Menu Close

Question-208782




Question Number 208782 by cherokeesay last updated on 22/Jun/24
Answered by mr W last updated on 23/Jun/24
Commented by cherokeesay last updated on 23/Jun/24
thank you master !
thankyoumaster!
Commented by Tawa11 last updated on 23/Jun/24
Weldone sir.
Weldonesir.
Commented by mr W last updated on 23/Jun/24
(r/R)=(√((25)/(49)))=(5/7) ⇒R=((7r)/5)  d=2r  (R−r)^2 +((√(15)))^2 =(3r−R)^2   25=4r^2    ⇒r=(5/2) ⇒R=(7/2)  (√(R^2 −b^2 ))+(√(r^2 −b^2 ))=3r−R  (√(R^2 −b^2 ))=4−(√(r^2 −b^2 ))  5=4(√(r^2 −b^2 ))  b^2 =r^2 −((25)/(16)) ⇒b=((5(√3))/4)  sin α_1 =(b/R)=((5(√3))/(14))  sin α_2 =(b/r)=((√3)/2) ⇒α_2 =(π/3)  green area:  (R^2 /4)(2α_1 −sin 2α_1 )+(r^2 /4)(2α_2 −sin 2α_2 )  =(7^2 /(4×4))(2sin^(−1) ((5(√3))/(14))−((2×5(√3)×11)/(14^2 )))+(5^2 /(4×4))(((2π)/3)−((√3)/2))  =((49)/8) sin^(−1) ((5(√3))/(14))+((25π)/(24))−((5(√3))/2)≈3.027
rR=2549=57R=7r5d=2r(Rr)2+(15)2=(3rR)225=4r2r=52R=72R2b2+r2b2=3rRR2b2=4r2b25=4r2b2b2=r22516b=534sinα1=bR=5314sinα2=br=32α2=π3greenarea:R24(2α1sin2α1)+r24(2α2sin2α2)=724×4(2sin153142×53×11142)+524×4(2π332)=498sin15314+25π245323.027

Leave a Reply

Your email address will not be published. Required fields are marked *