Bài 1 :
\(BC^2=AB^2+AC^2\left(Pitago\right)\)
\(\Leftrightarrow BC^2=9+16=25\)
\(\Leftrightarrow BC=5\left(cm\right)\)
\(AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{9}{5}\left(cm\right)\)
\(\Rightarrow CH=BC-BH=5-\dfrac{9}{5}=\dfrac{16}{5}\left(cm\right)\)
\(AH^2=BH.CH=\dfrac{9}{5}.\dfrac{16}{5}=\dfrac{3^2.4^2}{5^2}\)
\(\Leftrightarrow AH=\dfrac{3.4}{5}=\dfrac{12}{5}\left(cm\right)\)
Bài 4:
\(BC=BH+CH=25+144=169\left(cm\right)\)
\(AH^2=BH.CH=25.144=5^2.12^2\)
\(\Leftrightarrow AH=5.12=60\left(cm\right)\)
\(AB^2=BH.BC=25.169=5^2.13^2\)
\(\Leftrightarrow AB=5.13=65\left(cm\right)\)
\(AC^2=CH.BC=144.169=12^2.13^2\)
\(\Leftrightarrow AC=12.13=156\left(cm\right)\)
Bài 2 : \(AH=4cm;AM=5cm\)
Ta có :
\(AM=\sqrt[]{\dfrac{AB^2+AC^2}{2}-\dfrac{BC^2}{4}}\)
\(\Leftrightarrow AM=\sqrt[]{\dfrac{BC^2}{2}-\dfrac{BC^2}{4}}\)
\(\Leftrightarrow AM=\sqrt[]{\dfrac{BC^2}{4}}\)
\(\Leftrightarrow AM=\dfrac{BC}{2}\)
\(\Leftrightarrow BC=2AM=2.5=10\left(cm\right)\)
\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)
\(\Leftrightarrow\dfrac{1}{AH^2}=\dfrac{AB^2+AC^2}{AB^2.AC^2}\)
\(\Leftrightarrow\dfrac{1}{AH^2}=\dfrac{BC^2}{\left(AB.AC\right)^2}\)
\(\Leftrightarrow\left(AB.AC\right)^2=AH^2.BC^2=4^2.10^2\)
\(\Leftrightarrow AB.AC=4.10=40\left(cm^2\right)\)
\(AB^2+AC^2=BC^2\left(Pitago\right)\)
\(\Leftrightarrow\left(AB+AC\right)^2-2AB.AC=BC^2\)
\(\Leftrightarrow\left(AB+AC\right)^2=BC^2+2AB.AC\)
\(\Leftrightarrow\left(AB+AC\right)^2=10^2+2.40=180\)
\(\Rightarrow AB+AC=6\sqrt[]{5}\left(cm\right)\)
mà \(AB.AC=40\)
\(\Leftrightarrow AB.\left(6\sqrt[]{5}-AB\right)=40\)
\(\Leftrightarrow-AB^2+6\sqrt[]{5}AB=40\)
\(\Leftrightarrow AB^2-6\sqrt[]{5}AB+40=0\)
\(\Delta'=45-40=5\)
\(\Rightarrow\left[{}\begin{matrix}AB=3\sqrt[]{5}+\sqrt[]{5}=4\sqrt[]{5}\\AB=3\sqrt[]{5}-\sqrt[]{5}=2\sqrt[]{5}\end{matrix}\right.\)
- Với \(AB=4\sqrt[]{5}\Rightarrow AC=\dfrac{40}{4\sqrt[]{5}}=\dfrac{10}{\sqrt[]{5}}=2\sqrt[]{5}\left(cm\right)\)
- Với \(AB=2\sqrt[]{5}\Rightarrow AC=\dfrac{40}{2\sqrt[]{5}}=\dfrac{20}{\sqrt[]{5}}=4\sqrt[]{5}\left(cm\right)\)
