Three identical circles of radius 30 cm are tangent to each other externally. A fourth circle of the same radius was drawn so that its center is coincidental with the center of the space bounded by the three tangent circles. Find the area of the region inside the fourth circle but outside the first three circles. It is the shaded region shown in the figure below.
My goal is to find the area of a segment BGC. Let's get started.
AJ = sqrt(AK2 - JK2) = 30√3
AH = 2/3(AJ) = 20√3
AE = HJ = 1/3(AJ) = 10√3
Angle BAC = 2[arccos(AB / AE)] = 109.4712206º
Area of sector ABGC = 302pi * (∠BAC / 360) = 859.784956 u2
Area of triangle ABC = BE * AE = 424.2640687 u2
Area of segment BGC = [ABGC] - [ABC] = 435.5208873 u2
Area of circle H = 302pi = 2827.433388
Shaded area = 2827.433388 - (6 * 435.5208873) = 214.3080644 u2
I've posted this question nearly 6 years ago, and my son answered it. He told me about that 2 days ago. (Beautiful communication)
His answer is very close to mine. (Dragan ≅ jugoslav ) Here's his answer:
This one was a bit tricky, but I think it's because I made it more complicated than it needed to be.
The most important first step was to determine the center of our fourth circle. How far away is it from the centers of the other three circles?
I set the center of the bottom left circle as my "origin." Then, creating a triangle between the three centers, I solved for the centroid of the triangle (which had an effective coordinate of (30,17.32)). This means that the distance between the fourth circle and the other three circles was 34.64 cm.
Now, knowing this, I generated an independent problem: what is an overlapping area of two circles 34.64 cm apart, with radii of 30 cm? Once I had this number, it was simply a matter of multiplying it by three and subtracting it from the overall area of one circle.
You probably don't want to see my math...I used an integral in the cartesian coordinate system. I basically found the midpoint of their intersecting area and set it as my lower limit (17.32), set 30 as my upper limit, and integrated sqrt(30^2-x^2)dx. Then I multiplied that area by 4, since integrating circles in cartesian coordinates is weird. It was messy, but the answer: the overlapping area was 871.08 cm^2.
Since the area of the fourth circle is pi*r^2 (or 2827.43 cm^2), you simply subtract 871.08*3 (or 2613.24 cm^2, which is the area of intersection between the fourth circle and each of the other three circles).
This yields the final answer of 214.22 cm^2...give or take for rounding error. ;)
-KMM (May 19, 2015)