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