BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATBootcamp Starts Sep 28
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE BOOTCAMP

Live Online Bootcamp Class with Top GMAT Expert Chris Peckover

15 live classes from Sep 28, 2026

Schedule
Mon to Fri · 7:00 to 10:00 PM ET
Included
Live classes + 6 months of TTP OnDemand
  • Boost your GMAT score in less than one month in a live online class
  • 6 months access to TTP OnDemand video courses included
View bootcamp & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Hexagon

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 358
Joined: Thu Apr 18, 2013 9:46 amFollowed by: 7 members
Location: Jeddah, Saudi Arabia
Thanked: 42 times
GMAT Score:730
by faraz_jeddah » Fri Nov 01, 2013 9:39 am
My approach

r = 2

Interior angle = ((6-2)/6) * 180 = 120

each arc represents 120/360 of the circle i.e. 1/3 rd

Length of one arc = 2*pi*2 * (1/3) = 4*pi / 3

Each vertice has 2 arcs.
=> 6 vertices have 12 arcs

Total length = 12 * 4*pi / 3 = 16*pi

OA is [spoiler]8*pi[/spoiler]
Attachments
arcs.jpg
A good question also deserves a Thanks.

Messenger Boy: The Thesselonian you're fighting... he's the biggest man i've ever seen. I wouldn't want to fight him.
Achilles: That's why no-one will remember your name.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1556
Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members
Thanked: 448 times
GMAT Score:650

by theCodeToGMAT » Fri Nov 01, 2013 9:50 am
I got 8*pi

Angle of Hexagon = (n-2)*180 = (6-2)*180 = 720

Measure of each angle = 720/6 = 120

Measure of one arc of 120 degree = 120/360 * 2 * pi * (2) = 4/3 * pi

Total sum of arcs = 6 [4/3 * pi] = 8 * pi
R A H U L
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 358
Joined: Thu Apr 18, 2013 9:46 amFollowed by: 7 members
Location: Jeddah, Saudi Arabia
Thanked: 42 times
GMAT Score:730

by faraz_jeddah » Fri Nov 01, 2013 10:00 am
theCodeToGMAT wrote:I got 8*pi

Angle of Hexagon = (n-2)*180 = (6-2)*180 = 720

Measure of each angle = 720/6 = 120

Measure of one arc of 120 degree = 120/360 * 2 * pi * (2) = 4/3 * pi

Total sum of arcs = 6 [4/3 * pi] = 8 * pi
I see you edited your answer :)

Mind telling me how you got 6 arcs and not 12.
A good question also deserves a Thanks.

Messenger Boy: The Thesselonian you're fighting... he's the biggest man i've ever seen. I wouldn't want to fight him.
Achilles: That's why no-one will remember your name.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1556
Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members
Thanked: 448 times
GMAT Score:650

by theCodeToGMAT » Fri Nov 01, 2013 10:06 am
faraz_jeddah wrote:
theCodeToGMAT wrote:I got 8*pi

Angle of Hexagon = (n-2)*180 = (6-2)*180 = 720

Measure of each angle = 720/6 = 120

Measure of one arc of 120 degree = 120/360 * 2 * pi * (2) = 4/3 * pi

Total sum of arcs = 6 [4/3 * pi] = 8 * pi
I see you edited your answer :)

Mind telling me how you got 6 arcs and not 12.
Yep, Earlier I made a silly mistake while solving .. I realized & corrected it

Check the attachment for the reason :) .. Each arc is made with different color..
Attachments
solution.JPG
R A H U L
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Fri Nov 01, 2013 10:56 am
theCode's figure is great, but on the GMAT we won't have access to different colors.
To avoid careless errors in geometry problems:

1. DRAW your own figure.
2. Label EVERYTHING.

Image

Hexagon ABCDEF:
In a polygon with n sides, the sum of the interior angles = (n-2)180.
In a hexagon, the sum of the interior angles = (6-2)180 = 720.
In a REGULAR hexagon, all of the sides are equal, as are all of the interior angles.
Thus, each of the interior angles of hexagon ABCDEF = 720/6 = 120.

Circle with center B:
Since r=2, the circumference = 2Ï€r = 2*Ï€*2 = 4Ï€.
Since ∠ABC = 120, and 120/360 = 1/3, the arc intercepted by ∠ABC -- arc AC -- constitutes 1/3 of the circumference.
Thus, arc AC = (1/3) * 4Ï€ = (4/3)Ï€.

Sum of the arc lengths:
List all of the arcs.
Each vertex is the endpoint of 2 different arcs.
Thus, each vertex must appear in our list EXACTLY 2 TIMES:
AC
AE
BD
BF
CE
DF
Total = 6.
Sum of the arc lengths = 6 * (4/3)Ï€ = 8Ï€.

The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 269
Joined: Thu Sep 19, 2013 12:46 amFollowed by: 7 members
Thanked: 94 times

by mevicks » Fri Nov 01, 2013 8:08 pm
Image
Each vertex of the regular hexagon shown above is the center of a circle. If the side of the hexagon is 2, what is the total length of the arcs?
A) 4Ï€
B) 6Ï€
C) 8Ï€
D) 12Ï€
E) 16Ï€
Regular hexagon can be broken up into 6 equilateral triangles.

This property is very helpful during time crunch situations and thus we can play around with regular hexagons very easily and reach a conclusion just by looking at the symmetry (the mathematical proof can be found out but is not necessary)

Focus on any one equilateral triangle and its corresponding arc:
Image

Length of the arc = (60/360)2π2 = π(2/3)
We have 12 such arcs encompassing the equilateral triangles, thus the total length = 12Ï€(2/3) = [spoiler]8Ï€[/spoiler]

Answer C
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 269
Joined: Thu Sep 19, 2013 12:46 amFollowed by: 7 members
Thanked: 94 times

by mevicks » Fri Nov 01, 2013 8:12 pm
mevicks wrote:
Regular hexagon can be broken up into 6 equilateral triangles.

This property is very helpful during time crunch situations and thus we can play around with regular hexagons very easily and reach a conclusion just by looking at the symmetry (the mathematical proof can be found out but is not necessary)
Also, such properties are useful on complex problems:
https://www.beatthegmat.com/hexagon-prob ... 06401.html
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1