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Question Number 190405 by cortano12 last updated on 02/Apr/23
Answered by cherokeesay last updated on 02/Apr/23
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Answered by mr W last updated on 02/Apr/23
((p−q)/(p+q))=((q−r)/(q+r))=k q=((1+k)/(1−k))r p=((1+k)/(1−k))q=(((1+k)/(1−k)))^2 r (p+q)^2 −(p−q)^2 =a^2 4pq=a^2 4(((1+k)/(1−k)))^3 r^2 =a^2 ...(i) (q+r)^2 −(q−r)^2 =b^2 4qr=b^2 4(((1+k)/(1−k)))r^2 =b^2 ...(ii) ⇒((1+k)/(1−k))=(a/b) 4×(a/b)×r^2 =b^2 ⇒r=(b/2)(√(b/a)) ⇒q=(b/2)((a/b))(√(b/a)) ⇒p=(b/2)((a/b))^2 (√(b/a)) R=p+q+r=[((a/b))^2 +((a/b))+1](b/2)(√(b/a)) A_(red) =(π/2)(R^2 −p^2 −q^2 −r^2 ) =(π/2){[((a/b))^2 +((a/b))+1]^2 −((a/b))^4 −((a/b))^2 −1}(b^2 /4)×(b/a) =π{((a/b))^3 +((a/b))^2 +((a/b))}(b^2 /4)×(b/a) ={((a/b))^2 +(a/b)+1}((πb^2 )/4) ={((8/4))^2 +(8/4)+1}((π×4^2 )/4) =28π
$$\frac{{p}−{q}}{{p}+{q}}=\frac{{q}−{r}}{{q}+{r}}={k} \\ $$$${q}=\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}{r} \\ $$$${p}=\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}{q}=\left(\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}\right)^{\mathrm{2}} {r} \\ $$$$\left({p}+{q}\right)^{\mathrm{2}} −\left({p}−{q}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\mathrm{4}{pq}={a}^{\mathrm{2}} \\ $$$$\mathrm{4}\left(\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}\right)^{\mathrm{3}} {r}^{\mathrm{2}} ={a}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\left({q}+{r}\right)^{\mathrm{2}} −\left({q}−{r}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\mathrm{4}{qr}={b}^{\mathrm{2}} \\ $$$$\mathrm{4}\left(\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}\right){r}^{\mathrm{2}} ={b}^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}=\frac{{a}}{{b}} \\ $$$$\mathrm{4}×\frac{{a}}{{b}}×{r}^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{{b}}{\mathrm{2}}\sqrt{\frac{{b}}{{a}}} \\ $$$$\Rightarrow{q}=\frac{{b}}{\mathrm{2}}\left(\frac{{a}}{{b}}\right)\sqrt{\frac{{b}}{{a}}} \\ $$$$\Rightarrow{p}=\frac{{b}}{\mathrm{2}}\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} \sqrt{\frac{{b}}{{a}}} \\ $$$${R}={p}+{q}+{r}=\left[\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} +\left(\frac{{a}}{{b}}\right)+\mathrm{1}\right]\frac{{b}}{\mathrm{2}}\sqrt{\frac{{b}}{{a}}} \\ $$$${A}_{{red}} =\frac{\pi}{\mathrm{2}}\left({R}^{\mathrm{2}} −{p}^{\mathrm{2}} −{q}^{\mathrm{2}} −{r}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:=\frac{\pi}{\mathrm{2}}\left\{\left[\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} +\left(\frac{{a}}{{b}}\right)+\mathrm{1}\right]^{\mathrm{2}} −\left(\frac{{a}}{{b}}\right)^{\mathrm{4}} −\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} −\mathrm{1}\right\}\frac{{b}^{\mathrm{2}} }{\mathrm{4}}×\frac{{b}}{{a}} \\ $$$$\:\:\:\:=\pi\left\{\left(\frac{{a}}{{b}}\right)^{\mathrm{3}} +\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} +\left(\frac{{a}}{{b}}\right)\right\}\frac{{b}^{\mathrm{2}} }{\mathrm{4}}×\frac{{b}}{{a}} \\ $$$$\:\:\:\:=\left\{\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} +\frac{{a}}{{b}}+\mathrm{1}\right\}\frac{\pi{b}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:=\left\{\left(\frac{\mathrm{8}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{8}}{\mathrm{4}}+\mathrm{1}\right\}\frac{\pi×\mathrm{4}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:=\mathrm{28}\pi \\ $$
