The Quest for 700: Weekly GMAT Challenge (Answer)
November 26th, 2009
Yesterday, Manhattan GMAT posted a 700 level GMAT question on our blog. Today, they have followed up with the answer:
If polygon ABCDEF is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments “from symmetry” – that is, recognizing that many parts of the diagram are equivalent.
We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to 180(n – 2)° = 180(6 – 2)° = 720°, and there are 6 interior, equal angles, then each of those angles must measure 720°/6 = 120°.
Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle’s center O:
Consider small triangle AXB. The hypotenuse of this right triangle, AB, has length r. Angle ABO is 60°, so AXB is a 30-60-90 triangle. This means that XB is r/2 in length, and AX is 
in length. Since AX and XB are the same length (by symmetry), AC is 
in length.
This means that the area of triangle ABC is 
.Since there are three shaded triangles in all (including ABC), the total shaded area is 
.The area of the circle is 
, so the ratio of the shaded area to the area of the circle is given by 
.
The closest answer is 42%.
The correct answer is D.
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Jason Says:
October 31st, 2011 at 2:09 pm
My way is much longer!
Determine area of circle by finding its diameter or radius first. How do we determine the length EB, FC, AD? A hexagon has 6 sides, so the sum of the angles is (6-2) x 180 = 720. Therefore, each angle of the hexagon is 120* degrees. Because all sides of the hexagon are equal, each triangle must be an isoscoles triangle (30 + 30 + 120* = 180). Split each triangle by bisecting the 120 degree angle to form 2x 30,60,90 triangles. Because these shapes can be any size in relative proportions, invent a number for one of the sides and compute from there. Let’s choose “4″ for AC. Therfore, “2″ is the height of one of the 30,60,90 triangles. The base is therefore 2/?3 and its hypotenous 4/?3. “2/?3″ is important for determining the diameter.
Now, what is the height of the big center triangle? The triangle must be an equalateral triangle if all its sides are the same (60,60,60). We must split this triangle the same as before to determine its height. It becomes another 30,60,90 triangle. If AC = 4, then AG & GC = 2, GE = 2?3, and obviously AE=4.
Now we have all the information necessary to determine the diameter of the circle: the base of the small triangle plus the height of the big triangle (2/?3 + 2?3 = 2/?3 + 6/?3 = 8/?3). The radius is therefore 4/?3, the area (4/?3)^2 = 16/3 x ? = 16?/3.
Next step is to determine the percentage of the area that is shaded by finding total area of the isocoles triangles and making it the numerator (Small Triangle Area / [16?/3]). The area of the small triangles is easy to find. Use our old information regarding the 30,60,90 small triangles and compute 0.5BH = 0.5(2/?3)(2) = 2/?3; therefore, the area of each isocoles triangles is 2 x 2/?3 = 4/?3 and the sum area of all small triangles is 12/?3.
Now we can solve. (12/?3) / (16?/3) = 36 / 16??3 = 16 x (~3.1 x ~1.7 = ~5.3) = ~91… So, 36 / 90 = (36 x 10/9) / 100 = 360/9 /100 = 40/100 = 40%. Answer choice D is closest.