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![](https://rs.olm.vn/images/avt/0.png?1311)
Theo đề, ta có:
\(HB\left(13-HB\right)=36\)
\(\Leftrightarrow HB^2-13HB+36=0\)
\(\Leftrightarrow HB=4\left(cm\right)\)
hay HC=9(cm)
Áp dụng HTL:
\(AH\cdot BC=AB\cdot AC\Rightarrow AB\cdot AC=78\Rightarrow AB=\dfrac{78}{AC}\)
\(AB^2+AC^2=BC^2=169\\ \Leftrightarrow\dfrac{6084}{AC^2}+AC^2=169\\ \Leftrightarrow\dfrac{6084+AC^4}{AC^2}=\dfrac{169AC^2}{AC^2}\\ \Leftrightarrow AC^4-169AC^2+6084=0\\ \Leftrightarrow AC^4-117AC^2-52AC^2+6084=0\\ \Leftrightarrow AC^2\left(AC^2-117\right)-52\left(AC^2-117\right)=0\\ \Leftrightarrow\left(AC^2-52\right)\left(AC^2-117\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}AC^2=52\\AC^2=117\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}AC=2\sqrt{13}\\AC=3\sqrt{13}\end{matrix}\right.\left(AC>0\right)\)
Mà AC là cạnh lớn nên \(AC=3\sqrt{13}\left(cm\right)\) và \(AB=2\sqrt{13}\left(cm\right)\)
Tiếp tục áp dụng HTL:
\(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=4\left(cm\right)\\CH=\dfrac{AC^2}{BC}=9\left(cm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: AB=căn 4,5*12,5=7,5cm
AC=căn 8*12,5=10cm
b: HB=(13+5)/2=9cm
HC=13-9=4cm
AB=căn 9*13=3 căn 13cm
AC=căn 4*13=2căn 13cm
![](https://rs.olm.vn/images/avt/0.png?1311)
AB/AC=4/3
=>HB/HC=16/9
=>HB/16=HC/9=k
=>HB=16k; HC=9k
AH^2=HB*HC
=>144k^2=24^2=576
=>k=2
=>HB=32cm; HC=18cm
AB=căn 32*50=40cm
AC=căn 18*50=30cm
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\dfrac{AB}{AC}=\dfrac{5}{6}\Leftrightarrow AB=\dfrac{5}{6}AC\)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\left\{{}\begin{matrix}AB^2=HB\cdot BC\\AC^2=HC\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=\sqrt{5}\left(cm\right)\\AC=2\sqrt{5}\left(cm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(AH^2=HB.HC=50.8=400\)
\(\Rightarrow AH=20\left(cm\right)\)
\(S_{ABC}=\dfrac{1}{2}AH.BC=\dfrac{1}{2}.20\left(50+8\right)=\dfrac{1}{2}.20.58\left(cm^2\right)\)
mà \(S_{ABC}=\dfrac{1}{2}AB.AC\)
\(\Rightarrow AB.AC=20.58=1160\)
Theo Pitago cho tam giác vuông ABC :
\(AB^2+AC^2=BC^2\)
\(\Rightarrow\left(AB+AC\right)^2-2AB.AC=BC^2\)
\(\Rightarrow\left(AB+AC\right)^2=BC^2+2AB.AC\)
\(\Rightarrow\left(AB+AC\right)^2=58^2+2.1160=5684\)
\(\Rightarrow AB+AC=\sqrt[]{5684}=2\sqrt[]{1421}\left(cm\right)\)
Chu vi Δ ABC :
\(AB+AC+BC=2\sqrt[]{1421}+58=2\left(\sqrt[]{1421}+29\right)\left(cm\right)\)