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
$$\frac{{p}−{q}}{{p}+{q}}=\frac{{q}−{r}}{{q}+{r}}={k} \\ $$$${q}=\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}{r} \\ $$$${p}=\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}{q}=\left(\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}\right)^{\mathrm{2}} {r} \\ $$$$\left({p}+{q}\right)^{\mathrm{2}} −\left({p}−{q}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\mathrm{4}{pq}={a}^{\mathrm{2}} \\ $$$$\mathrm{4}\left(\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}\right)^{\mathrm{3}} {r}^{\mathrm{2}} ={a}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\left({q}+{r}\right)^{\mathrm{2}} −\left({q}−{r}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\mathrm{4}{qr}={b}^{\mathrm{2}} \\ $$$$\mathrm{4}\left(\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}\right){r}^{\mathrm{2}} ={b}^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}=\frac{{a}}{{b}} \\ $$$$\mathrm{4}×\frac{{a}}{{b}}×{r}^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{{b}}{\mathrm{2}}\sqrt{\frac{{b}}{{a}}} \\ $$$$\Rightarrow{q}=\frac{{b}}{\mathrm{2}}\left(\frac{{a}}{{b}}\right)\sqrt{\frac{{b}}{{a}}} \\ $$$$\Rightarrow{p}=\frac{{b}}{\mathrm{2}}\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} \sqrt{\frac{{b}}{{a}}} \\ $$$${R}={p}+{q}+{r}=\left[\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} +\left(\frac{{a}}{{b}}\right)+\mathrm{1}\right]\frac{{b}}{\mathrm{2}}\sqrt{\frac{{b}}{{a}}} \\ $$$${A}_{{red}} =\frac{\pi}{\mathrm{2}}\left({R}^{\mathrm{2}} −{p}^{\mathrm{2}} −{q}^{\mathrm{2}} −{r}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:=\frac{\pi}{\mathrm{2}}\left\{\left[\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} +\left(\frac{{a}}{{b}}\right)+\mathrm{1}\right]^{\mathrm{2}} −\left(\frac{{a}}{{b}}\right)^{\mathrm{4}} −\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} −\mathrm{1}\right\}\frac{{b}^{\mathrm{2}} }{\mathrm{4}}×\frac{{b}}{{a}} \\ $$$$\:\:\:\:=\pi\left\{\left(\frac{{a}}{{b}}\right)^{\mathrm{3}} +\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} +\left(\frac{{a}}{{b}}\right)\right\}\frac{{b}^{\mathrm{2}} }{\mathrm{4}}×\frac{{b}}{{a}} \\ $$$$\:\:\:\:=\left\{\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} +\frac{{a}}{{b}}+\mathrm{1}\right\}\frac{\pi{b}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:=\left\{\left(\frac{\mathrm{8}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{8}}{\mathrm{4}}+\mathrm{1}\right\}\frac{\pi×\mathrm{4}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:=\mathrm{28}\pi \\ $$
Commented by mr W last updated on 02/Apr/23
Answered by a.lgnaoui last updated on 03/Apr/23
$${Soit}\::{C}_{\mathrm{1}} \left({O}\mathrm{1},{R}_{\mathrm{1}} \right)\:;\:\:\:{C}_{\mathrm{2}} \left({O}_{\mathrm{2}} ,{R}_{\mathrm{2}} \right)\:\:\:;{C}_{\mathrm{3}} \left({O}_{\mathrm{3}} ,{R}_{\mathrm{3}} \right) \\ $$$${R}_{\mathrm{1}} ={O}_{\mathrm{1}} {C}\:\:\:\:\:{R}_{\mathrm{2}} ={O}_{\mathrm{2}} {B}\:\:\:\:\:{R}_{\mathrm{3}} ={O}_{\mathrm{3}} {A} \\ $$$$\mathrm{1}\bullet\mathrm{tan}\:\theta=\frac{{r}_{\mathrm{2}} −{r}_{\mathrm{1}} }{\mathrm{4}}=\frac{{r}_{\mathrm{3}} −{r}_{\mathrm{2}} }{\mathrm{8}}=\frac{{r}_{\mathrm{3}} −{r}_{\mathrm{1}} }{\mathrm{12}}\left(\mathrm{1}\right) \\ $$$$\mathrm{2}\bullet\begin{cases}{\mathrm{4}^{\mathrm{2}} +_{\mathrm{1}} \left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)^{\mathrm{2}} =\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} \:\:\:\:\:\:\left(\mathrm{2}\right)}\\{\mathrm{8}^{\mathrm{2}} +\left({r}_{\mathrm{3}} −{r}_{\mathrm{2}} \right)^{\mathrm{2}} =\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} \:\:\:\:\:\:\:\:\left(\mathrm{3}\right)}\end{cases} \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow\begin{cases}{\mathrm{2}\left({r}_{\mathrm{2}} −{r}_{\mathrm{1}} \right)=\left({r}_{\mathrm{3}} −{r}_{\mathrm{2}} \right)}\\{\mathrm{3}\left({r}_{\mathrm{3}} −{r}_{\mathrm{2}} \right)=\mathrm{2}\left({r}_{\mathrm{3}} −{r}_{\mathrm{1}} \right)}\end{cases}\:\:\: \\ $$$$\Rightarrow\:\:\:\:\:\:\:\boldsymbol{{r}}_{\mathrm{3}} =\mathrm{3}\boldsymbol{{r}}_{\mathrm{2}} −\mathrm{2}\boldsymbol{{r}}_{\mathrm{1}} \:\:\:\:\:\:\:\:\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{2}\right)\begin{cases}{{r}_{\mathrm{1}} {r}_{\mathrm{2}} =\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{5}\right)}\\{{r}_{\mathrm{2}} {r}_{\mathrm{3}} =\mathrm{16}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{6}\right)}\end{cases} \\ $$$$\left(\mathrm{5}\right){et}\left(\mathrm{6}\right)\Rightarrow\:\:\boldsymbol{{r}}_{\mathrm{3}} =\mathrm{4}\boldsymbol{{r}}_{\mathrm{1}} \:\:\:\left(\mathrm{7}\right) \\ $$$$\left(\mathrm{4}\right)\Rightarrow\mathrm{4}{r}_{\mathrm{1}} =\mathrm{3}{r}_{\mathrm{2}} −\mathrm{2}{r}_{\mathrm{1}} \:\:\:\:\:\:\: \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{r}}_{\mathrm{2}} =\mathrm{2}\boldsymbol{{r}}_{\mathrm{1}} \:\:\:\:\left(\mathrm{8}\right) \\ $$$$ \\ $$$$\mathrm{3}\bullet\:\:\:\left(\mathrm{1}\right)\Rightarrow\mathrm{12}^{\mathrm{2}} +\left({r}_{\mathrm{3}} −{r}_{\mathrm{1}} \right)^{\mathrm{2}} =\left({r}_{\mathrm{1}} +\mathrm{2}{r}_{\mathrm{2}} +{r}_{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{36}={r}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{1}} {r}_{\mathrm{3}} +{r}_{\mathrm{2}} \left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right) \\ $$$$=\mathrm{4}{r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{4}{r}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}{r}_{\mathrm{1}} \left(\mathrm{5}{r}_{\mathrm{1}} \right)=\mathrm{18}{r}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\mathrm{36}=\mathrm{18}{r}_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\Rightarrow{r}_{\mathrm{1}} =\sqrt{\mathrm{2}}\: \\ $$$${Rayon}\:{R}={r}_{\mathrm{1}} +\mathrm{2}{r}_{\mathrm{1}} +\mathrm{4}{r}_{\mathrm{1}} =\mathrm{7}{r}_{\mathrm{1}} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\boldsymbol{{R}}=\mathrm{7}\sqrt{\mathrm{2}} \\ $$$${Aire}\:{totale}=\pi{R}^{\mathrm{2}} \\ $$$${Aire}\:\left({Rouge}\right)=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}−\pi\left(\frac{{r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{3}} ^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{3}} ^{\mathrm{2}} =\mathrm{42} \\ $$$$\begin{cases}{{Aire}=\mathrm{28}\pi}\\{\boldsymbol{{Aire}}\:\:\:\cong\mathrm{87},\mathrm{96}}\end{cases} \\ $$
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
$$\mathrm{2}\sqrt{{pq}}={a} \\ $$$$\mathrm{2}\sqrt{{qr}}={b} \\ $$$$\frac{{p}}{{q}}=\frac{{q}}{{r}} \\ $$$$\Rightarrow\:\:\mathrm{4}{rp}=\mathrm{4}{q}^{\mathrm{2}} ={ab} \\ $$$${R}={p}+{q}+{r} \\ $$$$\frac{\mathrm{2}{S}}{\pi}=\left({p}+{q}+{r}\right)^{\mathrm{2}} −{p}^{\mathrm{2}} −{q}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\mathrm{2}\left({pq}+{qr}+{rp}\right) \\ $$$$\:\:\:\:\:=\mathrm{2}\left(\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}}+\frac{{ab}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\:{S}=\frac{\pi}{\mathrm{4}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\pi}{\mathrm{4}}\left(\mathrm{64}+\mathrm{16}+\mathrm{32}\right)=\mathrm{28}\pi \\ $$