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\(1,\dfrac{AB}{AC}=\dfrac{5}{6}\Leftrightarrow AB=\dfrac{5}{6}AC\)
Áp dụng HTL tam giác
\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\Leftrightarrow\dfrac{1}{900}=\dfrac{1}{\dfrac{25}{36}AC^2}+\dfrac{1}{AC^2}\\ \Leftrightarrow\dfrac{1}{900}=\dfrac{36}{25AC^2}+\dfrac{1}{AC^2}\\ \Leftrightarrow\dfrac{1}{900}=\dfrac{36+25}{25AC^2}\Leftrightarrow\dfrac{1}{900}=\dfrac{61}{25AC^2}\\ \Leftrightarrow25AC^2=54900\Leftrightarrow AC^2=2196\Leftrightarrow AC=6\sqrt{61}\left(cm\right)\\ \Leftrightarrow AB=\dfrac{5}{6}\cdot6\sqrt{61}=5\sqrt{61}\\ \Leftrightarrow BC=\sqrt{AB^2+AC^2}=61\left(cm\right)\)
Áp dụng HTL tam giác:
\(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=...\\CH=\dfrac{AC^2}{BC}=...\end{matrix}\right.\)
Bài 1:
Ta có: \(\dfrac{AB}{AC}=\dfrac{5}{6}\)
\(\Leftrightarrow HB=\dfrac{25}{36}HC\)
Ta có: \(AH^2=HB\cdot HC\)
\(\Leftrightarrow HC^2\cdot\dfrac{25}{36}=900\)
\(\Leftrightarrow HC=36\left(cm\right)\)
hay HB=25(cm)
Bài 2:
Ta có: \(\dfrac{HB}{HC}=\dfrac{1}{3}\)
nên HC=3HB
Ta có: \(AH^2=HB\cdot HC\)
\(\Leftrightarrow HB^2=48\)
\(\Leftrightarrow HB=4\sqrt{3}\left(cm\right)\)
\(\Leftrightarrow BC=4\cdot HB=16\sqrt{3}\left(cm\right)\)
Bài 1:
ta có: \(AB=\dfrac{1}{2}AC\)
\(\Leftrightarrow\dfrac{HB}{HC}=\dfrac{1}{4}\)
\(\Leftrightarrow HC=4HB\)
Ta có: \(AH^2=HB\cdot HC\)
\(\Leftrightarrow HB=1\left(cm\right)\)
\(\Leftrightarrow HC=4\left(cm\right)\)
hay BC=5(cm)
Xét ΔBAC vuông tại A có AH là đường cao ứng với cạnh huyền BC
nên \(\left\{{}\begin{matrix}AB^2=HB\cdot BC\\AC^2=HC\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=\sqrt{5}\left(cm\right)\\AC=2\sqrt{5}\left(cm\right)\end{matrix}\right.\)
Ta có: \(\dfrac{HB}{HC}=\dfrac{9}{16}\Rightarrow HB=\dfrac{9}{16}HC\)
Ta có: \(AB^2=BH.BC=BH\left(BH+HC\right)=\dfrac{9}{16}HC\left(\dfrac{9}{16}HC+HC\right)\)
\(=\dfrac{9}{16}HC.\dfrac{25}{16}HC=\dfrac{225}{256}HC^2\)
\(\Rightarrow HC^2=\dfrac{256AB^2}{225}=\dfrac{16384}{25}\Rightarrow HC=\dfrac{128}{5}\left(cm\right)\)
\(\Rightarrow HB=\dfrac{72}{5}\Rightarrow BC=\dfrac{128+72}{5}=40\left(cm\right)\)
\(\Rightarrow AC=\sqrt{BC ^2-AB^2}=\sqrt{40^2-24^2}=32\)
Ta có: \(AB.AC=AH.BC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{24.32}{40}=\dfrac{96}{5}\left(cm\right)\)
\(\dfrac{HB}{HC}=\dfrac{9}{16}\Rightarrow HC=\dfrac{16}{9}HB\)
Áp dụng hệ thức lượng:
\(AB^2=HB.BC=HB\left(HB+HC\right)\)
\(\Leftrightarrow24^2=HB.\left(HB+\dfrac{16}{9}HB\right)\)
\(\Rightarrow HB^2=\dfrac{5184}{25}\Rightarrow HB=\dfrac{72}{5}\left(cm\right)\)
\(HC=\dfrac{16}{9}HB=\dfrac{128}{5}\) (cm)
\(BC=HB+HC=40\) (cm)
\(AC=\sqrt{BC^2-AB^2}=32\) (cm)
\(AH=\dfrac{AB.AC}{BC}=\dfrac{96}{5}\left(cm\right)\)
Ta có: \(\dfrac{AB}{AC}=\dfrac{5}{6}\Rightarrow AB=\dfrac{5}{6}AC\)
Áp dụng hệ thức lượng: \(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)
\(\Leftrightarrow\dfrac{1}{30^2}=\dfrac{1}{\left(\dfrac{5}{6}AC\right)^2}+\dfrac{1}{AC^2}=\dfrac{1}{AC^2}\left(\dfrac{1}{\left(\dfrac{5}{6}\right)^2}+1\right)=\dfrac{61}{25}.\dfrac{1}{AC^2}\)
\(\Rightarrow AC=6\sqrt{61}\)
\(AB=\dfrac{5}{6}AC=5\sqrt{61}\)
\(BC=\sqrt{AB^2+AC^2}=61\)
Áp dụng hệ thức lượng:
\(AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}=25\)
\(CH=BC-BH=36\)
Ta có: \(\dfrac{AB}{AC}=\dfrac{5}{7}\)
nên \(\dfrac{HB}{HC}=\dfrac{25}{49}\)
hay \(HB=\dfrac{25}{49}HC\)
Ta có: \(AH^2=HB\cdot HC\)
\(\Leftrightarrow HC^2=15^2:\dfrac{25}{49}=441\)
\(\Leftrightarrow HC=21\left(cm\right)\)
\(\Leftrightarrow HB=\dfrac{75}{7}\left(cm\right)\)
Bài 2:
Xét ΔABC có
\(BC^2=AB^2+AC^2\)
nên ΔABC vuông tại A
Xét ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC, ta được:
\(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH=\dfrac{25}{13}\left(cm\right)\\CH=\dfrac{144}{13}\left(cm\right)\end{matrix}\right.\)
Bài 1:
Ta có: \(\dfrac{AB}{AC}=\dfrac{5}{6}\)
\(\Leftrightarrow HB=\dfrac{25}{36}HC\)
Ta có: \(AH^2=HB\cdot HC\)
\(\Leftrightarrow HC^2\cdot\dfrac{25}{36}=900\)
\(\Leftrightarrow HC=36\left(cm\right)\)
hay HB=25(cm)