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Regular hexagon ABCDEF has a perimeter of 36. O is the cente
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22 Jan 2010, 11:10
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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. \(10818\pi\) B. \(54\sqrt{3}9\pi\) C. \(54\sqrt{3}18\pi\) D. \(10827\pi\) 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
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Regular hexagon ABCDEF has a perimeter of 36. O is the cente
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22 Jan 2010, 12:06
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. \(10818\pi\) B. \(54\sqrt{3}9\pi\) C. \(54\sqrt{3}18\pi\) D. \(10827\pi\) 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 [ 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 306090 right triangle. Height would be the opposite side to the 60 degrees angle (half of 120 degrees). In 306090 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.
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Re: Manhattan GMAT test
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Updated on: 22 Jan 2010, 12:10
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
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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.



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Re: Manhattan GMAT test
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25 Jan 2010, 16:41
Now, I got it. I was making mistake in calculating the height of the trapezoid. Many Thanks.
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PS: Geometry hexagon with circles
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20 Jun 2010, 09:25
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. \(36*(123\pi)\) D. \(36*(2*sqrt(3)3\pi)\) E. None of the above. Attachment:
P1.JPG [ 4.95 KiB  Viewed 21713 times ]
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Re: PS: Geometry hexagon with circles
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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. \(36*(123\pi)\) 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.
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Re: Manhattan GMAT test
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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}



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Re: Manhattan GMAT test
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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: \(6*\sqrt{3}*r^23\pi{r^2}\).
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Re: Manhattan GMAT test
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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



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Re: Regular hexagon ABCDEF has a perimeter of 36
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01 Mar 2013, 23:47
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. 10818pi b. 54sqrt(3)9pi c. 54sqrt(3)18pi d. 10827pi 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
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Re: Regular hexagon ABCDEF has a perimeter of 36
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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. 10818pi b. 54sqrt(3)9pi c. 54sqrt(3)18pi d. 10827pi 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 ?
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Re: Regular hexagon ABCDEF has a perimeter of 36
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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. 10818pi b. 54sqrt(3)9pi c. 54sqrt(3)18pi d. 10827pi 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.....
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Re: Regular hexagon ABCDEF has a perimeter of 36
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02 Mar 2013, 09:32
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*(n2). This means the sum of these angles is \(180*(62)=180*4=360*2\). 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.



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Re: Regular hexagon ABCDEF has a perimeter of 36
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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



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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente
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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?
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente
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10 Jun 2013, 05:22
WarriorGmat wrote: 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 Yes, that's correct.
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Re: Manhattan GMAT test
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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. \(10818\pi\) B. \(54\sqrt{3}9\pi\) C. \(54\sqrt{3}18\pi\) D. \(10827\pi\) 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 306090 right triangle. Height would be the opposite side to the 60 degrees angle (half of 120 degrees). In 306090 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?



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Re: Manhattan GMAT test
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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. \(10818\pi\) B. \(54\sqrt{3}9\pi\) C. \(54\sqrt{3}18\pi\) D. \(10827\pi\) 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 306090 right triangle. Height would be the opposite side to the 60 degrees angle (half of 120 degrees). In 306090 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?
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente
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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: \(6*\sqrt{3}*r^23\pi{r^2}\). 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.



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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente
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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. \(10818\pi\) B. \(54\sqrt{3}9\pi\) C. \(54\sqrt{3}18\pi\) D. \(10827\pi\) 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 306090 right triangle . Height would be the opposite side to the 60 degrees angle (half of 120 degrees). In 306090 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!




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