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Re: GRE Quant 1 minute challenge [#permalink]
1
The probability is directly related to how much of the square's area is occupied by the triangle.
This is a longer approach, but I'm trying to PROVE why the answer is 1/2


Let each side of square have length x
So, area of square = x^2

Area of triangle = (1/2)(base)(height)
= (1/2)(x)(x)
= (1/2)x^2

We can see that the triangle takes 1/2 of the area of the square.
So, P(point is INSIDE triangle) = 1/2
AND, P(point is OUTSIDE triangle) = 1/2

Cheers,
Brent - Greenlight Test Prep
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Re: GRE Quant 1 minute challenge [#permalink]
Thanks
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Re: GRE Quant 1 minute challenge [#permalink]
1/2
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Re: GRE Quant 1 minute challenge [#permalink]
Why isn't the answer E? We don't know if E is the midpoint of CD.
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Re: GRE Quant 1 minute challenge [#permalink]
2
Nishan wrote:
Why isn't the answer E? We don't know if E is the midpoint of CD.


E need not be the midpoint of CD.
Regardless of where point E is located on side CD, the base of the triangle will be x (if we let x = length of each side) and the height of the triangle will still be x.

So, regardless of where point E is located on side CD, the area of the triangle will always = (base)(height)/2 = (x)(x)/2 = x²/2

Cheers,
Brent
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Re: GRE Quant 1 minute challenge [#permalink]
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