Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

It is currently 22 Jul 2019, 18:31

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Senior Manager
Senior Manager
avatar
Joined: 18 Aug 2009
Posts: 325
Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0
Regular hexagon ABCDEF has a perimeter of 36. O is the cente  [#permalink]

Show Tags

New post 22 Jan 2010, 11:10
3
30
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

64% (02:50) correct 36% (03:01) wrong based on 807 sessions

HideShow timer Statistics


Attachment:
1.JPG
1.JPG [ 32.32 KiB | Viewed 26552 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. \(54sqrt{3}-27\pi\)

Guys,
I did not understand why we can not form two trapezoids of this hexagon, and solve through it. The official solution says, that we have to form 6 equiaterial triangles in the hexagon and form through it. Forming trapezoids, I got 108 as a square of the hexagon.
Please. elaborate

_________________
Never give up,,,
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56357
Regular hexagon ABCDEF has a perimeter of 36. O is the cente  [#permalink]

Show Tags

New post 22 Jan 2010, 12:06
5
5
mirzohidjon 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. \(54sqrt{3}-27\pi\)


Guys,
I did not understand why we can not form two trapezoids of this hexagon, and solve through it. The official solution says, that we have to form 6 equiaterial triangles in the hexagon and form through it. Forming trapezoids, I got 108 as a square of the hexagon.
Please. elaborate


You can do this by calculating the area of the trapezoid as well.
Attachment:
hexagon20and20circles.jpg
hexagon20and20circles.jpg [ 7.6 KiB | Viewed 17025 times ]

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 \(\frac{2}{\sqrt{3}}=\frac{6}{h}\) --> \(h=3\sqrt{3}\). 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 \(54\sqrt{3}-27\pi\).

Answer: E.
_________________
General Discussion
CEO
CEO
User avatar
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2552
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Reviews Badge
Re: Manhattan GMAT test  [#permalink]

Show Tags

New post Updated on: 22 Jan 2010, 12:10
2
1
you can use trapezoids as well..

total perimeter = 36...thus side = 36/6 = 6

radii of circle = 6/2 = 3

now area of 2 trapeziods = 2* 1/2 (sum of parallel sides) height.

=(6+12)h = 18h

Now to find the height....consider any triangle and u will find the height is 6sin60 = 6*\sqrt{3}/2
= 3\sqrt{3}

Thus total area of 2 trapezoids = 18*3\sqrt{3} = 54\sqrt{3} --1

We have to subtract area of one circle + 6*(area by arc which is subtending 120* at each centre)

Thus we have area of 1+ 2 circles = 3circle = 3π r^2 = 3π * 3*3 = 27π ----2

Subtract 2 from 1

we get ans =E
_________________
Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned :)

Jo Bole So Nihaal , Sat Shri Akaal

:thanks Support GMAT Club by putting a GMAT Club badge on your blog/Facebook :thanks

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Originally posted by gurpreetsingh on 22 Jan 2010, 11:54.
Last edited by gurpreetsingh on 22 Jan 2010, 12:10, edited 1 time in total.
Senior Manager
Senior Manager
avatar
Joined: 18 Aug 2009
Posts: 325
Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0
Re: Manhattan GMAT test  [#permalink]

Show Tags

New post 25 Jan 2010, 16:41
Now, I got it.
I was making mistake in calculating the height of the trapezoid.
Many Thanks.
_________________
Never give up,,,
Intern
Intern
avatar
Joined: 30 May 2010
Posts: 26
Schools: YALE SOM
PS: Geometry hexagon with circles  [#permalink]

Show Tags

New post 20 Jun 2010, 09:25
1
In the figure above, all the circles are congruent, all the vertices of the regular hexagon coincide with the center of a circle, and all the circles have at most 1 point in common with any other circle. If a side of the hexagon is 12, what is the area of the shaded region?

Answers:
A. \(108*(2*sqrt(3)-\pi)\)
B. \(108*(4*sqrt(3)-3\pi)\)
C. \(36*(12-3\pi)\)
D. \(36*(2*sqrt(3)-3\pi)\)
E. None of the above.

Attachment:
P1.JPG
P1.JPG [ 4.95 KiB | Viewed 21713 times ]


Thanks!
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56357
Re: PS: Geometry hexagon with circles  [#permalink]

Show Tags

New post 20 Jun 2010, 09:36
Nusa84 wrote:
In the figure above, all the circles are congruent, all the vertices of the regular hexagon coincide with the center of a circle, and all the circles have at most 1 point in common with any other circle. If a side of the hexagon is 12, what is the area of the shaded region?

Answers:
A. \(108*(2*sqrt(3)-\pi)\)
B. \(108*(4*sqrt(3)-3\pi)\)
C. \(36*(12-3\pi)\)
D. \(36*(2*sqrt(3)-3\pi)\)
E. None of the above.

