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Question Number 210072 by peter frank last updated on 29/Jul/24

Answered by Frix last updated on 29/Jul/24

$$\mathrm{The}\mathrm{incircle}\mathrm{of}\mathrm{a}\mathrm{rectangular}\mathrm{triangle}\mathrm{with}$$$$\mathrm{sides}a,b,\sqrt{a^2+b^2}\mathrm{is}\frac{a+b-\sqrt{a^2+b^2}}{2}(\ast )$$$$\mathrm{We}\mathrm{have}\frac{h+10-\sqrt{h^2+100}}{2}=2$$$$\Rightarrow h+6=\sqrt{h^2+100}\Rightarrow h=\frac{16}{3}$$$$$$$$(\ast )\mathrm{Incircle}\mathrm{of}\mathrm{a}\mathrm{triangle}=\frac{2\times \mathrm{area}}{\mathrm{circumference}}$$$$\mathrm{Area}\mathrm{of}\mathrm{a}\mathrm{rectangular}\mathrm{triangle}=\frac{ab}{2}$$$$\Rightarrow \mathrm{Incircle}=\frac{ab}{2(a+b+\sqrt{a^2+b^2})}=\mathrm{the}\mathrm{above}$$

Answered by cherokeesay last updated on 29/Jul/24

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