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Question Number 98149 by me2love2math last updated on 11/Jun/20

Commented by me2love2math last updated on 11/Jun/20

Q1 and 2 pls

$${Q}\mathrm{1}\:{and}\:\mathrm{2}\:{pls} \\ $$

Answered by mr W last updated on 11/Jun/20

Commented by mr W last updated on 11/Jun/20

let′s look at the area of triangle ABC, which is Δ_(ABC) =(√(s(s−a)(s−b)(s−c))) with s=((a+b+c)/2) we see also Δ_(ABC) =Δ_(ABO) +Δ_(ACO) −Δ_(BCO) and Δ_(ABO) =((cr)/2) Δ_(ACO) =((br)/2) Δ_(BCO) =((ar)/2) ⇒Δ_(ABC) =(((b+c−a)r)/2)=(((a+b+c)/2)−a)r=(s−a)r ⇒(s−a)r=(√(s(s−a)(s−b)(s−c))) ⇒r=(√((s(s−b)(s−c))/((s−a)))) with c=10, b=6, a=7 ⇒s=((10+6+7)/2)=((23)/2) ⇒r=(√((((23)/2)(((23)/2)−10)(((23)/2)−6))/((((23)/2)−7))))=((√(759))/6)≈4.59

$${let}'{s}\:{look}\:{at}\:{the}\:{area}\:{of}\:{triangle}\:{ABC}, \\ $$$${which}\:{is} \\ $$$$\Delta_{{ABC}} =\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$${with}\:{s}=\frac{{a}+{b}+{c}}{\mathrm{2}} \\ $$$${we}\:{see}\:{also} \\ $$$$\Delta_{{ABC}} =\Delta_{{ABO}} +\Delta_{{ACO}} −\Delta_{{BCO}} \\ $$$${and} \\ $$$$\Delta_{{ABO}} =\frac{{cr}}{\mathrm{2}} \\ $$$$\Delta_{{ACO}} =\frac{{br}}{\mathrm{2}} \\ $$$$\Delta_{{BCO}} =\frac{{ar}}{\mathrm{2}} \\ $$$$\Rightarrow\Delta_{{ABC}} =\frac{\left({b}+{c}−{a}\right){r}}{\mathrm{2}}=\left(\frac{{a}+{b}+{c}}{\mathrm{2}}−{a}\right){r}=\left({s}−{a}\right){r} \\ $$$$\Rightarrow\left({s}−{a}\right){r}=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$$\Rightarrow{r}=\sqrt{\frac{{s}\left({s}−{b}\right)\left({s}−{c}\right)}{\left({s}−{a}\right)}} \\ $$$$ \\ $$$${with}\:{c}=\mathrm{10},\:{b}=\mathrm{6},\:{a}=\mathrm{7} \\ $$$$\Rightarrow{s}=\frac{\mathrm{10}+\mathrm{6}+\mathrm{7}}{\mathrm{2}}=\frac{\mathrm{23}}{\mathrm{2}} \\ $$$$\Rightarrow{r}=\sqrt{\frac{\frac{\mathrm{23}}{\mathrm{2}}\left(\frac{\mathrm{23}}{\mathrm{2}}−\mathrm{10}\right)\left(\frac{\mathrm{23}}{\mathrm{2}}−\mathrm{6}\right)}{\left(\frac{\mathrm{23}}{\mathrm{2}}−\mathrm{7}\right)}}=\frac{\sqrt{\mathrm{759}}}{\mathrm{6}}\approx\mathrm{4}.\mathrm{59} \\ $$

Commented by me2love2math last updated on 12/Jun/20

many thx

$${many}\:{thx} \\ $$

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