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# Two identical circles of area 36π overlap as shown above. If the dista

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Math Expert
Joined: 02 Sep 2009
Posts: 50619
Two identical circles of area 36π overlap as shown above. If the dista  [#permalink]

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14 Sep 2018, 00:16
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45% (medium)

Question Stats:

83% (01:45) correct 17% (01:46) wrong based on 18 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. $$12π – 18\sqrt{3}$$

Attachment:

image020.jpg [ 2.53 KiB | Viewed 521 times ]

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Joined: 22 Oct 2016
Posts: 15
GMAT 1: 590 Q47 V25
Re: Two identical circles of area 36π overlap as shown above. If the dista  [#permalink]

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14 Sep 2018, 01:03
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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Joined: 04 Oct 2016
Posts: 8
Re: Two identical circles of area 36π overlap as shown above. If the dista  [#permalink]

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16 Sep 2018, 07:11
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swapnilce.nitdgp wrote:
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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Joined: 16 Sep 2018
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Re: Two identical circles of area 36π overlap as shown above. If the dista  [#permalink]

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16 Sep 2018, 09:16
mischiefmanaged wrote:
swapnilce.nitdgp wrote:
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...

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  [#permalink]

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16 Sep 2018, 11:39
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$$\sqrt{3}$$
Two identical circles of area 36π overlap as shown above. If the dista &nbs [#permalink] 16 Sep 2018, 11:39
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# Two identical circles of area 36π overlap as shown above. If the dista

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