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# Three semicircles of radius 1 are constructed on diameter AB of a semi

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Math Expert
Joined: 02 Sep 2009
Posts: 54372
Three semicircles of radius 1 are constructed on diameter AB of a semi  [#permalink]

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20 Mar 2019, 23:21
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Difficulty:

55% (hard)

Question Stats:

42% (02:42) correct 58% (02:32) wrong based on 12 sessions

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Three semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2. The centers of the small semicircles divide AB into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?

(A) $$\pi - \sqrt{3}$$

(B) $$\pi - \sqrt{2}$$

(C) $$\frac{\pi + \sqrt{2}}{2}$$

(D) $$\frac{\pi + \sqrt{3}}{2}$$

(E) $$\frac{7\pi}{6} - \frac{\sqrt{3}}{2}$$

Attachment:

bc697df5c9078d8ccf0708f353185c707d3275d4.png [ 12.92 KiB | Viewed 306 times ]

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Re: Three semicircles of radius 1 are constructed on diameter AB of a semi  [#permalink]

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21 Mar 2019, 01:15
Bunuel wrote:

Three semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2. The centers of the small semicircles divide AB into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?

(A) $$\pi - \sqrt{3}$$

(B) $$\pi - \sqrt{2}$$

(C) $$\frac{\pi + \sqrt{2}}{2}$$

(D) $$\frac{\pi + \sqrt{3}}{2}$$

(E) $$\frac{7\pi}{6} - \frac{\sqrt{3}}{2}$$

Attachment:
bc697df5c9078d8ccf0708f353185c707d3275d4.png

not able to completely solve the question well ..
have managed to get 2 equilateral triangles and area of 1 semicircle , but not able to move forward then
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Joined: 09 Jun 2017
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Three semicircles of radius 1 are constructed on diameter AB of a semi  [#permalink]

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26 Mar 2019, 23:01
i spend too much time on this , it's very simple though
after getting 2 equilateral tringles , we can see that the white area consist of 2 x and 2y and m
x is 1/3 the area of one of the small circles ( since we found a 120 angle )
each y is equilateral tringle, so y= √3 /4 , 2y=√3 /2
same for m
so , now we have : white area =5/6 π +( √3 /2 )
the shaded area is 2π - 5π/6 -( √3 /2 ) =7π/6 -( √3 /2 )
Attachments

.PNG [ 160.07 KiB | Viewed 141 times ]

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Three semicircles of radius 1 are constructed on diameter AB of a semi   [#permalink] 26 Mar 2019, 23:01
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