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Equilateral Triangle

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by neelgandham » Wed Nov 02, 2011 7:53 am
h = ((square root(3))/2)*(S-R)

(((square root(3))/2)*s) -(h))/(r/2) = h/((s/2)-(r/2)) Similar triangles
Last edited by neelgandham on Wed Nov 02, 2011 8:01 am, edited 1 time in total.
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by frumpychic » Wed Nov 02, 2011 7:58 am
Could you show the steps?
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by satishchandra » Wed Nov 02, 2011 8:01 am
I think the data is not fully complete.
Is it a rectangle in the middle of triangle?
If not What about the angles inside the triangle?

Without these, I think answer can not be calculated in my opinion.

@above person
Can you explain in detail how you arrive at that?
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by neelgandham » Wed Nov 02, 2011 8:08 am
frumpychic wrote:Could you show the steps?
From the diagram in the attachment

AB/BC = CD/DE (Similar Triangles)

Replacing the values of AB,BC,CD,DE you get the answer mentioned above.
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by GmatMathPro » Wed Nov 02, 2011 8:14 am
I agree with satishchandra. We don't know for sure that the figure inside is a rectangle or that its vertices lie on the triangle. However, if we DO assume that it's a rectangle with all 4 vertices on the triangle....

We have an equilateral triangle with side length s. So every side has length s. The triangle formed on top is an equilateral triangle of length r, so every side has length r. The right side of the big triangle is divided into lengths r and s-r by the vertex of the rectangle. s-r is the hypotenuse of a 30-60-90 triangle with length h across from the 60 degree angle. We can use 30-60-90 triangle ratios to set up the following equation:

2h/√3=s-r

h=√3(s-r)/2
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by satishchandra » Wed Nov 02, 2011 8:18 am
neelgandham wrote:
From the diagram in the attachment

AB/BC = CD/DE (Similar Triangles)
I think you have made a few assumptions. The diagram did not mention anything about the angles. They are not necessarily rightangles
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by neelgandham » Wed Nov 02, 2011 8:18 am
I completely agree with GmatMathPro and Satish. The solution provided above is on an assumption that the figure enclosed in the triangle is a rectangle that has all the four vertices on the triangle.
If the same question is posed in GMAT(which will not happen anyway), I would assume the same as above and solve the problem !
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by frumpychic » Wed Nov 02, 2011 8:38 am
the rectangle is enclosed in the triangle. I couldnt figure out if the side of the triangle was "s" or if the width of the rectangle on the base was "s". If the width of the rectangle was s, then that would mean r and s would be equal. I saw this question, so my memory isn't exact.
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