Area between Circles

Aquafina · #1 $Old$ November 2nd, 2009, 06:54 AM

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. *

earboth · #2 $Old$ November 2nd, 2009, 07:25 AM

Quote:

Originally Posted by Aquafina $View Post$

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. *

One quarter of the area in question is the difference between a sector (radius = 1, central angle = 60°) and a right triangle. (see attachment)

Aquafina · #3 $Old$ November 2nd, 2009, 08:45 AM

Quote:

Originally Posted by earboth $View Post$

One quarter of the area in question is the difference between a sector (radius = 1, central angle = 60°) and a right triangle. (see attachment)

Don't you mean the Area of the triangle + the sector?

ukorov · #4 $Old$ November 2nd, 2009, 11:50 AM

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

earboth · #5 $Old$ November 2nd, 2009, 02:00 PM

Quote:

Originally Posted by Aquafina $View Post$

Don't you mean the Area of the triangle + the sector?<<<<< No

1. The area of the sector (That's the area with the blue borderline) is

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

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

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

3. The white area is

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

4. The area in question is

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