Either of two unit circles passes through the center of the other. How do you prove that the common area is $\frac{2}{3} π - \frac{\sqrt{3}}{2}$ $2/3pi-sqrt 3/2$ areal units?

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Answer 1 · 2016-09-20T05:11:23

Proved in the explanation

The common area is enclosed by unit circle arcs subtending

$∠ 120^{o} = \frac{2}{3} π$ $angle120^o=2/3pi$ radian, at the respective centers.

For this arc, sector area of a unit circle

$= \frac{\frac{2}{3} π}{2 π}$ $=(2/3pi)/(2pi)$ (area of unit circle)=(2/3pi)/(2pi)(pi)=pi/3#

One half of the common area

= this sector area less area of the inner triangle, of sides $[1, \sqrt{3}, 1]$ $[1 ,sqrt 3, 1]$

$= \frac{π}{3} - (\frac{1}{2}) (\sqrt{3}) (\frac{1}{2})$ $=pi/3-(1/2)(sqrt3)(1/2)$

Twice this is the common area = $\frac{2}{3} π - \frac{\sqrt{3}}{2}$ $2/3pi-sqrt 3/2$ areal units.

I welcome a graphical depiction,.from an interested reader.

enter image source here

Answer 2 · 2016-09-20T07:10:53

see explanation.

enter image source here

1) Equilateral triangle area $A_{e} = \sqrt{\frac{3}{4}} \times 1^{2} = \frac{\sqrt{3}}{4}$ $A_e = sqrt((3)/4)xx1^2=sqrt3/4$

2) yellow area $A_{y} = \frac{π}{6} - \frac{\sqrt{3}}{4} = \frac{2 π - 3 \sqrt{3}}{12}$ $A_y=pi/6-sqrt3/4 = (2pi-3sqrt3)/12$

3) Common area $A_{c} = 2 A_{e} + 4 A_{y}$ $A_c=2A_e+4A_y$

$A_{c} = 2 \times \frac{\sqrt{3}}{4} + 4 (\frac{2 π - 3 \sqrt{3}}{12})$ $A_c=2xxsqrt3/4+4((2pi-3sqrt3)/12)$

$= \frac{\sqrt{3}}{2} + \frac{2 π - 3 \sqrt{3}}{3}$ $=sqrt3/2+(2pi-3sqrt3)/3$

$= \frac{3 \sqrt{3} + 4 π - 6 \sqrt{3}}{6}$ $=(3sqrt3+4pi-6sqrt3)/6$

$= \frac{4 π - 3 \sqrt{3}}{6} = 2 \frac{π}{3} - \frac{\sqrt{3}}{2}$ $=(4pi-3sqrt3)/6 =2pi/3-sqrt3/2$

Either of two unit circles passes through the center of the other. How do you prove that the common area is 23π−√322/3pi-sqrt 3/2 areal units?