- \(\Delta AHB\) ; \(\widehat{H}=90^o\)
\(AB^2=BH^2+HA^2\) ( ĐL pitago )
\(\Rightarrow AH^2=AB^2-BH^2=30^2-BH^2=900-BH^2\) (1)
- \(\Delta AHC\); \(\widehat{H}=90^o\) ; \(AH\perp BC\)
\(AH^2=BH.HC\) ( hệ thức liên quan tới đường cao )
\(\Leftrightarrow AH^2=32.BH\) (1)
- Từ (1) và (2) => \(900-BH^2=32.BH\)
\(\Leftrightarrow BH^2+32.BH-900=0\)
\(\Leftrightarrow\left(BH^2+50.BH\right)-\left(18.BH+900\right)=0\)
\(\Leftrightarrow BH.\left(BH+50\right)-18\left(BH+50\right)=0\)
\(\Leftrightarrow\left(BH+50\right)\left(BH-18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}BH+50=0\\BH-18=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}BH=-50\left(loai\right)\\BH=18\left(nhan\right)\end{matrix}\right.\)
BC = BH + HC = 18 + 32 = 50 cm
- \(\Delta ABC\) ; \(\widehat{A}=90^o\)
\(BC^2=AB^2+AC^2\) ( ĐL pitago )
\(\Rightarrow AC^2=BC^2-AB^2=50^2-32^2=1600\)
\(\Rightarrow AC=\sqrt{1600}=40\left(cm\right)\)