Commented by mr W last updated on 02/Apr/23
Answered by a.lgnaoui last updated on 03/Apr/23
Soit :C_1 (O1,R_1 ) ; C_2 (O_2 ,R_2 ) ;C_3 (O_3 ,R_3 ) R_1 =O_1 C R_2 =O_2 B R_3 =O_3 A 1•tan θ=((r_2 −r_1 )/4)=((r_3 −r_2 )/8)=((r_3 −r_1 )/(12))(1) 2• { ((4^2 +_1 (r_2 −r_1 )^2 =(r_1 +r_2 )^2 (2))),((8^2 +(r_3 −r_2 )^2 =(r_3 +r_2 )^2 (3))) :} (1) ⇒ { ((2(r_2 −r_1 )=(r_3 −r_2 ))),((3(r_3 −r_2 )=2(r_3 −r_1 ))) :} ⇒ r_3 =3r_2 −2r_1 (4) (2) { ((r_1 r_2 =4 (5))),((r_2 r_3 =16 (6))) :} (5)et(6)⇒ r_3 =4r_1 (7) (4)⇒4r_1 =3r_2 −2r_1 ⇒ r_2 =2r_1 (8) 3• (1)⇒12^2 +(r_3 −r_1 )^2 =(r_1 +2r_2 +r_3 )^2 36=r_2 ^2 +r_1 r_3 +r_2 (r_1 +r_3 ) =4r_1 ^2 +4r_1 ^2 +2r_1 (5r_1 )=18r_1 ^2 36=18r_1 ^2 ⇒r_1 =(√2) Rayon R=r_1 +2r_1 +4r_1 =7r_1 ⇒ R=7(√2) Aire totale=πR^2 Aire (Rouge)=((πR^2 )/2)−π(((r_1 ^2 +r_2 ^2 +r_3 ^2 )/2)) r_1 ^2 +r_2 ^2 +r_3 ^2 =42 { ((Aire=28π)),((Aire ≅87,96)) :}
$${Soit}\::{C}_{\mathrm{1}} \left({O}\mathrm{1},{R}_{\mathrm{1}} \right)\:;\:\:\:{C}_{\mathrm{2}} \left({O}_{\mathrm{2}} ,{R}_{\mathrm{2}} \right)\:\:\:;{C}_{\mathrm{3}} \left({O}_{\mathrm{3}} ,{R}_{\mathrm{3}} \right) \\ $$$${R}_{\mathrm{1}} ={O}_{\mathrm{1}} {C}\:\:\:\:\:{R}_{\mathrm{2}} ={O}_{\mathrm{2}} {B}\:\:\:\:\:{R}_{\mathrm{3}} ={O}_{\mathrm{3}} {A} \\ $$$$\mathrm{1}\bullet\mathrm{tan}\:\theta=\frac{{r}_{\mathrm{2}} −{r}_{\mathrm{1}} }{\mathrm{4}}=\frac{{r}_{\mathrm{3}} −{r}_{\mathrm{2}} }{\mathrm{8}}=\frac{{r}_{\mathrm{3}} −{r}_{\mathrm{1}} }{\mathrm{12}}\left(\mathrm{1}\right) \\ $$$$\mathrm{2}\bullet\begin{cases}{\mathrm{4}^{\mathrm{2}} +_{\mathrm{1}} \left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)^{\mathrm{2}} =\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} \:\:\:\:\:\:\left(\mathrm{2}\right)}\\{\mathrm{8}^{\mathrm{2}} +\left({r}_{\mathrm{3}} −{r}_{\mathrm{2}} \right)^{\mathrm{2}} =\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} \:\:\:\:\:\:\:\:\left(\mathrm{3}\right)}\end{cases} \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow\begin{cases}{\mathrm{2}\left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)=\left({r}_{\mathrm{3}} −{r}_{\mathrm{2}} \right)}\\{\mathrm{3}\left({r}_{\mathrm{3}} −{r}_{\mathrm{2}} \right)=\mathrm{2}\left({r}_{\mathrm{3}} −{r}_{\mathrm{1}} \right)}\end{cases}\:\:\: \\ $$$$\Rightarrow\:\:\:\:\:\:\:\boldsymbol{{r}}_{\mathrm{3}} =\mathrm{3}\boldsymbol{{r}}_{\mathrm{2}} −\mathrm{2}\boldsymbol{{r}}_{\mathrm{1}} \:\:\:\:\:\:\:\:\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{2}\right)\begin{cases}{{r}_{\mathrm{1}} {r}_{\mathrm{2}} =\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{5}\right)}\\{{r}_{\mathrm{2}} {r}_{\mathrm{3}} =\mathrm{16}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{6}\right)}\end{cases} \\ $$$$\left(\mathrm{5}\right){et}\left(\mathrm{6}\right)\Rightarrow\:\:\boldsymbol{{r}}_{\mathrm{3}} =\mathrm{4}\boldsymbol{{r}}_{\mathrm{1}} \:\:\:\left(\mathrm{7}\right) \\ $$$$\left(\mathrm{4}\right)\Rightarrow\mathrm{4}{r}_{\mathrm{1}} =\mathrm{3}{r}_{\mathrm{2}} −\mathrm{2}{r}_{\mathrm{1}} \:\:\:\:\:\:\: \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{r}}_{\mathrm{2}} =\mathrm{2}\boldsymbol{{r}}_{\mathrm{1}} \:\:\:\:\left(\mathrm{8}\right) \\ $$$$ \\ $$$$\mathrm{3}\bullet\:\:\:\left(\mathrm{1}\right)\Rightarrow\mathrm{12}^{\mathrm{2}} +\left({r}_{\mathrm{3}} −{r}_{\mathrm{1}} \right)^{\mathrm{2}} =\left({r}_{\mathrm{1}} +\mathrm{2}{r}_{\mathrm{2}} +{r}_{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{36}={r}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{1}} {r}_{\mathrm{3}} +{r}_{\mathrm{2}} \left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right) \\ $$$$=\mathrm{4}{r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{4}{r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}{r}_{\mathrm{1}} \left(\mathrm{5}{r}_{\mathrm{1}} \right)=\mathrm{18}{r}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\mathrm{36}=\mathrm{18}{r}_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\Rightarrow{r}_{\mathrm{1}} =\sqrt{\mathrm{2}}\: \\ $$$${Rayon}\:{R}={r}_{\mathrm{1}} +\mathrm{2}{r}_{\mathrm{1}} +\mathrm{4}{r}_{\mathrm{1}} =\mathrm{7}{r}_{\mathrm{1}} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\boldsymbol{{R}}=\mathrm{7}\sqrt{\mathrm{2}} \\ $$$${Aire}\:{totale}=\pi{R}^{\mathrm{2}} \\ $$$${Aire}\:\left({Rouge}\right)=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}−\pi\left(\frac{{r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{3}} ^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{3}} ^{\mathrm{2}} =\mathrm{42} \\ $$$$\begin{cases}{{Aire}=\mathrm{28}\pi}\\{\boldsymbol{{Aire}}\:\:\:\cong\mathrm{87},\mathrm{96}}\end{cases} \\ $$
Answered by Skabetix last updated on 10/Apr/23
Commented by Skabetix last updated on 10/Apr/23
Answered by ajfour last updated on 10/Apr/23
2(√(pq))=a 2(√(qr))=b (p/q)=(q/r) ⇒ 4rp=4q^2 =ab R=p+q+r ((2S)/π)=(p+q+r)^2 −p^2 −q^2 −r^2 =2(pq+qr+rp) =2((a^2 /4)+(b^2 /4)+((ab)/4)) ⇒ S=(π/4)(a^2 +b^2 +ab) =(π/4)(64+16+32)=28π
$$\mathrm{2}\sqrt{{pq}}={a} \\ $$$$\mathrm{2}\sqrt{{qr}}={b} \\ $$$$\frac{{p}}{{q}}=\frac{{q}}{{r}} \\ $$$$\Rightarrow\:\:\mathrm{4}{rp}=\mathrm{4}{q}^{\mathrm{2}} ={ab} \\ $$$${R}={p}+{q}+{r} \\ $$$$\frac{\mathrm{2}{S}}{\pi}=\left({p}+{q}+{r}\right)^{\mathrm{2}} −{p}^{\mathrm{2}} −{q}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\mathrm{2}\left({pq}+{qr}+{rp}\right) \\ $$$$\:\:\:\:\:=\mathrm{2}\left(\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}}+\frac{{ab}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\:{S}=\frac{\pi}{\mathrm{4}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{4}}\left(\mathrm{64}+\mathrm{16}+\mathrm{32}\right)=\mathrm{28}\pi \\ $$

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