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In the diagram above, A & B are the centers of the two circles, each w

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In the diagram above, A & B are the centers of the two circles, each w  [#permalink]

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New post 08 Mar 2015, 19:29
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In the diagram above, A & B are the centers of the two circles, each with radius r = 6, and ∠A = ∠B = 60°. What is the area of the shaded region?

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Attachment:
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Re: In the diagram above, A & B are the centers of the two circles, each w  [#permalink]

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New post 08 Mar 2015, 21:33
1
area of circle= pi r2
angle 60 degree , radius=6
then area of sector =1/6 pi r2
=36/6pi =6pi
area of equilateral triangle.=root 3/4a2=
root3/4*36=9root3
Area of shaded region= area of sector- area of equilateral triangle.
=6pi-9rrot3.
here two circles inscribed and equal radius 6
so area of unshaded region=2*(6pi-9root3)
=12pi-18root3
option A sufficient
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Re: In the diagram above, A & B are the centers of the two circles, each w  [#permalink]

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New post 09 Mar 2015, 00:18
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Hi Everyone

And dear Psiva00734 your have calculated the area of unshaded region whilst the question asks for the shaded

As shown in the diagram below triangles ACD & BCD are equilateral.

No considering circle 1
Area of the sector= (60/360)*πr^2 = 6π
Area of equilateral triangle ACD = (√3a^2)4=9√3
Area of unshaded(yellow) part = 6π - 9√3

Now total unshaded area = twice of 6π - 9√3 =12π - 18√3

Area of quadrilateral ABCD= Sum of areas of both the triangles = 9√3 + 9√3 =18√3

Now Area of shaded part = Area of quadrilateral - unshaded area = 18√3 - (12π - 18√3) = 36√3 -12π
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Re: In the diagram above, A & B are the centers of the two circles, each w  [#permalink]

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New post 09 Mar 2015, 02:58
D for me .. It took me almost 5 min to complete
Good question
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Re: In the diagram above, A & B are the centers of the two circles, each w  [#permalink]

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New post Updated on: 09 Mar 2015, 18:15
1
Sorry, didn't follow the shaded region properly.

Option D.
Required area
=2*triangle area-2*(area of sector-area of triangle in each circle)
=2*sqrt(3)/4*side^2-2*(sector angle/total circle angle*pi*side^2 - sqrt(3)/4*side^2)
=2*sqrt(3)/4*side^2-2*(60/360*pi*6^2-sqrt(3)/4*6^2)
=18*sqrt(3)-(12*pi-18*sqrt(3))
=36*sqrt(3)-12*pi

Originally posted by VSabc on 09 Mar 2015, 14:16.
Last edited by VSabc on 09 Mar 2015, 18:15, edited 1 time in total.
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Re: In the diagram above, A & B are the centers of the two circles, each w  [#permalink]

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New post 09 Mar 2015, 17:52
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Bunuel wrote:
Attachment:
gpp_img5.png
In the diagram above, A & B are the centers of the two circles, each with radius r = 6, and ∠A = ∠B = 60°. What is the area of the shaded region?
Attachment:
gpp_img6.png


The area of 1/6 of a circle is 6pi.

The area of the equilateral triangle is 9sqrt(3) which is found by (1/2)*(6)*sqrt(27)

subtracting the equilateral triangle from the one-sixth portion of the circle gives us the small unshaded portion, which is 6pi-9sqrt(3).

The area of shaded region can be found by:
2 equilateral triangles - 2 small unshaded portions.
2*9sqrt(3)-2(6pi-9sqrt(3))
=36sqrt(3)-12pi

Answer: D
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Re: In the diagram above, A & B are the centers of the two circles, each w  [#permalink]

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New post 09 Mar 2015, 21:28
2
Bunuel wrote:
Attachment:
gpp_img5.png
In the diagram above, A & B are the centers of the two circles, each with radius r = 6, and ∠A = ∠B = 60°. What is the area of the shaded region?
Attachment:
gpp_img6.png


