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Question Number 204505 by cherokeesay last updated on 19/Feb/24

Answered by mr W last updated on 19/Feb/24

[(√((2−r)^2 −r^2 ))−1]^2 +(1−r)^2 =r^2 [2(√(1−r))−1]^2 =2r−1 4(1−r)−4(√(1−r))+1=2r−1 3−3r=2(√(1−r)) 9−18r+9r^2 =4−4r 9r^2 −14r+5=0 ⇒r=((7−2)/9)=(5/9) ✓

[(2−r)2−r2−1]2+(1−r)2=r2[21−r−1]2=2r−14(1−r)−41−r+1=2r−13−3r=21−r9−18r+9r2=4−4r9r2−14r+5=0⇒r=7−29=59✓

Commented by cherokeesay last updated on 19/Feb/24

nice,thank you sir.

nice,thankyousir.

Commented by Panav last updated on 02/Mar/24

how you did this

howyoudidthis

Commented by mr W last updated on 02/Mar/24

Commented by mr W last updated on 02/Mar/24

AG=1+1=2 AB=AG−r=2−r AE=(√(AB^2 −EB^2 ))=(√((2−r)^2 −r^2 )) DC=AF−EB=1−r BD=AE−CF=(√((2−r)^2 −r^2 ))−1 BD^2 +DC^2 =BC^2 ⇒[(√((2−r)^2 −r^2 ))−1]^2 +(1−r)^2 =r^2 the rest see above.

AG=1+1=2AB=AG−r=2−rAE=AB2−EB2=(2−r)2−r2DC=AF−EB=1−rBD=AE−CF=(2−r)2−r2−1BD2+DC2=BC2⇒[(2−r)2−r2−1]2+(1−r)2=r2therestseeabove.

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