What is the area between two circles, radius one, that go through each other's centres?Not sure if this requires calculus or not. I couldn't work it out. *

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- Nov 2nd 2009, 04:54 AMAquafinaArea between Circles
**What is the area between two circles, radius one, that go through each other's centres?**Not sure if this requires calculus or not. I couldn't work it out. *

- Nov 2nd 2009, 05:25 AMearboth
- Nov 2nd 2009, 06:45 AMAquafina
- Nov 2nd 2009, 09:50 AMukorov
referring to the attached pic:

both triangles ABC and DBC can be proved equilateral and congruent to each other.

the area of__sector__ABC

= (1/6)(pi)(1^2)

= pi/6

and the area of triangle ABC

= (1/2)(AB)(BC)(sin60)

= (1/2)(1)(1)(3^0.5 / 2)

= 3^0.5 / 4

the difference between these two areas is the area of each one red region (in attachment), let A be it.

hence, the total area of overlapped region of the two circles

= area of triangle ABC x 2 + 4A

= 3^0.5 /4 x 2 + 4(pi/6 - 3^0.5 / 4)

= 1.2284 sq. units - Nov 2nd 2009, 12:00 PMearboth
1. The area of the sector (That's the area with the blue borderline) is

$\displaystyle a_{sector} = \dfrac16 \cdot \pi \cdot 1^2 = \dfrac16 \cdot \pi$

2. The area of the right triangle (That's the lightgrey area) is

$\displaystyle a_{triangle} = \dfrac12 \cdot \dfrac12 \cdot \dfrac12 \cdot \sqrt{3} = \dfrac18 \cdot \sqrt{3}$

3. The white area is

$\displaystyle a_{quart} = a_{sector} - a_{triangle} = \dfrac16 \pi - \dfrac18 \sqrt{3}$

4. The area in question is

$\displaystyle a = 4 \cdot a_{quart} = \dfrac23 \pi - \dfrac12 \sqrt{3} \approx 1.22837$