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ABC-AB-a-BC-b-AC-c-and-C-90-r-




Question Number 84304 by Wepa last updated on 11/Mar/20
ΔABC AB=a BC=b AC=c   and ∠C=90°  r=?
$$\Delta\mathrm{ABC}\:\mathrm{AB}=\mathrm{a}\:\mathrm{BC}=\mathrm{b}\:\mathrm{AC}=\mathrm{c}\: \\ $$$$\mathrm{and}\:\angle\mathrm{C}=\mathrm{90}° \\ $$$$\mathrm{r}=? \\ $$
Commented by mr W last updated on 11/Mar/20
r=((ab)/(a+b+c))
$${r}=\frac{{ab}}{{a}+{b}+{c}} \\ $$
Answered by $@ty@m123 last updated on 11/Mar/20
General formula:  r=(△/s), △=(1/2)absin C  Here C=90^o   ∴ r=((absin 90^o )/(2×((a+b+c)/2)))  ⇒ r=((ab)/(a+b+c))
$${General}\:{formula}: \\ $$$${r}=\frac{\bigtriangleup}{{s}},\:\bigtriangleup=\frac{\mathrm{1}}{\mathrm{2}}{ab}\mathrm{sin}\:{C} \\ $$$${Here}\:{C}=\mathrm{90}^{\mathrm{o}} \\ $$$$\therefore\:{r}=\frac{{ab}\mathrm{sin}\:\mathrm{90}^{\mathrm{o}} }{\mathrm{2}×\frac{{a}+{b}+{c}}{\mathrm{2}}} \\ $$$$\Rightarrow\:{r}=\frac{{ab}}{{a}+{b}+{c}} \\ $$

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