Attachment:
P1.JPG


Thanks!


Basically the same question. The only difference is that in the first question the side is 36/6=6 and in the second problem 12.
_________________
Intern
Intern
avatar
Joined: 27 Jan 2010
Posts: 4
Re: Manhattan GMAT test  [#permalink]

Show Tags

New post 21 Jun 2010, 05:13
Can we calculate the area of the hexagon by calculating the area of 1 equilateral triangle andmultiply by 6 to get the area of the hexagon?

Area of Equilateral Triangle = (s^2/4)*\sqrt{3}
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56357
Re: Manhattan GMAT test  [#permalink]

Show Tags

New post 21 Jun 2010, 05:45
undernet wrote:
Can we calculate the area of the hexagon by calculating the area of 1 equilateral triangle andmultiply by 6 to get the area of the hexagon?

Area of Equilateral Triangle = (s^2/4)*\sqrt{3}


Sure we can: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}\), as side=2*raius, \(Area_{equilateral}=4r^2*\frac{\sqrt{3}}{4}=r^2*\sqrt{3}\) --> \(Area_{hexagon}=6*Area_{equilateral}=6*\sqrt{3}*r^2\);

Not shaded area inside the hexagon is the size of 3 circles = \(3\pi{r^2}\).

Area of the shaded region: \(6*\sqrt{3}*r^2-3\pi{r^2}\).
_________________
Intern
Intern
avatar
Joined: 06 Feb 2010
Posts: 1
Re: Manhattan GMAT test  [#permalink]

Show Tags

New post 23 Jun 2010, 10:29
mirzohidjon wrote:
Guys,
I did not understand why we can not form two trapezoids of this hexagon, and solve through it. The official solution says, that we have to form 6 equiaterial triangles in the hexagon and form through it. Forming trapezoids, I got 108 as a square of the hexagon.
Please. elaborate


The simplest way to calculate above problem:-
Area of shaded region= Area of hexagon - area of circumscribed circle - area of each arcs inscribed by the hexagon.

Area of hexagon= (3√3/2) (side)2
= (3√3/2) x 36
= 54√3

Area of circle=9π
Combined area of six arcs inscribed= 6x (120/360) π 32
= 18 π

Thus, Area of shaded region = 54√3 - 9π - 18 π
Area of shaded region = 54√3 - 27 π
VP
VP
User avatar
Joined: 02 Jul 2012
Posts: 1153
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Regular hexagon ABCDEF has a perimeter of 36  [#permalink]

Show Tags

New post 01 Mar 2013, 23:47
2
mun23 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)?(see attached file)

a. 108-18pi
b. 54sqrt(3)-9pi
c. 54sqrt(3)-18pi
d. 108-27pi
e. 54sqrt(3)-27p


Area of a regular hexagon (6 equilateral triangles) = \(6a^2\)\(\frac{\sqrt{3}}{4}\) = \(6*36*\)\(\frac{\sqrt{3}}{4}\)

= \(54\sqrt{3}\)

Internal angles of regular hexagon = 120 degrees. So each circle has a third of its area inside the hexagon. So 6 circles = \(6*\frac{1}{3}*\pi*r^2\) = \(18\pi\)

Area of the circle entirely inside = \(9\pi\)

So required area = \(54\sqrt{3}\) - \(18\pi\) - \(9\pi\)

= \(54\sqrt{3}\) - \(27\pi\)

Answer is E
_________________
Did you find this post helpful?... Please let me know through the Kudos button.

Thanks To The Almighty - My GMAT Debrief

GMAT Reading Comprehension: 7 Most Common Passage Types
Manager
Manager
avatar
Joined: 04 Sep 2012
Posts: 127
GMAT ToolKit User
Re: Regular hexagon ABCDEF has a perimeter of 36  [#permalink]

Show Tags

New post 02 Mar 2013, 06:36
MacFauz wrote:
mun23 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)?(see attached file)

a. 108-18pi
b. 54sqrt(3)-9pi
c. 54sqrt(3)-18pi
d. 108-27pi
e. 54sqrt(3)-27p


Area of a regular hexagon (6 equilateral triangles) = \(6a^2\)\(\frac{\sqrt{3}}{4}\) = \(6*36*\)\(\frac{\sqrt{3}}{4}\)

= \(54\sqrt{3}\)

Internal angles of regular hexagon = 120 degrees. So each circle has a third of its area inside the hexagon. So 6 circles = \(6*\frac{1}{3}*\pi*r^2\) = \(18\pi\)

Area of the circle entirely inside = \(9\pi\)

So required area = \(54\sqrt{3}\) - \(18\pi\) - \(9\pi\)

= \(54\sqrt{3}\) - \(27\pi\)

Answer is E


When you say , Each triangle is 1/3 rd inside do you mean as angle of arc if 120degree and 120/360 = 1/3 that means that sector of circle is inside Hexagon ???

