Blog

# The Quest for 700: Weekly GMAT Challenge (Answer)

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 .

A first-of-its-kind, on-demand MBA application experience that delivers a personalized curriculum for you and leverages interactive tools to guide you through the entire MBA application process.

### Upcoming Events

• Columbia J-Term (Round 2)
• Cambridge Judge (Round 1)
• HBS (Round 1)
• Penn Wharton (Round 1)
• Notre Dame Mendoza (Early Decision)
• Virginia Darden (Early Decision)
• Michigan Ross (Round 1)
• Columbia (Round 1)
• Stanford GSB (Round 1)
• Yale SOM (Round 1)
• Northwestern Kellogg (Round 1)
• Berkeley Haas (Round 1)