✔ 最佳答案
As we do not know what the question in part a is, we can only proceed without it.
The problem can be readily solved with the help of a well drawn diagram.
Let O be the centre of the circles,
H1 the mid-point of D1-E1, so that
O-D1-H1 is a right triangle with angles equal to 60-30-90 degrees respectively.
In the circumscribed triangle D2-E2-F2,
O-D2 is the radius of the circumscribed circle C1,
O-H1 is the radius of the inscribed circle C1, so
O-D2 = O-H1
Solving the right triangle,
O-D2=O-H1=O-D1*sin(30)=O-D1/2
So the perimeter of triangle D2-E2-F2=perimeter of triangle D1-E1-F1 / 2
By repeating the process,
perimeter of D6-E6-F6
=D1-E1-F1/(2^(6-1))
=D1-E1-F1/32
Thus the perimeter of triangle D6-E6-F6 is 1/32 th of that of triangle D1-E1-F1.