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Question Number 208049 by mr W last updated on 03/Jun/24

Commented by Frix last updated on 03/Jun/24

a=1∧b=3  r=a+b−(√(a^2 +b^2 ))=4−(√(10))

$${a}=\mathrm{1}\wedge{b}=\mathrm{3} \\ $$$${r}={a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{4}−\sqrt{\mathrm{10}} \\ $$

Answered by mr W last updated on 03/Jun/24

Commented by mr W last updated on 03/Jun/24

AB=c=(√(a^2 +b^2 ))  R=(c/2)=((√(a^2 +b^2 ))/2)  (r−(a/2))^2 +(r−(b/2))^2 =(R−r)^2     r−(a+b−2R)=0  ⇒r=a+b−(√(a^2 +b^2 ))

$${AB}={c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${R}=\frac{{c}}{\mathrm{2}}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\left({r}−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({r}−\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${r}−\left({a}+{b}−\mathrm{2}{R}\right)=\mathrm{0} \\ $$$$\Rightarrow{r}={a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$

Commented by Tawa11 last updated on 21/Jun/24

weldone sir

$$\mathrm{weldone}\:\mathrm{sir} \\ $$

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