emmak wrote:
Three congruent circles overlap in such a way that each circle intersects the centers of both of the other circles, as shown below. If the radius of each of the circles is 8, what is the area of the central section where all three circles overlap?
A. \(16\sqrt{3}\)
B. \(32(\pi-\sqrt{3})\)
C. \(16(\pi+\sqrt{3})\)
D. \(32\pi\)
E. \(32(\pi+\sqrt{3})\)
My solution (presented below) follows Bunuel´s idea and one of VeritasKarishma´s comments.
I have decided to post it so that readers may study it explicitly:
\(? = {S_{\Delta {\rm{equil}}}} + \,\,3 \cdot \,{S_{{\rm{blue}}}}\)
\({S_{\Delta {\rm{equil}}}} = {{{r^{\,2}}\sqrt 3 } \over 4}\,\,\,\mathop = \limits^{r\, = \,8} \,\,16\sqrt 3\)
\({S_{{\rm{blue}}}} = {{60} \over {360}}\left( {\pi \cdot {8^2}} \right) - {S_{\Delta {\rm{equil}}}} = {{\pi \cdot {8^2}} \over 6} - \,16\sqrt 3 = 8\left( {{{4\pi } \over 3} - 2\sqrt 3 } \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,3 \cdot \,{S_{{\rm{blue}}}} = 8\left( {4\pi - 6\sqrt 3 } \right)\)
\(? = \,\,16\sqrt 3 + 8\left( {4\pi - 6\sqrt 3 } \right) = 32\left( {\pi - \sqrt 3 } \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\rm{B}} \right)\)
Regards,
Fabio.
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