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# 6 circles exactly surround a circle and all of them are

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6 circles exactly surround a circle and all of them are [#permalink]

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06 May 2006, 05:20
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

6 circles exactly surround a circle and all of them are identical. A right hexagon can be gotten by linking the 6 centres. The raduis of the circle is 3. What is the area of the none-circle part in the hexagon?

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13 May 2006, 20:53
getzgetzu wrote:
6 circles exactly surround a circle and all of them are identical. A right hexagon can be gotten by linking the 6 centres. The raduis of the circle is 3. What is the area of the none-circle part in the hexagon?

i understand the question as "what is the are of circles that are not included in the hexagon".

= 6 (pi r^2)(120/360) = 18 pi

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14 May 2006, 09:15
Professor wrote:
getzgetzu wrote:
6 circles exactly surround a circle and all of them are identical. A right hexagon can be gotten by linking the 6 centres. The raduis of the circle is 3. What is the area of the none-circle part in the hexagon?

i understand the question as "what is the are of circles that are not included in the hexagon".

= 6 (pi r^2)(120/360) = 18 pi

how did you get this prof?

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14 May 2006, 11:36
getzgetzu

CAN U SHOW ME THE PICTURE PLZ?
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14 May 2006, 13:32
Yes without a picture im lost as well

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14 May 2006, 13:51
I understand this as what is the area inside the resuted hexagon
that is not covered by the circles.

I got

54sqrt(3) - 27pi

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14 May 2006, 15:35
trivikram wrote:
Professor wrote:
getzgetzu wrote:
6 circles exactly surround a circle and all of them are identical. A right hexagon can be gotten by linking the 6 centres. The raduis of the circle is 3. What is the area of the none-circle part in the hexagon?

i understand the question as "what is the areas of circles that are not included in the hexagon".
= 6 (pi r^2)(120/360) = 18 pi

how did you get this prof?

update(:wink:):

total angle of a regular hexagone = 180 (6-2) = 720
each angle of a regular hexagone = 180 (6-2)/6 = 120

the covered area of each of 6 circles=120/360(pi r^2)= 1/3(pi 3^2) = 3 pi
the un-covered area of each of 6 circles=240/360(pi r^2)=2/3(pi 3^2)=6 pi
the un-covered area of 6 circles = 6(240/360)(pi 3^2) = 4(pi 3^2) = 36pi

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14 May 2006, 20:29
Area of circle not covered by single corner of hexagon
= pi*r^2 - (120/360)pi*r^2
=2/3 pi*9
= 6pi
Therefore area not covered by the 6 circles = 6 x 6pi
= 36 pi

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15 May 2006, 11:46
deowl wrote:
I understand this as what is the area inside the resuted hexagon
that is not covered by the circles.

I got

54sqrt(3) - 27pi

I understand the question as you do - what is the area of that part of the the hexagon that is not covered by circles. And my answer is approximately the same as yours.

1) The area of each of those circles is 9 pi
We have only one circle that is fully inside the hexagon. The other 6 circles are partially in the hexagon. We have to find what part of those 6 circles is in the hexagon.
2)"6 circles exactly surround a circle" - that means that the 6 circles are tangent to the circle in the middle. Thus, the distance from the center of the middle circle to the centers of the other circles is 2r=6. If we connect the center of the middle circle with the centers of the other 6 circles, we'll see that those segments divide the hexagon into 6 pieces. Since the hexagon is right, those segments are equal.
3) 360/6=60 The angle of vertex O of each of the 6 isosceles triangles is 60 degrees. That means that those triangular parts are actually equilateral triangles. Hence, the sides of the hexagon=6
4)60+60=120 degrees =1/3 of 360
The part of each of the 6 circles that is inside the hexagon is 1/3 of its area=1/3*9pi
5)We have 6 of those circles, thus 6*1/3*9pi=18pi
6)The area of the hexagon that is covvered by circles is
18pi(the area of the parts of the six circles)+9pi(the area of the circle in the middle) = 27pi
7)We can find the area of a right hexagon by finding the area of the six triangles. The altitude of an equilateral triangle divides its side to two equal parts. Hence, we cal find the altitude with the Pitagorean theorem. 3^2 + a^2=6^2
a=5
The area of each of those 6 triangles is 6*5/2=15
since we have 6 of those, the area of hexagon is 15*6=90

8)The area of the hexagon that is not coverred by circles is

90-27pi

or

90-27*22/7=36/7

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17 May 2006, 07:24
What I understood is: The area of the Hexagon which is not part of any circle.

Here I go:

Area of each circle = 9 PI
Interior angle of hexagon = 120
Area of each outer circle that is part of hexagon = (1/3) * 9 PI = 3 PI
Area of hexagon = 6 * 9 * SQRT(3)/4 = 27 * SQRT(3)/2

Area of hexagon which is not part of any circle = 27 * SQRT(3)/2 - 6 * 3 PI - 9 PI

= 27 * ((SQRT(3)/2) - PI)
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17 May 2006, 10:33
deowl wrote:
I understand this as what is the area inside the resuted hexagon
that is not covered by the circles.

