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Three congruent circles overlap in such a way that each circ

by Anaira Mitch » Wed Dec 12, 2018 3:21 pm

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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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by Anaira Mitch » Wed Dec 12, 2018 6:58 pm
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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by GMATGuruNY » Thu Dec 13, 2018 4:19 am
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.

Image

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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by fskilnik@GMATH » Thu Dec 13, 2018 4:46 am
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)

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


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

Regards,
Fabio.
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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by Anaira Mitch » Thu Dec 13, 2018 6:06 am
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.

Image

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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circles

by GMATGuruNY » Thu Dec 13, 2018 7:16 am
Anaira Mitch wrote:The circle is the circumscribed circle of an equilateral triangle.
The statement above is incorrect.
Image
The shaded region encompassing the equilateral triangle is not a circle.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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