When you place three equal circles such that they’re all touching each other, there’s a little triangular (sorta) shape in the center. Also, what is a quick proof of the area.
(I’m not a high school student looking for a homework answer. The question just popped into my head out of the blue, and I couldn’t find it in my “Universal Encyclopedia of Mathematics.”)
I’m not going to do it but I think the answer is pretty simple. You have an equilateral traingle with side = 2*r . Calculating the part of one circle contained in the traingle is pretty straightforward. Subtract three times this value form the area of the triangle and that’s your answer.
If these circles aren’t of equal radius, I’ve got no clue. However, if they are. the area isn’t bad. Find the equilateral triangle that this circle can be inscribed in, and compute the area of that. Subtract the area of the circle off, and you’ve got it.
And yes, I can spell Triangle if I type slowly. maybe I should learn to preview more often.
I don’t know a name for the figure, but I’d get the area this way:
Take the area of the equilateral triangle connecting the centers of the circles, and subtract the area of the three 60 degree sectors.
So (1/2)base altitude - (3/6) Pi r^2
or (sqrt(3)/2)*r - (1/2)pir^2
If the circles have different radii you can use the same method of finding the area of the triangle connecting the centers of the cirlces and subtracting the areas of the sectors. The sides of the triangle are r1 + r2, r1 + r3, and r2 + r3, and you can use the law of sines and the law of cosines to get the angles. There are a bunch of formulas for the triangle’s area, I’m not sure which one is best here (maybe .5absin(angle C)), and the area of each sector is pir^2*angle/360. I don’t know what you end up with, but you can get it this way. I also don’t know what the shape is called.