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Question-157392




Question Number 157392 by ajfour last updated on 22/Oct/21
Commented by ajfour last updated on 22/Oct/21
They are released as shown.
$${They}\:{are}\:{released}\:{as}\:{shown}. \\ $$
Commented by mr W last updated on 23/Oct/21
hard question...
$${hard}\:{question}… \\ $$
Answered by ajfour last updated on 23/Oct/21
(b+r)^2 +(a+r)^2 =(a+b)^2   ⇒  r^2 +(a+b)r−ab=0        (r+((a+b)/2))^2 =(((a+b)/2))^2 +ab  r=(√((a^2 +b^2 +6ab)/4))−((a+b)/2)  ...
$$\left({b}+{r}\right)^{\mathrm{2}} +\left({a}+{r}\right)^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{r}^{\mathrm{2}} +\left({a}+{b}\right){r}−{ab}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\left({r}+\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{ab} \\ $$$${r}=\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{6}{ab}}{\mathrm{4}}}−\frac{{a}+{b}}{\mathrm{2}} \\ $$$$… \\ $$

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