Two identical circles of radius 5cm touch each other externally and a larger circle with Centre O and radius 15cm internally.

Question

Show that the perimeter of the region between the two identical circles and the internal surface of the larger circle is 11 2/3pi cm.

Answer 1

This problem is not hard to solve, but requires a drawing to make the solution understandable.

 

Draw the large circle with radius 15.

 

Draw the two radius 5 circles touching each other and the inner surface of the large circle.

 

Draw a radius from the center of the large circle through the center of each small circle.

 

The distance from the center of the large circle to each of the centers of the small circles is 15-5=10.

 

Draw a line between the centers of the two small circles.  The length of this line is 5+5=10.

 

The 10 cm line and the two lines from the center of the large circle to the centers of the small circles form an equilateral triangle with sides equal to 10.

 

We now know:

 

The central angle defined by the two large circle radii is 60 degrees.  This implies that the portion of the required perimeter contributed by the large circle's arc is 1/6 of the large circle's perimeter.

 

The central angles defined by the the line between the centers of the two small circles and the continuation of the two radii from the large circle's center are 180-60=120.  This implies that the portion of the required perimeter contributed by each of the small circles' arcs is 1/3 of the small circles' perimeter.

 

Total perimeter required is therefore:

 

(1/6)(2pi15)+2((1/3)(2pi5)=5pi+(20/3)pi=11 2/3 pi