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

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

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New post 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. \(108-18\pi\)
B. \(54\sqrt{3}-9\pi\)
C. \(54\sqrt{3}-18\pi\)
D. \(108-27\pi\)
E. \(54\sqrt{3}-27\pi\)

Attachment:
1.JPG
1.JPG [ 32.32 KiB | Viewed 29925 times ]

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  [#permalink]

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


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.
Image
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.

Attachment:
hexagon20and20circles.jpg
hexagon20and20circles.jpg [ 7.6 KiB | Viewed 20348 times ]

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

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New post 28 Aug 2014, 23:26
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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.
= \(54\sqrt{3} - 27 * pie\)

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

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New post Updated on: 22 Jan 2010, 12:10
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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
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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: Regular hexagon ABCDEF has a perimeter of 36. O is the cente  [#permalink]

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

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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: \(6*\sqrt{3}*r^2-3\pi{r^2}\).
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente  [#permalink]

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New post 01 Mar 2013, 23:47
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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
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente  [#permalink]

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

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New post 03 Nov 2014, 02:11
russ9 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.



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!


1. Check the diagram below:
Attachment:
Untitled.png
Untitled.png [ 25.77 KiB | Viewed 4220 times ]

2. Cannot follow you. How did you get that the hypotenuse, which is the side of the hexagon, equals to \(\sqrt{72}\)?
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente  [#permalink]

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New post 01 Aug 2016, 10:15
gmatjon wrote:
Attachment:
The attachment 1.JPG is no longer available
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. \(54sqrt{3}-27\pi\)


Please check the explanation in the attached file
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente  [#permalink]

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New post 21 Oct 2016, 23:13
Hello ,

Just adding this as an alternative approach which i think would be pretty easy to get through :

Consider :

Regular Hexagon : ABCDEF

thus Equilateral triangle : ABO (See attached image )

we need to calculate the shaded portion so what we can do is Calculate the area of equilateral triangle ABO and deduct the area of smaller quadrants within that triangle (that are formed by extending the radius of all the triangle's) : so here we will deduct quadrant made from Circle A , Circle B and Circle O from the bigger equlilateral triangle and hence we can conclude the remaining area !

Now : as regular hexagon (ab=6) thus (a+b / 2) =3 the radius of all the circle

Area of ABO (bigger equilateral triangle ) = [[square_root]3 /4 ]s*s so ([square_root]3 / 4) *6 *6 = 9 [square_root]3

Area of 1 smaller quadrant : pie * r^2 / 6 ( because if we extend all the radius of adjacent circles to O six equal quadrants would be formed because ABCDEF is a regular hexagon )
thus : pie *3^2 /6 = 3pie /2

now as 3 quadrants the total area would be (3pie /2) * 3 = 9 pie /2

no subtract 3 quadrant area from the total area of equilateral triangle to get the area of shaded portion :

subtract : 9 [square_root]3 - 9pie /2 = [18 [square_root]3 - 9 pie /2 ] multiply this into 6 to get the area of all the shaded portions thus

54 [square_root]3 - 27 pie .... Option d

Sorry for pie ! :-D

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

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New post 22 Oct 2016, 01:39
While this Q can be solved in multiple ways one of which includes utilizing knowledge of trigonometry, the most intuitive way is visualizing the diagram and solving for it.

(1) Area of a sector = Pi *R^2 (Arc angle /360)
= Pi *3^3 (120/360)
= 3*Pi

Area of 6 sectors = 6 * ( 3*Pi) = 18*Pi

(2) Area of innermost circle = Pi * R^2 = Pi * 3^2 = 9* Pi

(3) Area of hexagon = 6 * (Area of equilateral triangles)
= 6 * (Sq Rt 3/4 * (side)^2 )
= 6 * (Sq Rt 3/4 * 36 )
= 6 * (9 Sq Rt 3)
= 54 Sq Rt 3

Therefore , area of shaded part = Area of hexagon - (area of 6 sectors + Area of innermost circle)
= 54 Sq Rt 3 - (18*Pi +9* Pi )
= 54 Sq Rt 3 - 27* Pi

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

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New post 31 Jul 2019, 04:36
Bunuel wrote:
mirzohidjon wrote:
Image
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. \(54\sqrt{3}-27\pi\)


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

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.

Bunuel
I have a doubt
here ab=6cm
so why we are taking both circles or rather all to be 3cm radius
why cannot it be like circle a has radius 2 and b has 4
also what is use of the information given in qsn " If each circle is tangent to the two circles adjacent to it and to circle O," plz reply
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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente  [#permalink]

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New post 14 Sep 2019, 11:57
given hexagaon of perimeter 36 has side =6
and a hexagon has 6 equilateral ∆'s so the area of the hexagon ; 6*36*√3/4 ; 54√3
to find the area of shaded region we see that for the 6 circles each has radius of 3 and is at 120* so each circle has area coverage under hexagon of 120/360 * pi * 3^2 = 3pi
since 6 circles are present ; 6*3Pi = 18 pi
and area of central circle of radius 3 (O) ; 9pi
we get \(54\sqrt{3}-27\pi\) shaded region area.
IMO E


gmatjon wrote:
Image
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. \(54\sqrt{3}-27\pi\)

Attachment:
1.JPG

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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Re: Regular hexagon ABCDEF has a perimeter of 36. O is the cente   [#permalink] 14 Sep 2019, 11:57
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