Question Number 210072 by peter frank last updated on 29/Jul/24
Answered by Frix last updated on 29/Jul/24
$$\mathrm{The}\:\mathrm{incircle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangular}\:\mathrm{triangle}\:\mathrm{with} \\ $$$$\mathrm{sides}\:{a},\:{b},\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\mathrm{is}\:\frac{{a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{2}}\:\:\:\:\:\left(\ast\right) \\ $$$$\mathrm{We}\:\mathrm{have}\:\frac{{h}+\mathrm{10}−\sqrt{{h}^{\mathrm{2}} +\mathrm{100}}}{\mathrm{2}}=\mathrm{2} \\ $$$$\Rightarrow\:{h}+\mathrm{6}=\sqrt{{h}^{\mathrm{2}} +\mathrm{100}}\:\Rightarrow\:{h}=\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$ \\ $$$$\left(\ast\right)\:\mathrm{Incircle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:=\frac{\mathrm{2}×\mathrm{area}}{\mathrm{circumference}} \\ $$$$\mathrm{Area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangular}\:\mathrm{triangle}\:=\frac{{ab}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{Incircle}\:=\frac{{ab}}{\mathrm{2}\left({a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)}=\mathrm{the}\:\mathrm{above} \\ $$
Answered by cherokeesay last updated on 29/Jul/24
