\(NP=NI+PI=8\left(cm\right)\)
Áp dụng HTL:
\(\left\{{}\begin{matrix}MP^2=PI\cdot PN=40\\MN^2=NI\cdot PN=24\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}MP=2\sqrt{10}\left(cm\right)\\NM=2\sqrt{6}\left(cm\right)\end{matrix}\right.\)
\(P_{MPN}=MN+NP+PM=2\sqrt{10}+2\sqrt{6}+8=2\left(\sqrt{10}+\sqrt{6}+4\right)\left(cm\right)\)
Xét ΔMPN vuông tại M có MI là đường cao ứng với cạnh huyền NP, ta được:
\(\left\{{}\begin{matrix}MP^2=PI\cdot PN\\MN^2=NI\cdot NP\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}MP=2\sqrt{10}\left(cm\right)\\MN=2\sqrt{6}\left(cm\right)\end{matrix}\right.\)
\(C_{MPN}=2\sqrt{10}+2\sqrt{6}+8\left(cm\right)\)
Ta có: \(NP=NI+PI=3+5=8\left(cm\right)\)
Áp dụng HTL:
\(\left\{{}\begin{matrix}MN^2=NI.NP=3.8=24\\MP^2=PI.NP=5.8=40\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}MN=2\sqrt{6}\left(cm\right)\\MP=2\sqrt{10}\left(cm\right)\end{matrix}\right.\)
\(P_{MNP}=8+2\sqrt{6}+2\sqrt{10}\approx19,2\left(cm\right)\)
