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by GmatMathPro » Mon Dec 19, 2011 3:28 pm
Overall strategy: Find the area of the big triangle and subtract the area of the right triangle to find the area of BCDE.

All of the sides of the big triangle are the same length, so it's an equilateral triangle with all three angles equal to 60 degrees. To find the area, we can use the equilateral triangle area formula A=s^2√3/4 to get A=9√3/4. Or, we can drop an altitude to one of the sides and use 30-60-90 triangle ratios to find the height.

If angle A=60 and angle ABE=90, then triangle ABE is a 30-60-90 right triangle. Using those ratios along with AB=1, we can deduce that BE=√3. Thus, the area of triangle ABE is √3/2

Subtracting the area of the small triangle from the area of the big triangle: 9√3/4 - √3/2 = 7√3/4

Ans: B
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by neelgandham » Mon Dec 19, 2011 3:32 pm
From the question, we know that triangle ACD is an equilateral triangle. So, angle CAD = angle BAE = 60 degrees. From right angled triangle ABE, angle BAE = 60 degrees, angle ABE = 90 degrees, so angle BEA = 30 degrees. BA = 1 given, So BE = √3( Sin 60 = BE/BA = √3).

Area of region BCDE = Area of equilateral triangle ACD - Area of right angled triangle ABE
Area of region BCDE = (√3/4)*a*a - 1/2 * base*height, where a = side length of the equilateral triangle and base, height are base and height of the right angled triangle respectively.
Area of region BCDE = (√3/4)*3*3 - 1/2 *√3*1
Area of region BCDE = (7/4)√3

IMO B
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by karthikpandian19 » Wed Dec 28, 2011 9:55 pm
OA is B
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