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) ✓](https://www.tinkutara.com/question/Q204507.png)
$$\left[\sqrt{\left(\mathrm{2}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }−\mathrm{1}\right]^{\mathrm{2}} +\left(\mathrm{1}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left[\mathrm{2}\sqrt{\mathrm{1}−{r}}−\mathrm{1}\right]^{\mathrm{2}} =\mathrm{2}{r}−\mathrm{1} \\ $$$$\mathrm{4}\left(\mathrm{1}−{r}\right)−\mathrm{4}\sqrt{\mathrm{1}−{r}}+\mathrm{1}=\mathrm{2}{r}−\mathrm{1} \\ $$$$\mathrm{3}−\mathrm{3}{r}=\mathrm{2}\sqrt{\mathrm{1}−{r}} \\ $$$$\mathrm{9}−\mathrm{18}{r}+\mathrm{9}{r}^{\mathrm{2}} =\mathrm{4}−\mathrm{4}{r} \\ $$$$\mathrm{9}{r}^{\mathrm{2}} −\mathrm{14}{r}+\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{\mathrm{7}−\mathrm{2}}{\mathrm{9}}=\frac{\mathrm{5}}{\mathrm{9}}\:\checkmark \\ $$
Commented by cherokeesay last updated on 19/Feb/24
$${nice},{thank}\:{you}\:{sir}. \\ $$
Commented by Panav last updated on 02/Mar/24
$${how}\:{you}\:{did}\:{this} \\ $$
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.](https://www.tinkutara.com/question/Q204925.png)
$${AG}=\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$$${AB}={AG}−{r}=\mathrm{2}−{r} \\ $$$${AE}=\sqrt{{AB}^{\mathrm{2}} −{EB}^{\mathrm{2}} }=\sqrt{\left(\mathrm{2}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${DC}={AF}−{EB}=\mathrm{1}−{r} \\ $$$${BD}={AE}−{CF}=\sqrt{\left(\mathrm{2}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }−\mathrm{1} \\ $$$${BD}^{\mathrm{2}} +{DC}^{\mathrm{2}} ={BC}^{\mathrm{2}} \\ $$$$\Rightarrow\left[\sqrt{\left(\mathrm{2}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }−\mathrm{1}\right]^{\mathrm{2}} +\left(\mathrm{1}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${the}\:{rest}\:{see}\:{above}. \\ $$