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Question-146000

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Question Number 146000 by mnjuly1970 last updated on 10/Jul/21
Commented by mr W last updated on 10/Jul/21
i think the area of ΔABC can′t be uniquely determined with the given condition. please recheck.
$${i}\:{think}\:{the}\:{area}\:{of}\:\Delta{ABC}\:{can}'{t}\:{be}\: \\ $$$${uniquely}\:{determined}\:{with}\:{the}\:{given}\: \\ $$$${condition}.\:{please}\:{recheck}. \\ $$
Commented by mr W last updated on 10/Jul/21
AC=a BC=b AB=c=(√(a^2 +b^2 )) r=((ab)/(a+b+c)) AD=a−r BD=b−r AD×BD=DE^2 =h^2 (here with h=1) (a−r)(b−r)=h^2 ab(((a+c)/(a+b+c)))(((b+c)/(a+b+c)))=h^2 ((ab(a+c)(b+c))/((a+b+c)^2 ))=h^2 ab=(1+(b/(a+c)))(1+(a/(b+c)))h^2 A_(ΔABC) =(1+(b/(a+c)))(1+(a/(b+c)))h^2 ≠unique
$${AC}={a} \\ $$$${BC}={b} \\ $$$${AB}={c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${r}=\frac{{ab}}{{a}+{b}+{c}} \\ $$$${AD}={a}−{r} \\ $$$${BD}={b}−{r} \\ $$$${AD}×{BD}={DE}^{\mathrm{2}} ={h}^{\mathrm{2}} \:\:\:\left({here}\:{with}\:{h}=\mathrm{1}\right) \\ $$$$\left({a}−{r}\right)\left({b}−{r}\right)={h}^{\mathrm{2}} \\ $$$${ab}\left(\frac{{a}+{c}}{{a}+{b}+{c}}\right)\left(\frac{{b}+{c}}{{a}+{b}+{c}}\right)={h}^{\mathrm{2}} \\ $$$$\frac{{ab}\left({a}+{c}\right)\left({b}+{c}\right)}{\left({a}+{b}+{c}\right)^{\mathrm{2}} }={h}^{\mathrm{2}} \\ $$$${ab}=\left(\mathrm{1}+\frac{{b}}{{a}+{c}}\right)\left(\mathrm{1}+\frac{{a}}{{b}+{c}}\right){h}^{\mathrm{2}} \\ $$$${A}_{\Delta{ABC}} =\left(\mathrm{1}+\frac{{b}}{{a}+{c}}\right)\left(\mathrm{1}+\frac{{a}}{{b}+{c}}\right){h}^{\mathrm{2}} \neq{unique} \\ $$
Commented by mr W last updated on 10/Jul/21

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