∵ Quadrilateral ABCD is a square, and the circle is tangent to both sides of the sector and the square.
∴C, O 1, a is in a straight line,
∴ac= 16√2,co 1= 16√2-﹙ 16+r﹚，
Point O 1 is the vertical line of CD, and the vertical foot is the G point.
Then CG=R, obtained by △CO 1G∽△CAD:
r∶ 16=[ 16√2-﹙ 16+r﹚]∶ 16√2，
Solution: r = 16 (3-2 √ 2),
∴ perimeter m = 2π r = 32π (3-2 √ 2),
Sector arc length N=? ×2π× 16=8π,
Obviously: M≠N, ∴ does not match.
(2) Scheme 2: Let the sector radius =R and the circle radius =r,
Take the point intersecting O2 as the vertical line of CD and the vertical foot as the H point.
∴CH=r,CO2= 16√2-﹙R+r﹚，
According to the similarity: ① r:16 = [16 √ 2-√ r+r √]:16 √ 2,
From sector arc length = circumference: ②? ×2πR=2πr，
Solutions of simultaneous equations;
R=32√2/3，r=8√2/3，
∴ Bus length of cone = r = 32 √ 2/3,
Radius of base circle = r = 8 √ 2/3.
Feasible.