Area of sector ACD =\(Q/360 * \pi * r^2 = 60/360 * \pi * 6^2 = 6\pi\)
Area of triangle ACD \(= \sqrt{3}/4 * side^2 = \sqrt{3}/4 * 6^2 = 9\sqrt{3}\)

Half of the unshaded region in middle \(= 6\pi - 9\sqrt{3}\)

Area of shaded region = 2 * (Area of triangle ACD - Half of the unshaded region in middle)
\(= 2 * 9\sqrt{3} - 2*6\pi + 2*9\sqrt{3} = 36\sqrt{3} - 12\pi\)

Answer (D)
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Re: In the diagram above, A & B are the centers of the two circles, each w  [#permalink]

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New post 15 Mar 2015, 19:44
Bunuel wrote:
Image
In the diagram above, A & B are the centers of the two circles, each with radius r = 6, and ∠A = ∠B = 60°. What is the area of the shaded region?
Attachment:
The attachment gpp_img6.png is no longer available


MAGOOSH OFFICIAL SOLUTION:

Step #1: Each circle has an area of \(A = \pi{r^2}=\pi{(6^2)}=36\pi\).

Step #2: One sector, one “slice of pie”, occupies 60°, one sixth of the circle.
Attachment:
gpp_img13.png
gpp_img13.png [ 11.52 KiB | Viewed 6272 times ]


Each sector has an area of \(A = 6\pi\).

Step #3: Now look at the equilateral triangle.
Attachment:
gpp_img15.png
gpp_img15.png [ 11.5 KiB | Viewed 6272 times ]


This has a side of s = 6, so it’s area is \(A = 9\sqrt{3}\).

Step #4: A circular segment is the name for that little leftover piece, the part of the sector that’s beyond the triangle:
Attachment:
gpp_img17.png
gpp_img17.png [ 12.02 KiB | Viewed 6272 times ]


As may be clear, the (area of segment) = (area of sector) – (area of triangle), so the sector has an area of \(A = 6\pi - 9\sqrt{3}\).

Step #5: Now, notice that the shaded region in the diagram is two equilateral triangles minus two circular segments. That’s exactly as we’ll calculate it.
\(A =2(9\sqrt{3})-2( 6\pi - 9\sqrt{3})=36\sqrt{3}-12\pi\).

Answer = (D)
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Re: In the diagram above, A & B are the centers of the two circles, each w  [#permalink]

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New post 24 Mar 2015, 14:25
Bunuel wrote:
Bunuel wrote:
Image
In the diagram above, A & B are the centers of the two circles, each with radius r = 6, and ∠A = ∠B = 60°. What is the area of the shaded region?
Attachment:
gpp_img6.png


MAGOOSH OFFICIAL SOLUTION:

Step #1: Each circle has an area of \(A = \pi{r^2}=\pi{(6^2)}=36\pi\).

Step #2: One sector, one “slice of pie”, occupies 60°, one sixth of the circle.
Attachment:
gpp_img13.png


Each sector has an area of \(A = 6\pi\).

Step #3: Now look at the equilateral triangle.
Attachment:
gpp_img15.png


This has a side of s = 6, so it’s area is \(A = 9\sqrt{3}\).

Step #4: A circular segment is the name for that little leftover piece, the part of the sector that’s beyond the triangle:
Attachment:
gpp_img17.png


As may be clear, the (area of segment) = (area of sector) – (area of triangle), so the sector has an area of \(A = 6\pi - 9\sqrt{3}\).

Step #5: Now, notice that the shaded region in the diagram is two equilateral triangles minus two circular segments. That’s exactly as we’ll calculate it.
\(A =2(9\sqrt{3})-2( 6\pi - 9\sqrt{3})=36\sqrt{3}-12\pi\).

Answer = (D)


I am clear about how we get the area of segment but the total area calculation (highlighted part) is still confusing. Why 2* area of segment subtracted from 2*Area of Triangles.
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Re: In the diagram above, A & B are the centers of the two circles, each w  [#permalink]

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New post 24 Mar 2015, 22:23
earnit wrote:
Bunuel wrote:
Bunuel wrote:
Image
In the diagram above, A & B are the centers of the two circles, each with radius r = 6, and ∠A = ∠B = 60°. What is the area of the shaded region?
Attachment:
gpp_img6.png


MAGOOSH OFFICIAL SOLUTION:

Step #1: Each circle has an area of \(A = \pi{r^2}=\pi{(6^2)}=36\pi\).

