — Problem Solving |

by AleksandrM » Tue Apr 22, 2008 9:57 am
I've only been able to arrive at the height of the triangle, which is 13.

This triangle will obviously have angles 60:60:60, and all three sides will be equal... hence, equilateral.

You can split the triangle into two right angle triangles with base 13, but we will then need one of the other two heights. The triangle will be 90:60:30 with sides x:xroot3:2x.

The answer might be 169sqroot3.

Hope someone can take this further.
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by Stuart@KaplanGMAT » Tue Apr 22, 2008 10:50 am
AleksandrM wrote:I've only been able to arrive at the height of the triangle, which is 13.

This triangle will obviously have angles 60:60:60, and all three sides will be equal... hence, equilateral.

You can split the triangle into two right angle triangles with base 13, but we will then need one of the other two heights. The triangle will be 90:60:30 with sides x:xroot3:2x.

Hope someone can take this further.
You're almost there!

The height of an equliateral triangle is the xroot3 side of the 30/60/90 triangle.

To calcuate the area of the triangle, we also need the base - so, we need to solve for x.

xroot3 = 13
x = 13/root3

The base of the equilateral triangle is actually the 2x side (since every side is the same), so the base is 2*(13/root3) or 26/root3

Area = 1/2 * base * height
Area = 1/2 (26/root3) (13)
Area = 169/root3

If that appears among the answers, we're done. However, most GMAT answer choices "rationalize the denominator", which basically means getting rid of the root. To do so, we multiply the top and bottom of the fraction by the root itself.

169/root3 = 169(root3)/root3(root3) = 169root3/3

In all likelihood, that's the actual correct answer.
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by GMATPR » Tue Apr 22, 2008 12:17 pm
Thanks AleksandrM and Stuart, I understand this now.
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