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Question Number 75104 by chess1 last updated on 07/Dec/19

Answered by mr W last updated on 07/Dec/19

Commented by mr W last updated on 08/Dec/19

let b=side length of pentagon α=54° 2r+2r cot α=b ⇒r=(b/(2(1+cot α))) AD=((b tan 72°)/2)−(r/(sin α))−a (((b tan 72°)/2)−(r/(sin α))−a−2r cos α)^2 +(2r sin α−(a/2))^2 =r^2 [r(1+cot α)tan 72°−(r/(sin α))−a−2r cos α]^2 +(2r sin α−(a/2))^2 =r^2 let λ=(a/r) [(1+cot 54)tan 72°−(1/(sin 54))−λ−2 cos 54]^2 +(2 sin 54−(λ/2))^2 =1 ⇒λ=(a/r)≈2.0845

letb=sidelengthofpentagonα=54°2r+2rcotα=b⇒r=b2(1+cotα)AD=btan72°2−rsinα−a(btan72°2−rsinα−a−2rcosα)2+(2rsinα−a2)2=r2[r(1+cotα)tan72°−rsinα−a−2rcosα]2+(2rsinα−a2)2=r2letλ=ar[(1+cot54)tan72°−1sin54−λ−2cos54]2+(2sin54−λ2)2=1⇒λ=ar≈2.0845

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