Regular hexagon ABCDEF has a perimeter of 36. O is the center of the : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 27 Feb 2017, 20:46

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Regular hexagon ABCDEF has a perimeter of 36. O is the center of the

Author Message
TAGS:

### Hide Tags

Intern
Joined: 22 Sep 2007
Posts: 29
Followers: 0

Kudos [?]: 10 [0], given: 0

Regular hexagon ABCDEF has a perimeter of 36. O is the center of the [#permalink]

### Show Tags

09 Dec 2007, 19:36
8
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

59% (03:02) correct 41% (01:52) wrong based on 199 sessions

### HideShow timer Statistics

Attachment:

hexagon%20and%20circles.jpg [ 48.78 KiB | Viewed 8644 times ]
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. $$108-18\pi$$
B. $$54\sqrt{3}-9\pi$$
C. $$54\sqrt{3}-18\pi$$
D. $$108-27\pi$$
E. $$54sqrt{3}-27\pi$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: regular-hexagon-abcdef-has-a-perimeter-of-36-o-is-the-cente-89544.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Sep 2014, 19:30, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
Manager
Joined: 03 Sep 2006
Posts: 233
Followers: 1

Kudos [?]: 19 [1] , given: 0

Re: Regular hexagon ABCDEF has a perimeter of 36. O is the center of the [#permalink]

### Show Tags

09 Dec 2007, 20:08
1
KUDOS
This is MGMAT Question =)

1) The side of a hex = 6 => r of the circle = 3.
2) Area of a hex = 6 * area of triangles with the side =6:
S = 1/2 height * base. The height divide triangle (with the sides = 6) to 2 triangles with the sides 6, 3*sqrt3 and 3 (as this is 1:sqrt3:2 triangles)
S of the triangle = 1/2 * (3*sqrt 3) * 6 = 9 * sqrt 3

So the area of hex = 6 * 9 * sqrt 3 = 54 * sqrt 3

The area of shaded region = 54 * sqrt 3 - 3 (Pi*3^2) = 54 * sqrt 3 - 27*Pi
MBA Blogger
Joined: 19 Apr 2014
Posts: 100
Location: India
Concentration: Strategy, Technology
WE: Analyst (Computer Software)
Followers: 2

Kudos [?]: 40 [1] , given: 52

Re: Regular hexagon ABCDEF has a perimeter of 36. O is the center of the [#permalink]

### Show Tags

28 Aug 2014, 22:26
1
KUDOS
kizito2001 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)?

My approach:
Divide the hexagon into 6 equilateral triangles:

1. Area of hexagon. = 6 * area of a triangle (As all triangles are equilateral with same side).
= $$6 * \sqrt{3} /4 * 6^2$$
= $$54\sqrt{3}$$

2. Area of orange part in the diagram:
$$pie*r^2* 120/360$$
$$pie*9* 1/3$$
$$3* pie$$

Total of 6 areas like this + the central circle = $$18 * pie + 9 * pie = 27 * pie$$

Now area of black shaded region = area point 1 - area point 2.
= $$54\sqrt{3} - 27 * pie$$

Attachments

File comment: hexagon

hexagon%2520and%2520circles.jpg [ 43.38 KiB | Viewed 5643 times ]

_________________

Warm Regards.
Visit My Blog

Intern
Joined: 25 Dec 2012
Posts: 5
Followers: 0

Kudos [?]: 5 [0], given: 5

Re: Regular hexagon ABCDEF has a perimeter of 36. O is the center of the [#permalink]

### Show Tags

02 Sep 2014, 09:56
My approach was as follows :

1. Area of hexagon : 6 * \sqrt{3} / 4 * 3 * 3

2. Now calculate the area of 6 sectors of circles of angle 120 degree using sector formula : 120/360 * 3.14 * 3 * 3

3. Area of step 2 + Area of another circle : 3.14 * 3 * 3

4. Final Solution : Area of hexagon(step 1) - (Area of sectors+Area of 7th circle)(Step 3)
Math Expert
Joined: 02 Sep 2009
Posts: 37144
Followers: 7274

Kudos [?]: 96812 [0], given: 10786

Re: Regular hexagon ABCDEF has a perimeter of 36. O is the center of the [#permalink]

### Show Tags

02 Sep 2014, 19:33
Expert's post
1
This post was
BOOKMARKED
kizito2001 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. $$108-18\pi$$
B. $$54\sqrt{3}-9\pi$$
C. $$54\sqrt{3}-18\pi$$
D. $$108-27\pi$$
E. $$54sqrt{3}-27\pi$$

You can do this by calculating the area of the trapezoid.

First of all the angle = 120 degrees. Non shaded area will consist of the areas of 6*120 degrees sectors plus one whole circle = 3 whole circles = $$3\pi{r^2}=27\pi$$, (as $$r=3$$, half of the side of the hexagon).

Now if we want to proceed with the trapezoid areas. We would have two trapezoids. Bases would be $$a=6$$ (the smallest base) and $$b=12$$ (the largest base). $$Area=\frac{a+b}{2}*h$$. The height of the trapezoid can be calculated as the side in 30-60-90 right triangle. Height would be the opposite side to the 60 degrees angle (half of 120 degrees). In 30-60-90 right triangle the ratio of the sides is $$1/\sqrt{3}/2$$, the height would correspond with \sqrt{3} and the hypotenuse (which on the other hand would be the side of the hexagon=6) would correspond with 2. so we'll have $$2/\sqrt{3}=\frac{6}{h}$$ --> $$h=3\sqrt{3}$$. Area of the trapezoid would be $$Area=\frac{a+b}{2}*h=\frac{6+12}{2}*3\sqrt{3}=27\sqrt{3}$$. As we have two trapezoids the whole area would be $$2*27\sqrt{3}=54\sqrt{3}$$.

Area of the shaded region $$54\sqrt{3}-27\pi$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: regular-hexagon-abcdef-has-a-perimeter-of-36-o-is-the-cente-89544.html
_________________
Re: Regular hexagon ABCDEF has a perimeter of 36. O is the center of the   [#permalink] 02 Sep 2014, 19:33
Similar topics Replies Last post
Similar
Topics:
3 Each vertex of the regular hexagon shown above is the center of a circ 3 03 Oct 2016, 22:22
4 The figure shown is a regular hexagon with center H 3 11 Sep 2016, 15:09
5 In the figure, JKLMNP is a regular hexagon . 5 06 Nov 2015, 09:06
20 A regular hexagon has a perimeter of 30 units. What is the 8 30 Apr 2012, 00:02
27 Regular hexagon ABCDEF has a perimeter of 36. O is the cente 24 22 Jan 2010, 10:10
Display posts from previous: Sort by