Three congruent circles overlap in such a way that each circ

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.

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.

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?