Practice GMAT Data Sufficiency Question
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If angle ABC is 30 degrees, what is the area of triangle BCE?
 Angle CDF is 120 degrees, lines L and M are parallel, and AC = 6, BC = 12, and EC = 2AC
 Angle DCG is 60 degrees, angle CDG is 30 degrees, angle FDG = 90, and GC = 6, CD = 12 and EC = 12
Correct Answer: D
 Even though lines L and M look parallel and angle BAC looks like a right angle, you cannot make these assumptions.
 The formula for the area of a triangle is .5bh

Evaluate Statement (1) alone.
 Since EC = 2AC, EA = CA, EC = 2(6) = 12 and line AB is an angle bisector of angle EBC. This means that angle ABC = angle ABE. Since we know that angle ABC = 30, we know that angle ABE = 30. Further, since lines L and M are parallel, we know that line AB is perpendicular to line EC, meaning angle BAC is 90.
 Since all the interior angles of a triangle must sum to 180:
angle ABC + angle BCA + angle BAC = 180
30 + angle BCA + 90 = 180
angle BCA = 60  Since all the interior angles of a triangle must sum to 180:
angle BCA + angle ABC + angle ABE + angle AEB = 180
60 + 30 + 30 + angle AEB = 180
angle AEB = 60  This means that triangle BCA is an equilateral triangle.
 To find the area of triangle BCE, we need the base (= 12 from above) and the height, i.e., line AB. Since we know BC and AC and triangle ABC is a right triangle, we can use the Pythagorean theorem on triangle ABC to find the length of AB.
6^{2} + (AB)^{2} = 12^{2}
AB^{2} = 144  36 = 108
AB = 108^{1/2}  Area = .5bh
Area = .5(12)(108^{1/2}) = 6*108^{1/2}  Statement (1) is SUFFICIENT

Evaluate Statement (2) alone.
 The sum of the interior angles of any triangle must be 180 degrees.
DCG + GDC + CGD = 180
60 + 30 + CGD = 180
CGD = 90
Triangle CGD is a right triangle.  Using the Pythagorean theorem, DG = 108^{1/2}
(CG)^{2} + (DG)^{2} = (CD)^{2}
6^{2} + (DG)^{2} = 12^{2}
DG = 108^{1/2}  At this point, it may be tempting to use DG = 108^{1/2} as the height of the triangle BCE, assuming that lines AB and DG are parallel and therefore AB = 108^{1/2} is the height of triangle BCE. However, we must show two things before we can use AB = 108^{1/2} as the height of triangle BCE: (1) lines L and M are parallel and (2) AB is the height of triangle BCE (i.e., angle BAC is 90 degrees).
 Lines L and M must be parallel since angles FDG and CGD are equal and these two angles are alternate interior angles formed by cutting two lines with a transversal. If two alternate interior angles are equal, we know that the two lines that form the angles (lines L and M) when cut by a transversal (line DG) must be parallel.
 Since lines L and M are parallel, DG = the height of triangle BCE = 108^{1/2}. Note that it is not essential to know whether AB is the height of triangle BCE. It is sufficient to know that the height is 108^{1/2}. To reiterate, we know that the height is 8 since the height of BCE is parallel to line DG, which is 108^{1/2}.
 Since we know both the height (108^{1/2}) and the base (CE = 12) of triangle BCE, we know that the area is: .5*12*108^{1/2} = 6*108^{1/2}
 Statement (2) alone is SUFFICIENT.
 The sum of the interior angles of any triangle must be 180 degrees.
 Since Statement (1) alone is SUFFICIENT and Statement (2) alone is SUFFICIENT, answer D is correct.
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