I got

54sqrt(3) - 27pi

Me too got this answer. The question asks the area in white color inside the hexagone (What is the area of the non-circle part in the hexagon?). I hope this figure will help.
Attachments

File comment: Hex

Hexagone.JPG [ 6.26 KiB | Viewed 569 times ]

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07 Jul 2006, 07:26
I get 54sqrt3 - 36 pi...

deowl or vivek ,c an you explain how u get 27 pi

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08 Jul 2006, 02:17
ps_dahiya wrote:
What I understood is: The area of the Hexagon which is not part of any circle.

Here I go:

Area of each circle = 9 PI
Interior angle of hexagon = 120
Area of each outer circle that is part of hexagon = (1/3) * 9 PI = 3 PI
Area of hexagon = 6 * 9 * SQRT(3)/4 = 27 * SQRT(3)/2

Area of hexagon which is not part of any circle = 27 * SQRT(3)/2 - 6 * 3 PI - 9 PI

= 27 * ((SQRT(3)/2) - PI)

Area of hexagon = 6 * 36 * SQRT(3)/4 = 27 * SQRT(3)/2
=54 * SQRT(3) as radius is 3 so length of side is 6

the answer should be = 54 * (SQRT(3) - 27PI

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08 Jul 2006, 02:34
bpattnaik wrote:
ps_dahiya wrote:
What I understood is: The area of the Hexagon which is not part of any circle.

Here I go:

Area of each circle = 9 PI
Interior angle of hexagon = 120
Area of each outer circle that is part of hexagon = (1/3) * 9 PI = 3 PI
Area of hexagon = 6 * 9 * SQRT(3)/4 = 27 * SQRT(3)/2

Area of hexagon which is not part of any circle = 27 * SQRT(3)/2 - 6 * 3 PI - 9 PI

= 27 * ((SQRT(3)/2) - PI)

Area of hexagon = 6 * 36 * SQRT(3)/4 = 27 * SQRT(3)/2
=54 * SQRT(3) as radius is 3 so length of side is 6

the answer should be = 54 * (SQRT(3) - 27PI

Yes the bol dpart is incorrect. This should be 6 * 9 * SQRT(3)/2

So answer is 54 * (SQRT(3) - 27PI
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08 Jul 2006, 02:54
ps_dahiya wrote:
bpattnaik wrote:
ps_dahiya wrote:
What I understood is: The area of the Hexagon which is not part of any circle.

Here I go:

Area of each circle = 9 PI
Interior angle of hexagon = 120
Area of each outer circle that is part of hexagon = (1/3) * 9 PI = 3 PI
Area of hexagon = 6 * 9 * SQRT(3)/4 = 27 * SQRT(3)/2

Area of hexagon which is not part of any circle = 27 * SQRT(3)/2 - 6 * 3 PI - 9 PI

= 27 * ((SQRT(3)/2) - PI)

Area of hexagon = 6 * 36 * SQRT(3)/4 = 27 * SQRT(3)/2
=54 * SQRT(3) as radius is 3 so length of side is 6

the answer should be = 54 * (SQRT(3) - 27PI

Yes the bol dpart is incorrect. This should be 6 * 9 * SQRT(3)/2

So answer is 54 * (SQRT(3) - 27PI

ok, I dont understand how you calculate 27pi, I'm sure I'm making a mistake somewhere.

I used the sector formula (pi *r^2 * measure of angle )/ 360 and then multiply by 6 for 6 circles, I get 18pi and not 27pi..can someone please explain if this is the right method and where I'm going wrong.

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08 Jul 2006, 20:10
shahnandan wrote:

ok, I dont understand how you calculate 27pi, I'm sure I'm making a mistake somewhere.

I used the sector formula (pi *r^2 * measure of angle )/ 360 and then multiply by 6 for 6 circles, I get 18pi and not 27pi..can someone please explain if this is the right method and where I'm going wrong.

Look at the picture posted by vivek.

Area of each circle = 9 PI
Interior angle of hexagon = 120
Area of 6 outer circles that are part of hexagon = 6* (1/3) * 9 PI = 18 PI
Area of hexagon = 6 * 36 * SQRT(3)/4 = 54 * SQRT(3)
Area of the circle that is fully inside the hexagon = 9PI

Area of hexagon which is not part of any circle = 54 * SQRT(3) - 18PI- 9PI = 54 * SQRT(3) - 27PI

= 27 * ((SQRT(3)/2) - PI)
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08 Jul 2006, 21:39
thanks dahiya I get it now .

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08 Jul 2006, 21:39
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