Menu Close

Question-51422

Tinku Tara 0 Comments
Pin




Question Number 51422 by ajfour last updated on 26/Dec/18
Commented by ajfour last updated on 26/Dec/18
Find R the radius of semicircle in terms of radii a and b of the circles.
FindRtheradiusofsemicircleintermsofradiiaandbofthecircles.
Answered by ajfour last updated on 26/Dec/18
let sin θ = (a/(R−a)) , sin φ = (b/(R−b)) (R−a)cos θ+(R−b)cos φ = (√(4ab)) ⇒ (√(R(R−2a)))+(√(R(R−2b))) = (√(4ab)) squaring 2R^2 −2(a+b)R +2(√(R^2 {R^2 −2(a+b)R+4ab})) = 4ab rearranging and squaring R^2 {R^2 −2(a+b)R+4ab} = {2ab−R^2 +(a+b)R}^2 ⇒ 4abR^2 = 4a^2 b^2 +(a+b)^2 R^2 −4abR^2 +4ab(a+b)R ⇒ (a^2 +b^2 −6ab)R^2 +4ab(a+b)R +4a^2 b^2 = 0 ⇒ R = ((−2ab(a+b)−(√(4a^2 b^2 (8ab))))/(a^2 +b^2 −6ab)) or R = [(√(8ab))+(a+b)](((2ab)/(6ab−a^2 −b^2 ))) .
letsinθ=aR−a,sinϕ=bR−b(R−a)cosθ+(R−b)cosϕ=4ab⇒R(R−2a)+R(R−2b)=4absquaring2R2−2(a+b)R+2R2{R2−2(a+b)R+4ab}=4abrearrangingandsquaringR2{R2−2(a+b)R+4ab}={2ab−R2+(a+b)R}2⇒4abR2=4a2b2+(a+b)2R2−4abR2+4ab(a+b)R⇒(a2+b2−6ab)R2+4ab(a+b)R+4a2b2=0⇒R=−2ab(a+b)−4a2b2(8ab)a2+b2−6aborR=[8ab+(a+b)](2ab6ab−a2−b2).
Answered by behi83417@gmail.com last updated on 27/Dec/18
⇒AB=(√(2(a^2 +b^2 ))) TO=OS=R OA=R−a,OB=R−b sinα=(a/(R−a)),sinβ=(b/(R−b)) cosα=((√((R−a)^2 −a^2 ))/(R−a)),cosβ=((√((R−b)^2 −b^2 ))/(R−b)) cosϕ=cos(π−α−β)=−cos(α+β)= =((ab−(√(R^2 −2Ra)).(√(R^2 −2Rb)))/((R−a)(R−b)))(i) cosϕ=((OA^2 +OB^2 −AB^2 )/(2OA.OB))=(((R−a)^2 +(R−b)^2 −2(a^2 +b^2 ))/(2(R−a)(R−b)))= =((2R^2 −2Ra−2Rb−a^2 −b^2 )/(2(R−a)(R−b)))(ii) i=ii⇒ 2R^2 −2Ra−2Rb−a^2 −b^2 = =ab−(√(R^2 −2Ra)).(√(R^2 −2Rb)) a^2 +b^2 +ab+2Ra+2Rb−2R^2 =(√(R^2 −2Ra)).(√(R^2 −2Rb)) R^2 −2Ra=p,R^2 −2Rb=q,a^2 +b^2 +ab=t a^2 +b^2 +ab−(p+q)=(√(pq)) t−(p+q)=(√(pq))⇒t^2 −2t(p+q)+(p+q)^2 =pq p+q=2R^2 −2Ra−2Rb (p+q)^2 =4R^4 +4R^2 a^2 +4R^2 b^2 −8R^3 a−8R^3 b+8R^2 ab pq=(R^2 −2Ra)(R^2 −2Rb)=R^4 −2R^3 b−2R^3 a+4R^2 ab t^2 −4R(R−a−b)t+3R^4 +4R^2 (a^2 +b^2 +ab) −6R^3 a−6R^3 b=0 t^2 +4R(a+b)t+3R^3 (R−2a−2b)=0 ⇒△′=4R^2 (a+b)^2 −3R^3 (R−2a−2b)= =R^2 [4(a+b)^2 −3R^2 +6R(a+b)]=R^2 m^2 a^2 +b^2 +ab=−2R(a+b)+Rm m=((a^2 +b^2 +ab)/R)+2(a+b)
⇒AB=2(a2+b2)TO=OS=ROA=R−a,OB=R−bsinα=aR−a,sinβ=bR−bcosα=(R−a)2−a2R−a,cosβ=(R−b)2−b2R−bcosφ=cos(π−α−β)=−cos(α+β)==ab−R2−2Ra.R2−2Rb(R−a)(R−b)(i)cosφ=OA2+OB2−AB22OA.OB=(R−a)2+(R−b)2−2(a2+b2)2(R−a)(R−b)==2R2−2Ra−2Rb−a2−b22(R−a)(R−b)(ii)i=ii⇒2R2−2Ra−2Rb−a2−b2==ab−R2−2Ra.R2−2Rba2+b2+ab+2Ra+2Rb−2R2=R2−2Ra.R2−2RbR2−2Ra=p,R2−2Rb=q,a2+b2+ab=ta2+b2+ab−(p+q)=pqt−(p+q)=pq⇒t2−2t(p+q)+(p+q)2=pqp+q=2R2−2Ra−2Rb(p+q)2=4R4+4R2a2+4R2b2−8R3a−8R3b+8R2abpq=(R2−2Ra)(R2−2Rb)=R4−2R3b−2R3a+4R2abt2−4R(R−a−b)t+3R4+4R2(a2+b2+ab)−6R3a−6R3b=0t2+4R(a+b)t+3R3(R−2a−2b)=0⇒△′=4R2(a+b)2−3R3(R−2a−2b)==R2[4(a+b)2−3R2+6R(a+b)]=R2m2a2+b2+ab=−2R(a+b)+Rmm=a2+b2+abR+2(a+b)
Commented by ajfour last updated on 27/Dec/18
for b=a , we must get R=a(1+(√2)) your solution gets m= ((3a^2 )/R)+4a 3a^2 =−4aR+3a^2 +4aR so we dont obtain R, please resolve Sir.
forb=a,wemustgetR=a(1+2)yoursolutiongetsm=3a2R+4a3a2=−4aR+3a2+4aRsowedontobtainR,pleaseresolveSir.

Pin

Leave a Reply

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