SOLUTION: C is the center of the large circle. Each of the smaller circles has a radius that is half the radius of the large circle. Find the ratio of area a to area b. Link to diagram :

Algebra ->  Surface-area -> SOLUTION: C is the center of the large circle. Each of the smaller circles has a radius that is half the radius of the large circle. Find the ratio of area a to area b. Link to diagram :      Log On


   

Question 1134258: C is the center of the large circle. Each of the smaller circles has a radius that is half the radius of the large circle. Find the ratio of area a to area b.
Link to diagram : http://tinypic.com/r/2vxhrv4/9
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


There is critical information missing in your statement of the problem and in the diagram. The problem is not fully defined unless you specify the positions of the two smaller circles.

The problem has a nice solution if the diameters of the two smaller circles from the center of the large circle are at right angles -- for example, if the centers of the two small circles are at "noon" and "3 o'clock".

I will assume that is the case, since otherwise the problem can't be solved.

Since the diameters of the two small circles from the center of the large circle form a 90 degree angle, the region between those two diameters is one quarter of the large circle.

Let the radius of the small circle be x; then the radius of the large circle is 2x.

Divide region a in half with a line joining the two points of intersection of the two small circles. Each of the halves of that region created by that line can be viewed as a quarter of the small circle, minus a 45-45-90 right triangle with legs of length x. So the area of each half of region a is

%281%2F4%29%28pi%29x%5E2+-+%281%2F2%29x%5E2

Then the area of region a is twice that:

%281%2F2%29%28pi%29x%5E2-x%5E2

Region b can be viewed as (1) one quarter of the large circle; (2) minus the two semicircles of the small circles; (3) plus the area of region a (since in subtracting the area of the two small semicircles you subtracted that part twice). The area of region b is then

%281%2F4%29%28pi%29%282x%29%5E2+-+2%28%281%2F2%29%28pi%29%28x%5E2%29%29

plus the area of region a.

But those two terms cancel; leaving the result that the area of region b is equal to the area of region a. So

ANSWER: the ratio of area a to area b is 1:1.