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Area of an equilateral triangle

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dreamz008
 Post subject: Area of an equilateral triangle
 Post Posted: Thu Mar 26, 2009 3:58 pm 
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Posts: 2
Hi,

In the strategy guide, it gives the area formula for an equilateral triangle as follows: 1/2(s) * (s{rad 3}/2)

If the height of a 30-60-90 triangle is rad 3, wouldn't that remain true for an equilateral triangle? If this is true, why do you divide the height (radical 3) by 2 when solving for the area? The 1/2 in front of the first (s) should apply to all that follows so it's confusing why you then divide by two again.
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esledge
 Post subject: Re: Area of an equilateral triangle
 Post Posted: Tue Mar 31, 2009 1:25 pm 
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ManhattanGMAT Staff


Posts: 901
Location: St. Louis, MO
Keep in mind that when you split an equilateral triangle into two 30-60-90 triangles, you halve the length of the base side.

For example, if we vertically split an equilateral triangle with sides of length s, we get 30-60-90 triangles of the following dimensions:

short leg = s/2
long leg = (to be determined)
hypotenuse = s

If the ratio of the sides of the 30-60-90 triangle are 1x : root(3)x : 2x, x must be (s/2), so the long leg is root(3)*(s/2).

Area of the original equilateral triangle is thus (1/2)bh = (1/2)(s)(long leg) = (1/2)(s)(root(3)*s/2)=(s^2)*root(3)/4.

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Emily Sledge
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ManhattanGMAT
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