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GMAT Prep: Geometry (i.e. Triangle and Square)

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by dnkcdnguy » Mon Feb 16, 2009 2:07 pm
Can someone please help out with this question. It took me awhile to figure out the second time around and I want to see how other people approach this question.

Thanks
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by tini » Mon Feb 16, 2009 2:57 pm
area of an equilateral triangle =( 3^1/2)/4 t*t
area of the rectangle = s*s
t*t/s*s=4/((3)^1/2)
t/s=2/(3)^1/4)
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by Stuart@KaplanGMAT » Mon Feb 16, 2009 3:09 pm
tini wrote:area of an equilateral triangle =( 3^1/2)/4 t*t
area of the rectangle = s*s
t*t/s*s=4/((3)^1/2)
t/s=2/(3)^1/4)
Correct! In a bit more detail:

Area of the square is s^2.

Area of a triangle is (1/2)base*height.

Our triangle has a base of t and we find the height using the 30/60/90 triangle ratio of x:x(root3):2x.

Since our triangle has a hypotenuse of t instead of 2t, it's one-half the size of a x/xroot3/2x triangle. Therefore, instead of a height of xroot3, it has a height of (t*root3/2).

So, our area is:

(1/2)(t)(t*root3/2)

Simplifying, we get:

(t^2)(root3)/4

Since the two areas are equal, we know that:

(t^2)(root3)/4 = s^2

(t^2)(root3) = 4*s^2

t^2 = (4*s^2)/root3

t^2/s^2 = 4/root3

Taking the root of both sides

t/s = 2/root(root3)

t/s = 2/(4throot)3
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by dnkcdnguy » Mon Feb 16, 2009 6:15 pm
Thanks Tini and Stuart. I rushed this question and understand it now that we're able to use special triangles.
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