Ta có BC=BH+CH⇒BH+CH=26⇒CH=26-BH
Ta lại có △ABC vuông tại A đường cao AH⇒AH2=BH.CH⇒\(144=BH\left(26-BH\right)\Rightarrow26BH-BH^2=144\Rightarrow BH^2-26BH+144=0\Rightarrow\left[{}\begin{matrix}BH=8\left(cm\right)\\BH=18\left(cm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}CH=26-BH=26-8=18\left(cm\right)\\CH=26-BH=26-18=8\left(cm\right)\end{matrix}\right.\)Ta có △ABC vuông tại A đường cao AH⇒\(\left[{}\begin{matrix}AB^2=BH.BC=8.26=208\\AB^2=BH.BC=18.26=468\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}AB=4\sqrt{13}\left(cm\right)\\AB=6\sqrt{13}\left(cm\right)\end{matrix}\right.\)
Ta có △ABC vuông tại A⇒\(\left[{}\begin{matrix}BC^2=AB^2+AC^2\\BC^2=AB^2+AC^2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}AC^2=BC^2-AB^2=676-208=468\\AC^2=BC^2-AB^2=676-468=208\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}AC=6\sqrt{13}\\AC=4\sqrt{13}\end{matrix}\right.\)
