Question Number 114906 by mr W last updated on 21/Sep/20

Commented byMJS_new last updated on 22/Sep/20

do you have a solution? and is it ♮niceε?  I get  a≈6.78750  b≈7.55680  c≈5.19953  R≈3.87939  r≈1.75877  area of triangle ≈17.1866

Answered by mr W last updated on 21/Sep/20

find the area of triangle

Answered by bobhans last updated on 22/Sep/20

(1) (a/2)×(a/2)=(2r−2).2           a = 2(√(4r−4))  (2) b=2(√(6r−9))  (3) c =2(√(2r−1))  cosine rule  cos A=((2r−3)/( (√(6r−9)).(√(2r−1)))) ; sin A=((2(√(3r)))/(3(√(2r−1))))  Area = (1/2)bc sin A  → 2r = (a/(sin A)) ; r^3 −6r^2 +9r−3=0  set r=q+2 →q^3 −3q−1=0  set q=2cos t → 8cos ^3 q−6cos q−1=0  q=(π/9), ((7π)/9),((13π)/9)  ⇒r=2+2cos ((13π)/9)≈ 1.65  area ≈ 17.19

Commented bymr W last updated on 22/Sep/20

r must >3.  but r=2+2cos ((13π)/9)≈ 1.65 <3

Answered by mr W last updated on 22/Sep/20

Commented bymr W last updated on 22/Sep/20

BD=DC=(√(R^2 −(R−2)^2 ))=2(√(R−1))  AE=EC=(√(R^2 −(R−3)^2 ))=(√(3(2R−3)))  AF=FB=(√(R^2 −(R−1)^2 ))=(√(2R−1))  cos α=((R−2)/R)  cos β=((R−3)/R)  cos γ=((R−1)/R)  α+β=π−γ  cos α cos β−sin α sin β=−cos γ  (((R−2)(R−3))/R^2 )−((2(√(3(R−1)(2R−3))))/R^2 )=−((R−1)/R)  R^2 −3R+3=(√(3(R−1)(2R−3)))  (R^3 −6R^2 +9R−3)R=0  R^3 −6R^2 +9R−3=0  (R−2)^3 −3(R−2)−1=0  ⇒R−2=2 sin (−(1/3)sin^(−1) (1/2)+((2kπ)/3))  ⇒R=2+2 sin (−(π/(18))+((2kπ)/3)),k=0,1,2  ⇒R=2(1+sin ((11π)/(18)))≈3.879  Δ=2(R−2)(√(R−1))+(R−3)(√(3(2R−3)))+(R−1)(√(2R−1))  ≈17.18