When you draw three circles of radius r such that all three are externally tangent, what is the area of the shape in the center in terms of r?

Sorry, but they don't seem to want to activate the solution link.

sqrt(3)r^2-pi(r^2)/2

This solution is achieved by drawing simple math.

Take the distance between the centers of two circles (2r) and make it into the sides of an equilateral triangle.

You now find the height of the triangle: sqrt((2r)^2-(r^2)) = sqrt(3)r.

Now find the area of the triangle by subbing in sqrt(3)r as h. b=2r, so A=2sqrt(3)r^2/2 so A=sqrt(3)r^2.

Subtract the three sixths of the circles(each angle is 60 degrees), to get sqrt(3)r^2-pi(r^2)/2.

Posted by Justin on 2006-01-20 09:09:02