hình bên trái:
Ta có: \(BC=\sqrt{AB^2+AC^2}=\sqrt{4^2+\left(7,5\right)^2}=\dfrac{17}{2}\left(cm\right)\)
Ta có: \(\left\{{}\begin{matrix}AB^2=BH.BC\\AC^2=CH.BC\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4^2=x.\dfrac{17}{2}\\\left(7,5\right)^2=y.\dfrac{17}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{32}{17}\left(cm\right)\\y=\dfrac{225}{34}\left(cm\right)\end{matrix}\right.\)
hình bên phải:
Ta có: \(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{6^2}+\dfrac{1}{8^2}=\dfrac{25}{276}\Rightarrow AH^2=\dfrac{276}{25}\)
\(\Rightarrow AH=\dfrac{24}{5}\Rightarrow z=\dfrac{24}{5}\)
Ta có: \(BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=10\left(cm\right)\)
\(\Rightarrow y=10\left(cm\right)\)
Ta có: \(AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{6^2}{10}=\dfrac{36}{10}\left(cm\right)\Rightarrow x=\dfrac{36}{10}\left(cm\right)\)
