Plane Geometry - Bruce E. Shapiro

232 SECTION 42. AREA AND CIRCUMFERENCE OF CIRCLESProof. Within each circle construct the sequence of 3×2 n -gons as describedabove. Then each triangle △P k OP k+1 ∼ △P ′ k OP ′ k+1By the similar triangles theoremfor each k and for each H n .Since OP k = r and OP ′ k = r′ ,Hence for each n,P k P k+1OP k= P ′ k P ′ k+1OP ′ kP k P k+1rL nr= P ′ k P ′ k+1r ′=L′ nr ′Since L n → C and L ′ n → C ′ as n → ∞, the result follows.Definition 42.8 π = C/2r.Definition 42.9 The interior of circle Γ = C(O, r) is the set of all pointsinside Γ. The circular region of Γ is the union of Γ with Int(Γ).Recall that we denote the area of polygon H n by α(H n ). Then we have thefollowing.Theorem 42.10 Let H 1 , H 2 , ... be the sequence of inscribed polygons asdefined before. Then the areas of the polygonal regions forms an increasingsequence:α(H n ) < α(H n+1 )Proof. Each polygon H n can be dissected into a union of non-overlappingtrianglesn⋃H n = (△P (n)kOP (n)k+1 )k=1where we define P (n)n+1 = P (n)1 , andH n+1 =n⋃k=1Ä(n) △PkOQ (n)kwhere the Q k are as defined previously.∪ △Q (n)k(n)OPk+1ä« CC BY-NC-ND 3.0. Revised: 18 Nov 2012