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# A semicircle of diameter 1 sits at the top of a semicircle of diameter

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A semicircle of diameter 1 sits at the top of a semicircle of diameter  [#permalink]

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19 Mar 2019, 21:34
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40% (02:53) correct 60% (03:22) wrong based on 5 sessions

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A semicircle of diameter 1 sits at the top of a semicircle of diameter 2, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.

(A) $$\frac{\pi}{6} - \frac{\sqrt{3}}{4}$$

(B) $$\frac{\sqrt{3}}{4} - \frac{\pi}{12}$$

(C) $$\frac{\sqrt{3}}{4} - \frac{\pi}{24}$$

(D) $$\frac{\sqrt{3}}{4} + \frac{\pi}{24}$$

(E) $$\frac{\sqrt{3}}{4} + \frac{\pi}{12}$$

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ee832dfa1925af1fe40d6af62243894f0614e222.png [ 9.22 KiB | Viewed 209 times ]

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Re: A semicircle of diameter 1 sits at the top of a semicircle of diameter  [#permalink]

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19 Mar 2019, 22:30
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Bunuel wrote:

A semicircle of diameter $1$ sits at the top of a semicircle of diameter $2$, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.

(A) $$\frac{\pi}{6} - \frac{\sqrt{3}}{4}$$

(B) $$\frac{\sqrt{3}}{4} - \frac{\pi}{12}$$

(C) $$\frac{\sqrt{3}}{4} - \frac{\pi}{24}$$

(D) $$\frac{\sqrt{3}}{4} + \frac{\pi}{24}$$

(E) $$\frac{\sqrt{3}}{4} + \frac{\pi}{12}$$

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The attachment ee832dfa1925af1fe40d6af62243894f0614e222.png is no longer available

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Re: A semicircle of diameter 1 sits at the top of a semicircle of diameter   [#permalink] 19 Mar 2019, 22:30
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