In the figure above, equilateral hexagon ABCDEF is

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Gmat_mission: Legendary Member; Posts: 1622; Joined: Thu Mar 01, 2018 7:22 amFollowed by: 2 members

by Gmat_mission » Mon Jun 10, 2019 5:08 am

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In the figure above, equilateral hexagon ABCDEF is inscribed in the circle with center G, which has a diameter of 16. What is the length of AB?

A. $16$

B. $8\sqrt{3}$

C. $6\sqrt{3}$

D. $8$

E. $4\sqrt{3}$

[spoiler]OA=D[/spoiler]

Source: Princeton Review

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Source: Beat The GMAT — Problem Solving | Mon Jun 10, 2019 5:08 am

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

by Ian Stewart » Mon Jun 10, 2019 6:31 am

"Equilateral" means "equal sides". It is not a synonym for "regular", which means "equal sides *and* equal angles". Now, it's true that an equilateral hexagon that you can inscribe in a circle must be regular, but the GMAT certainly would never expect a test taker to know that.

So if this were a GMAT question, it would tell you the hexagon is regular, not that it is "equilateral". If we connect the center G of a regular hexagon to each vertex (A, B, C, D, E and F) we divide the hexagon into six equilateral triangles. The edges GA, GB, etc, of each equilateral are each radii of the circle, so they are 8 units long, and since the triangles are equilateral, the edges of the hexagon must also be 8 units long.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8151; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Tue Jun 11, 2019 7:14 pm

Gmat_mission wrote:
In the figure above, equilateral hexagon ABCDEF is inscribed in the circle with center G, which has a diameter of 16. What is the length of AB?

A. $16$

B. $8\sqrt{3}$

C. $6\sqrt{3}$

D. $8$

E. $4\sqrt{3}$

[spoiler]OA=D[/spoiler]

An equilateral hexagon can be broken up into 6 equal, equilateral triangles. Since the diameter of the circle is 16, each side of the triangle must be 8, and thus side AB (which also makes up a side of one of the equilateral triangles must also be 8.

Anaswer: D

Source: Princeton Review

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