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655-705 (Hard)|   Geometry|      
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Bunuel
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Two identical circles of area 36π overlap as shown above. If the dista 14 Sep 2018, 01:16
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58% (02:18) correct 42% (02:45) wrong based on 31 sessions

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Two identical circles of area 36π overlap as shown above. If the distance from point A to point B is 6, what is the area of the shaded region?


A. 6π

B. 12π

C. \(18\sqrt{3} – 6π\)

D. \(6π – 9\sqrt{3}\)

E. \(12π – 18\sqrt{3}\)


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swapnilce.nitdgp
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Two identical circles of area 36π overlap as shown above. If the dista 14 Sep 2018, 02:03
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distance between point A and B is given to assume an equilateral triangle with 6m sides, sinch area = 6^2xpie

Hence, Area of shaded region = 2(area of 60 deg sector - Area of Equilateral Triangle )
Ans = 12pie-18sqrt3
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Two identical circles of area 36π overlap as shown above. If the dista 16 Sep 2018, 08:11
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swapnilce.nitdgp
distance between point A and B is given to assume an equilateral triangle with 6m sides, sinch area = 6^2xpie

Hence, Area of shaded region = 2(area of 60 deg sector - Area of Equilateral Triangle )
Ans = 12pie-18sqrt3
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Hi,
Can someone please provide a lil more detailed explanation...
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Derapage
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Two identical circles of area 36π overlap as shown above. If the dista 16 Sep 2018, 10:16
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mischiefmanaged
swapnilce.nitdgp
distance between point A and B is given to assume an equilateral triangle with 6m sides, sinch area = 6^2xpie

Hence, Area of shaded region = 2(area of 60 deg sector - Area of Equilateral Triangle )
Ans = 12pie-18sqrt3

Hi,
Can someone please provide a lil more detailed explanation...
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ibb.co/iyL3zz (copy-paste this to see the picture I drew for you, it won't let me attach it for some reason...)

Basically, AB = 6 which happens to be = r (you can reverse engineer "r" looking at the area that is 36pi. Area of a circle = pi*r^2, therefore 36*pi means that r = 6).

You can now build an equilateral triangle with side = 6, and find its area since it's a 30-60-90 triangle.

You can also compute the "sector's area" using the formula in the image I posted (it's just a proportional fraction of the whole circle's area).

Now you can find the difference between the two, which will give you the area of half the shaded region. Multiply that by 2 and you get the result.
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Two identical circles of area 36π overlap as shown above. If the dista 16 Sep 2018, 12:39
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For only one circle

Area = 36\(\pi\)
So Radius = 6
Distance between A to B = 6

So Origin,A & B form a equilateral Triangle
Area of \(\triangle\) = 9\(\sqrt{3}\)

\(\angle\) AOB= 60\(^{o}\) , Since Equilateral Triangle [ O= Origin]

Area of Circular segment OAB =(36\(\pi\) / \(360^{\circ}\)) x \(60^{\circ}\) = 6\(\pi\)

Shaded Portion by only one Circle = 6\(\pi\)-9\(\sqrt{3}\)

Shaded Portion by two circle = 12\(\pi\)-18\(\sqrt{3}\)
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