Am I right ?
_________________
Regards,
Abhinav

GMAT 1 - 580 (Q47 V23) http://gmatclub.com/forum/a-tight-slap-on-face-149457.html

GMAT 2 - 670 (Q48 V34) http://gmatclub.com/forum/670-one-month-off-from-office-and-2-months-hard-work-163761.html#p1297561

“If you don't change your life; your life will change you.”
VP
VP
User avatar
Joined: 02 Jul 2012
Posts: 1153
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Regular hexagon ABCDEF has a perimeter of 36  [#permalink]

Show Tags

New post 02 Mar 2013, 07:43
abhinav11 wrote:
MacFauz wrote:
mun23 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)?(see attached file)

a. 108-18pi
b. 54sqrt(3)-9pi
c. 54sqrt(3)-18pi
d. 108-27pi
e. 54sqrt(3)-27p


Area of a regular hexagon (6 equilateral triangles) = \(6a^2\)\(\frac{\sqrt{3}}{4}\) = \(6*36*\)\(\frac{\sqrt{3}}{4}\)

= \(54\sqrt{3}\)

Internal angles of regular hexagon = 120 degrees. So each circle has a third of its area inside the hexagon. So 6 circles = \(6*\frac{1}{3}*\pi*r^2\) = \(18\pi\)

Area of the circle entirely inside = \(9\pi\)

So required area = \(54\sqrt{3}\) - \(18\pi\) - \(9\pi\)

= \(54\sqrt{3}\) - \(27\pi\)

Answer is E


When you say , Each triangle is 1/3 rd inside do you mean as angle of arc if 120degree and 120/360 = 1/3 that means that sector of circle is inside Hexagon ???

Am I right ?

Yes..... I guess you meant circle.....
_________________
Did you find this post helpful?... Please let me know through the Kudos button.

Thanks To The Almighty - My GMAT Debrief

GMAT Reading Comprehension: 7 Most Common Passage Types
Manager
Manager
User avatar
Joined: 24 Jan 2013
Posts: 72
Re: Regular hexagon ABCDEF has a perimeter of 36  [#permalink]

Show Tags

New post 02 Mar 2013, 09:32
2
The Area we are looking for is equal to the area of the hexagon minus the area of the circle with center O and minus the slices of the other 6 circles.

The slices of these 6 circles have an equal angle, and the sum of these angles equals the sum of the angles of the hexagon. We know that the sum of the angles of a polygon of n=6 sides is 180*(n-2). This means the sum of these angles is \(180*(6-2)=180*4=360*2\). Therefore, the area of the 6 slices equals the area of 2 circles.

If we add the area of the circle in the middle, we have that the "white" area inside the hexagon equals the area of 3 circles.

So the area we are looking for is the area of the hexagon (A) minus the area of 3 circles (B).

(A) We split the hexagon in 6 isosceles triangles. In an isosceles triangle with side \(6=2*X\), the height must be \(\sqrt{3}*X\). Therefore, because X=3, the height must be \(\sqrt{3}*3\). And then, the area of one triangle is \(\frac{1}{2}\), times the base of 6, times the height of \(\sqrt{3}*3\). This equals \(3*\sqrt{3}*3\). Times 6 triangles to find the area of the hexagon, we have: \(54*\sqrt{3}\)

(B) The area of 1 circle of radius 3 is \(pi*3^2\). Times 3 circles to find the "white" area inside of the hexagon, is: \(27*pi\).

Then, the area we are looking for is (A) minus (B):

\((54*\sqrt{3})-(27*pi)\)

Solution E.
Intern
Intern
avatar
Joined: 08 Sep 2012
Posts: 10
GMAT 1: 770 Q50 V45
WE: General Management (Non-Profit and Government)
Re: Regular hexagon ABCDEF has a perimeter of 36  [#permalink]

Show Tags

New post 02 Mar 2013, 23:33
Perimeter f hexagon is 36 which means each side is 6 and radius of each circle with center at A,B,C,D,E and F is 3. Each angle of hexagon is 120⁰ hence area of each circle which is inside the hexagon is π*32*(120/360)=3π. Area of all such sectors is 6*3π= 18π.

