500 PS

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500 PS

by dunkin77 » Mon Apr 02, 2007 12:12 pm
Hi,

My answer to the attached question was (unfortunately) 3+ root3/2....
The correct answer is B)

Could anyone help?
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by 800GMAT » Mon Apr 02, 2007 1:47 pm
A(BCDE) = A(ACD) - A(ABE)

A(ACD) = [(SQRT3)/4] * (SIDE)^2
= [(SQRT3)/4] * 9

ACD is equilateral; every angle is 60.
That makes ABE 30-60-90 triangle

AB = side opposite to 30 degrees = 1
we need BE to find the area
BE = side opposite to 60 = (SQRT3)*1

A(ABE) = HALF * AB * BE
= HALF * 1 * [(SQRT3)*]

After solving A(ACD) - A(ABE), you get 'B'