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Three congruent circles overlap in such a way that each circ

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Three congruent circles overlap in such a way that each circ

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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. 16sqrt3

B. 32(pie - sqrt3)

C. 16(pie + sqrt3)

D. 32 pie

E. 32(pie + sqrt3)


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Anaira Mitch wrote:
The circle is the circumscribed circle of an equilateral triangle.
The statement above is incorrect.

The shaded region encompassing the equilateral triangle is not a circle.

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Anaira Mitch 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. 16sqrt3

B. 32(pie - sqrt3)

C. 16(pie + sqrt3)

D. 32 pie

E. 32(pie + sqrt3)
Area of an equilateral triangle = (s²/4)√3.
√3 ≈ 1.7.



Each circle has a radius of 8, yielding the equilateral triangle above.
Area of the equilateral triangle = (8²/4)√3 = 16√3 ≈ (16)(1.7) = 27.
The three circles form an overlap that extends a little beyond the equilateral triangle.
Thus, the area of the overlap must be a little more than 16√3 ≈ 27.
Eliminate A, since the overlap must have an area greater than 16√3.
Of the four remaining answer choices, only B yields a value a little more than 27:
32(π - √3) ≈ 32(3 - 1.7) = 32*1.3 ≈ 41.

The correct answer is B.

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Mitch Hunt
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GMATGuruNY@gmail.com

If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.

Available for tutoring in NYC and long-distance.
For more information, please email me at GMATGuruNY@gmail.com.
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My Approach:

The circle is the circumscribed circle of an equilateral triangle.
Try to find median of Equilateral triangle( median=altitude=bisector).
median square + half side square= full side square
median square=64-16=48
median= 4 underoot 3

no,we knowinan equilateral triangle the centroid is divided in ratio of 2/3.

therefore radius ofcircumscribed circle is 2/3 of 4 underroot 3

r=8 / underroot 3

Now area of that circle= pie*64/3

which is not an option.

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Anaira Mitch 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. 16sqrt3

B. 32(pi - sqrt3)

C. 16(pi + sqrt3)

D. 32 pi

E. 32(pi + sqrt3)




$$? = {S_{\Delta {\rm{equil}}}} + \,\,3 \cdot \,{S_{{\rm{blue}}}}$$
$${S_{\Delta {\rm{equil}}}} = {{{r^{\,2}}\sqrt 3 } \over 4}\,\,\,\mathop = \limits^{r\, = \,8} \,\,16\sqrt 3 $$
$${S_{{\rm{blue}}}} = {{60} \over {360}}\left( {\pi \cdot {8^2}} \right) - {S_{\Delta {\rm{equil}}}} = {{\pi \cdot {8^2}} \over 6} - \,16\sqrt 3 = 8\left( {{{4\pi } \over 3} - 2\sqrt 3 } \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,3 \cdot \,{S_{{\rm{blue}}}} = 8\left( {4\pi - 6\sqrt 3 } \right)$$
$$? = \,\,16\sqrt 3 + 8\left( {4\pi - 6\sqrt 3 } \right) = 32\left( {\pi - \sqrt 3 } \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\rm{B}} \right)$$


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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GMATGuruNY wrote:
Anaira Mitch 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. 16sqrt3

B. 32(pie - sqrt3)

C. 16(pie + sqrt3)

D. 32 pie

E. 32(pie + sqrt3)
Area of an equilateral triangle = (s²/4)√3.
√3 ≈ 1.7.



Each circle has a radius of 8, yielding the equilateral triangle above.
Area of the equilateral triangle = (8²/4)√3 = 16√3 ≈ (16)(1.7) = 27.
The three circles form an overlap that extends a little beyond the equilateral triangle.
Thus, the area of the overlap must be a little more than 16√3 ≈ 27.
Eliminate A, since the overlap must have an area greater than 16√3.
Of the four remaining answer choices, only B yields a value a little more than 27:
32(π - √3) ≈ 32(3 - 1.7) = 32*1.3 ≈ 41.

The correct answer is B.
Hello Mitch,

Can you tell me what went wrong with my approach?

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