Hình 1:
Áp dụng HTL:
\(AH^2=BH\cdot CH\\ \Rightarrow x=CH=\dfrac{AH^2}{BH}=4,5\left(cm\right)\\ AC^2=CH\cdot BC=4,5\left(4,5+2\right)=29,25\\ \Rightarrow y=AC=\dfrac{3\sqrt{13}}{2}\left(cm\right)\)
Hình 2:
\(\dfrac{AB}{AC}=\dfrac{3}{4}\Rightarrow AB=\dfrac{3}{4}AC\)
Áp dụng Pytago: \(AB^2+AC^2=BC^2=225\)
\(\Rightarrow\dfrac{9}{16}AC^2+AC^2=225\Rightarrow\dfrac{25}{16}AC^2=225\\ \Rightarrow AC^2=144\Rightarrow AC=12\left(cm\right)\\ \Rightarrow AB=12\cdot\dfrac{3}{4}=9\left(cm\right)\)
Áp dụng HTL:
\(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=BH=\dfrac{AB^2}{BC}=5,4\left(cm\right)\\y=CH=\dfrac{AC^2}{BC}=9,6\left(cm\right)\end{matrix}\right.\)
a)
tam giác ABC vuông tại A , đường cao AH có
*AH2=BH.CH
9=2.x
x=4,5cm
*AC2=CH.BC
y2=4,5(4,5+2)
y2=4,5.6,5
y2=29,25
y=\(\dfrac{3\sqrt{13}}{2}\)cm
