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In the diagram above, A & B are the centers of the two circles, each w [#permalink]
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08 Mar 2015, 20:29
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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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08 Mar 2015, 22:33
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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. =6pi9rrot3. here two circles inscribed and equal radius 6 so area of unshaded region=2*(6pi9root3) =12pi18root3 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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09 Mar 2015, 01:18
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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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09 Mar 2015, 03:58
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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Updated on: 09 Mar 2015, 19:15
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Sorry, didn't follow the shaded region properly.
Option D. Required area =2*triangle area2*(area of sectorarea of triangle in each circle) =2*sqrt(3)/4*side^22*(sector angle/total circle angle*pi*side^2  sqrt(3)/4*side^2) =2*sqrt(3)/4*side^22*(60/360*pi*6^2sqrt(3)/4*6^2) =18*sqrt(3)(12*pi18*sqrt(3)) =36*sqrt(3)12*pi
Originally posted by VSabc on 09 Mar 2015, 15:16.
Last edited by VSabc on 09 Mar 2015, 19:15, 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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09 Mar 2015, 18:52
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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 onesixth portion of the circle gives us the small unshaded portion, which is 6pi9sqrt(3). The area of shaded region can be found by: 2 equilateral triangles  2 small unshaded portions. 2*9sqrt(3)2(6pi9sqrt(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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09 Mar 2015, 22:28
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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15 Mar 2015, 20:44
Bunuel wrote: 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 [ 11.52 KiB  Viewed 4693 times ]
Each sector has an area of \(A = 6\pi\). Step #3: Now look at the equilateral triangle. Attachment:
gpp_img15.png [ 11.5 KiB  Viewed 4694 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 [ 12.02 KiB  Viewed 4693 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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24 Mar 2015, 15:25
Bunuel wrote: Bunuel wrote: 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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24 Mar 2015, 23:23
earnit wrote: Bunuel wrote: Bunuel wrote: 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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25 Mar 2015, 00:02
Attachment:
rhombus.png [ 29.04 KiB  Viewed 4563 times ]
Area of rhombus \(= \frac{\sqrt{3}}{2} * 6^2 = 18\sqrt{3}\) Attachment:
gpp_img5.png [ 53.59 KiB  Viewed 4555 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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04 Jan 2016, 19:33
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 306090 triangles. the hypotenuse is 6, thus, knowing that the sides of 306090 triangle are in xx 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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