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Geometry Problem!

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Geometry Problem!

by exhilaration » Thu May 28, 2009 10:20 am
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?

A) 169 (sqrt 3) / 3

B) 84.5

C) 75 sqrt 3

D) 169 (sqrt 3) / 4

E) 225 (sqrt 4) / 3

[spoiler]OA: A[/spoiler]

Thanks!
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by dumb.doofus » Thu May 28, 2009 10:28 am
Distance PQ = 13

root(3)a/2 = 12

a = 26/root(3)

area = root(3)a^2/4

area = 169/root(3) or 169root(3)/3
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by sureshbala » Fri May 29, 2009 10:04 pm
Length of PQ = 13.

So sqrt(3)/2 x q = 13 (where a is the side of the equilateral triangle)

Now instead of calculating "a" and then the area do the following calculation which is definitely quickler.


Squaring the above eqn, we get

3/4 x a^2 = 169

But we need sqrt(3)/4 x a^2, which will be 169/sqrt(3)
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