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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

The incircle of a rectangular triangle with sides a, b, (√(a^2 +b^2 )) is ((a+b−(√(a^2 +b^2 )))/2) (∗) We have ((h+10−(√(h^2 +100)))/2)=2 ⇒ h+6=(√(h^2 +100)) ⇒ h=((16)/3) (∗) Incircle of a triangle =((2×area)/(circumference)) Area of a rectangular triangle =((ab)/2) ⇒ Incircle =((ab)/(2(a+b+(√(a^2 +b^2 )))))=the above

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

Answered by cherokeesay last updated on 29/Jul/24

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