Step #2: One sector, one “slice of pie”, occupies 60°, one sixth of the circle.
Attachment:
gpp_img13.png


Each sector has an area of \(A = 6\pi\).

Step #3: Now look at the equilateral triangle.
Attachment:
gpp_img15.png


This has a side of s = 6, so it’s area is \(A = 9\sqrt{3}\).

Step #4: A circular segment is the name for that little leftover piece, the part of the sector that’s beyond the triangle:
Attachment:
gpp_img17.png


As may be clear, the (area of segment) = (area of sector) – (area of triangle), so the sector has an area of \(A = 6\pi - 9\sqrt{3}\).

Step #5: Now, notice that the shaded region in the diagram is two equilateral triangles minus two circular segments. That’s exactly as we’ll calculate it.
\(A =2(9\sqrt{3})-2( 6\pi - 9\sqrt{3})=36\sqrt{3}-12\pi\).

Answer = (D)


I am clear about how we get the area of segment but the total area calculation (highlighted part) is still confusing. Why 2* area of segment subtracted from 2*Area of Triangles.


Refer diagram of Step 4 don by Bunuel

Twice the shaded area has to be removed from the area of rhombus to get the desired answer
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Re: In the diagram above, A & B are the centers of the two circles, each w  [#permalink]

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New post 24 Mar 2015, 23:02
Attachment:
rhombus.png
rhombus.png [ 29.04 KiB | Viewed 6139 times ]


Area of rhombus \(= \frac{\sqrt{3}}{2} * 6^2 = 18\sqrt{3}\)

Attachment:
gpp_img5.png
gpp_img5.png [ 53.59 KiB | Viewed 6132 times ]


Area of shaded region (each) = Area of sector - Area of equilateral triangle

\(= \frac{\pi6^2}{6} - \frac{\sqrt{3}}{4} * 6^2 = 6\pi - 9\sqrt{3}\)

Required shaded region area \(= 18\sqrt{3} - 2(6\pi - 9\sqrt{3}) = 36\sqrt{3} - 12\pi\)

Answer = D
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Re: In the diagram above, A & B are the centers of the two circles, each w  [#permalink]

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New post 04 Jan 2016, 18:33
that's a tough mother F... :D
ok, so how I solved:

area of each circle is 36pi. the two radii from A to C and D form a 1/6 arc, with the length of 6pi. the same is true for the arc BCD.
so the total area of both regions is something below 12pi.
now, we can create 4 right triangles if we draw 2 lines : from C to D, and from A to B. Angles A and B are bisected, and since both angles are 60 degrees, we have created four angles of 30 degrees, or four right 30-60-90 triangles.
the hypotenuse is 6, thus, knowing that the sides of 30-60-90 triangle are in x-x sqrt3 -2x ratio, we can find the legs. thus, the legs are 3 and sqrt 3.
area of each triangle is 4.5 * sqrt(3). or for all four triangles, 18*sqrt(3).

now, the area of the not shaded region must be: 12pi - 18*sqrt(3).

therefore, the area of the shaded region must be:
the area of the triangles - the area of the unshaded region, which is an overlap.
thus, 18*sqrt(3) - [12pi - 18*sqrt(3)] = 18*sqrt(3)+18*sqrt(3) - 12pi. or 36*sqrt(3) - 12pi.

ps. we can see as well that we can create 2 equilateral triangles, since angles C and D, when the line from C to D is drawn, are 60 degrees. Knowing the property of the equilateral triangle: area = s^2 * sqrt(3) /4 = we can find the area of each equilateral triangle.
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Re: In the diagram above, A & B are the centers of the two circles, each w  [#permalink]

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Re: In the diagram above, A & B are the centers of the two circles, each w &nbs [#permalink] 08 Apr 2018, 14:52
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