Do not assume that radius of circle with center at O is 3. It is same as area of other circles because it is hexagon, it will be different for any other figure. Triangle OAB is a equilateral triangle thus radius of circle with center at O can be calculated as 3. Area of circle is 9π
Total area covered by circles is 27π

Area of a regular hexagon is (3 *sqrt3/2)*square of side. This can be simplified by dividing the hexagon in to six equilateral triangles. Area of hexagon is (3*sqrt3/2)*36=54*sqrt3

Thus area of shaded region is 54*sqrt3 -27π
Intern
Intern
avatar
Joined: 09 May 2013
Posts: 44
Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente  [#permalink]

Show Tags

New post 09 Jun 2013, 23:56
Hi Bunuel,

I have applied central angle theorem since two adjacent equilateral triangle forms 120 degree
then central angle of all the six circle would be 120 degree.
is it correct?

Thanks & Regards,
WarriorGmat
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56357
Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente  [#permalink]

Show Tags

New post 10 Jun 2013, 05:22
Manager
Manager
avatar
Joined: 26 Sep 2013
Posts: 190
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41
Re: Manhattan GMAT test  [#permalink]

Show Tags

New post 26 Oct 2013, 11:35
Bunuel wrote:
mirzohidjon 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. \(54sqrt{3}-27\pi\)


Guys,
I did not understand why we can not form two trapezoids of this hexagon, and solve through it. The official solution says, that we have to form 6 equiaterial triangles in the hexagon and form through it. Forming trapezoids, I got 108 as a square of the hexagon.
Please. elaborate


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

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 \(54\sqrt{3}-27\pi\).

Answer: E.


Is this true for any trapezoid then?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56357
Re: Manhattan GMAT test  [#permalink]

Show Tags

New post 27 Oct 2013, 06:06
AccipiterQ wrote:
Bunuel wrote:
mirzohidjon 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. \(54sqrt{3}-27\pi\)


Guys,
I did not understand why we can not form two trapezoids of this hexagon, and solve through it. The official solution says, that we have to form 6 equiaterial triangles in the hexagon and form through it. Forming trapezoids, I got 108 as a square of the hexagon.
Please. elaborate


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

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 \(54\sqrt{3}-27\pi\).

Answer: E.


Is this true for any trapezoid then?


How can it be true for all trapezoids?
_________________
Manager
Manager
avatar
B
Joined: 10 Mar 2014
Posts: 185
Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente  [#permalink]

Show Tags

New post 03 Aug 2014, 04:13
Bunuel wrote:
undernet wrote:
Can we calculate the area of the hexagon by calculating the area of 1 equilateral triangle andmultiply by 6 to get the area of the hexagon?

Area of Equilateral Triangle = (s^2/4)*\sqrt{3}


Sure we can: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}\), as side=2*raius, \(Area_{equilateral}=4r^2*\frac{\sqrt{3}}{4}=r^2*\sqrt{3}\) --> \(Area_{hexagon}=6*Area_{equilateral}=6*\sqrt{3}*r^2\);

Not shaded area inside the hexagon is the size of 3 circles = \(3\pi{r^2}\).

Area of the shaded region: \(6*\sqrt{3}*r^2-3\pi{r^2}\).


HI Bunnel,

I did it using equilateral triangle. But I got a query in following statement.

Not shaded area inside the hexagon is the size of 3 circles = \(3\pi{r^2}\).

Here I consider hexagon as a 6 equilateral triangle. so each angle of the triangle is 60. Now if I will multiply 60*6 + 1 circle inside hexagon it becomes 2 circles.

How you are considering it as 3 circles inside the hexagon.

Please clarify

Thanks.
Manager
Manager
avatar
Joined: 15 Aug 2013
Posts: 239
Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente  [#permalink]

Show Tags

New post 02 Nov 2014, 18:33
Bunuel wrote:
mirzohidjon 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. \(54sqrt{3}-27\pi\)


Guys,
I did not understand why we can not form two trapezoids of this hexagon, and solve through it. The official solution says, that we have to form 6 equiaterial triangles in the hexagon and form through it. Forming trapezoids, I got 108 as a square of the hexagon.
Please. elaborate


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

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 \(54\sqrt{3}-27\pi\).

Answer: E.



Hi Bunuel,

Two questions:

1) Regarding your method above, it's still a little blurry as to how you calculated the area of the trapezoid, namely the height. Can you please elaborate a little?

2) The way I went about calculating the area of the trapezoid was 2 traingles plus a rectangle. To find the area of the triangle, I used pyt theorem -- Square Root of 72 was the hypotenuse of the triangle and the length of the rectangle. So I had (2*1/2*6*6) + (Root 72 * 6) = 36 + 36root 2. Why is that wrong? Doesn't that encapsulate the total hexagon?

Thanks!
GMAT Club Bot
Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente   [#permalink] 02 Nov 2014, 18:33

Go to page    1   2    Next  [ 26 posts ] 

Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





cron

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne