a) Xét tam giác ABC có:
\(AB^2+BC^2=5^2+12^2=169=13^2=AC\)
=> Tam giác ABC vuông tại B
b) Áp dụng HTL:
\(AB^2=AH.AC\)
\(\Rightarrow AH=\dfrac{AB^2}{AC}=\dfrac{5^2}{13}\approx1,9\left(cm\right)\)
\(BC^2=CH.AC\)
\(\Rightarrow CH=\dfrac{BC^2}{AC}=\dfrac{12^2}{13}\approx11,1\left(cm\right)\)
\(BH^2=AH.CH\Rightarrow BH=\sqrt{AH.CH}\approx\sqrt{1,9.11,1}\approx4,6\left(cm\right)\)
\(1,\)
\(a,\)Vì \(AB^2+BC^2=5^2+12^2=169=13^2=AC^2\) nên tam giác ABC vuông tại B
\(b,\) Áp dụng HTL:
\(\left\{{}\begin{matrix}AB^2=AH\cdot AC\\BC^2=CH\cdot AC\\BH^2=AH\cdot HC\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}AH=\dfrac{AB^2}{AC}=\dfrac{25}{13}\left(cm\right)\\CH=\dfrac{BC^2}{AC}=\dfrac{144}{13}\left(cm\right)\\BH=\sqrt{\dfrac{25}{13}\cdot\dfrac{144}{13}}=\dfrac{60}{13}\left(cm\right)\end{matrix}\right.\)
\(2,\dfrac{AB}{AC}=\dfrac{3}{4}\Rightarrow AB=\dfrac{3}{4}AC\)
Áp dụng HTL:
\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{\dfrac{9}{16}AC^2}+\dfrac{1}{AC^2}\\ \Rightarrow\dfrac{1}{100}=\dfrac{16}{9AC^2}+\dfrac{1}{AC^2}\\ \Rightarrow9AC^2=1600+900\\ \Rightarrow AC^2=\dfrac{2500}{9}\Rightarrow AC=\dfrac{50}{3}\left(cm\right)\\ \Rightarrow AB=\dfrac{3}{4}\cdot\dfrac{50}{3}=\dfrac{50}{4}\left(cm\right)\\ \Rightarrow BC=\sqrt{AB^2+BC^2}=\dfrac{125}{6}\left(cm\right)\)
Áp dụng tiếp HTL:
\(\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=\dfrac{2500}{16}:\dfrac{125}{6}=7,5\left(cm\right)\\CH=\dfrac{AC^2}{BC}=\dfrac{40}{3}\left(cm\right)\end{matrix}\right.\)
