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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

regular hexagon

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 324
Joined: Thu Dec 24, 2009 6:29 amFollowed by: 1 member
Thanked: 17 times
by rahul.s » Sun Feb 14, 2010 2:53 am
Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F, respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?

(A) 108 - 18pi
(B) 54√3 - 9pi
(C) 54√3 - 18pi
(D) 108 - 27pi
(E) 54√3 - 27pi

OA: E
Source: MGMAT
Level: 700 - 800
Attachments
Hexagon.doc
(69.5 KiB) Downloaded 145 times
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1275
Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members
Location: Arabian Sea
Thanked: 125 times

by ajith » Sun Feb 14, 2010 3:11 am
rahul.s wrote:Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F, respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?

(A) 108 - 18pi
(B) 54√3 - 9pi
(C) 54√3 - 18pi
(D) 108 - 27pi
(E) 54√3 - 27pi

OA: E
Source: MGMAT
Level: 700 - 800
Radius of the circles A B C D E F and O = 3cm
Area of the circles A B C D E F inside the hexagon = 6*(120/360)* pi*3*3 = 18pi
Area of the circle O inside the hexagon = pi*3*3 = 9 pi
Total Area of the hexagon = 3*sqrt(3)/2*6*6 = 54 √3
Area of the shaded region (inside the hexagon but outside the circles) = 54 √3- (18pi+9pi) = 54 √3- 27pi
Always borrow money from a pessimist, he doesn't expect to be paid back.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 27
Joined: Sat Feb 06, 2010 1:55 pm
Thanked: 5 times

by DeepthiRajan » Sun Feb 14, 2010 2:46 pm
ajith wrote:
rahul.s wrote:Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F, respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?

(A) 108 - 18pi
(B) 54√3 - 9pi
(C) 54√3 - 18pi
(D) 108 - 27pi
(E) 54√3 - 27pi

OA: E
Source: MGMAT
Level: 700 - 800
Radius of the circles A B C D E F and O = 3cm
Area of the circles A B C D E F inside the hexagon = 6*(120/360)* pi*3*3 = 18pi
Area of the circle O inside the hexagon = pi*3*3 = 9 pi
Total Area of the hexagon = 3*sqrt(3)/2*6*6 = 54 √3
Area of the shaded region (inside the hexagon but outside the circles) = 54 √3- (18pi+9pi) = 54 √3- 27pi

Ajith,

Could you please explain why angle=120 was used while calculating the area of the circles.

Also, it would be helpful if you could elaborate on the calculation of area of hexagon.

Thanks in advance.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1275
Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members
Location: Arabian Sea
Thanked: 125 times

by ajith » Sun Feb 14, 2010 9:38 pm
DeepthiRajan wrote:
Ajith,

Could you please explain why angle=120 was used while calculating the area of the circles.

Also, it would be helpful if you could elaborate on the calculation of area of hexagon.

Thanks in advance.
The internal angle of a regular polygon with "n" sides can be calculated using:

(n-2) × 180° / n

for hexagon n=6 so, internal angle = 180*4/6 =120

In this case the angle subtended at the center of the circle is the same as the internal angle of the regular hexagon

Hence 120 degree is used in the area calculation.


Area of a regular hexagon is best remembered as a standard formula = 3*sqrt(3)/2*a^2 [where a is the side]

[it is 6*area of an isosceles triangle with side a]


Hope it helps
Always borrow money from a pessimist, he doesn't expect to be paid back.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1560
Joined: Tue Nov 17, 2009 2:38 amFollowed by: 5 members
Thanked: 137 times

by thephoenix » Sun Feb 14, 2010 10:27 pm
area of hexagon:-6* area if an eq trianglewith side 6=54v3

angle of hexagon=(6-2)*180/6=120

area of a sector with radius=3 and angle=120 is 120/360(pi) *(3)^2=3 pi for 6 circles=18 pi

area of a circle=9pi

reqd area=54v3-18 pi-9pi=54v3-27pi
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 941
Joined: Sun Dec 27, 2009 12:28 amFollowed by: 1 member
Thanked: 20 times

by bhumika.k.shah » Mon Feb 15, 2010 2:47 am
ajith wrote:
DeepthiRajan wrote:
Ajith,

Could you please explain why angle=120 was used while calculating the area of the circles.

Also, it would be helpful if you could elaborate on the calculation of area of hexagon.

Thanks in advance.
The internal angle of a regular polygon with "n" sides can be calculated using:

(n-2) × 180° / n

for hexagon n=6 so, internal angle = 180*4/6 =120

In this case the angle subtended at the center of the circle is the same as the internal angle of the regular hexagon

Hence 120 degree is used in the area calculation.


Area of a regular hexagon is best remembered as a standard formula = 3*sqrt(3)/2*a^2 [where a is the side]

[it is 6*area of an isosceles triangle with side a]


Hope it helps
Ajith , could you elaborate a l'l more on how u got the area of the hexagon. i am not quite getting the 6 equilaterla triangles thingy ...:(
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1275
Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members
Location: Arabian Sea
Thanked: 125 times

by ajith » Mon Feb 15, 2010 3:33 am
bhumika.k.shah wrote: Ajith , could you elaborate a l'l more on how u got the area of the hexagon. i am not quite getting the 6 equilaterla triangles thingy ...:(
Image

That is illustrated well in this page

Source: https://homepage.mac.com/terhorab/iblog/ ... onArea.pdf
Always borrow money from a pessimist, he doesn't expect to be paid back.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Mon Jul 12, 2010 5:19 pm

by Atrain13gm » Sat Oct 09, 2010 6:37 pm
I understand the formula for figuring out the area of the hexagon and getting 54 sqrt3, but why do I get 108 if I try calculating the area of the hexagon as two trapezoids?

Trapezoid Area=
1/2 * (6) * (12+6) = 54

54*2 = 108 area for the hexagon
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1