Question Number 188048 by Rupesh123 last updated on 25/Feb/23 | ||
Answered by mr W last updated on 25/Feb/23 | ||
$${say}\:{A}={area},\:{r}={inradius} \\ $$$${A}=\frac{{ab}}{\mathrm{2}} \\ $$$${c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${r}=\frac{\mathrm{2}{A}}{{a}+{b}+{c}}=\frac{{ab}}{{a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\:\:\:\leqslant\frac{{ab}}{\mathrm{2}\sqrt{{ab}}+\sqrt{\mathrm{2}{ab}}}=\frac{\sqrt{{ab}}}{\:\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)} \\ $$$${r}^{\mathrm{2}} \leqslant\frac{{ab}}{\mathrm{2}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\frac{{A}}{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{{A}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}<\frac{{A}}{\mathrm{3}+\mathrm{2}}=\frac{{A}}{\mathrm{5}} \\ $$$$\Rightarrow{A}>\mathrm{5}{r}^{\mathrm{2}} \\ $$ | ||
Commented by Rupesh123 last updated on 25/Feb/23 | ||
Great! | ||