Two identical circles overlap such that the center point of one circle is on the edge of the other circle If the overlapping area is exactly 20000 square centimetres what is the radius of the circles?

Answer 1

One way to do this problem is as follows:

Let X represent the area of the overlap (which we know is 20000. Let S represent the area in one circle that is outside of the overlapping area (it is best to draw yourself a picture). Because the circles are identical, there is a symmetry such that the S is the same in each circle.

Now create a two equilateral triangles with sides of length r in the overlap, one going up and one going down. If you draw the figure correctly, in the overlap portion will comprise the two equilateral triangles, as well as four identical regions outside the triangles but in the overlap. Let Y represent the area of each of those regions.

The area of each of the equilateral triangles is (sqrt(3)/4)*r2 (For an equilateral triangle the height is (sqrt(3)/2) times the length of one of the sides, which can be found using the Pythagorean theorem and observing that the base of a right triangle is half the length of a side, and they hypotenuse is the length of the side.)

Now, for two equations that can be solved to find the radius, r:

1. As mentioned above, the overlapping portion comprises the two equilateral triangles that I mentioned plus the four additional regions, each with area Y. Thus,

4Y+((sqrt(3)/4)(r^2)*2=20000

2. In addition, consider the 120 degree sector in one of the circles that comprises the two triangles and two of the regions with area Y. 120 degrees is one third of a circle, so the area of this sector is (1/3)*pi*r^2. Thus,

2Y+((sqrt(3)/4)(r^2)*2= (1/3)*pi*(r^2)

Now you have two equations and two unknowns. Solving for r, yields:

r=sqrt((20000/((2/3)*pi-(sqrt(3))/2), which equals about 127.6 cm.