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)
Three congruent circles overlap in such a way that each circ
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- Anaira Mitch
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- Anaira Mitch
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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.
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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Area of an equilateral triangle = (s²/4)√3.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)
√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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- fskilnik@GMATH
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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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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- Anaira Mitch
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Hello Mitch,GMATGuruNY wrote:Area of an equilateral triangle = (s²/4)√3.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)
√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.
Can you tell me what went wrong with my approach?
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The statement above is incorrect.Anaira Mitch wrote:The circle is the circumscribed circle of an equilateral triangle.
The shaded region encompassing the equilateral triangle is not a circle.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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