Question

Answer 1

\(\dfrac{AB}{AC}=\dfrac{3}{4}\Leftrightarrow AB=\dfrac{3}{4}AC\)

Áp dụng HTL tam giác

\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\Leftrightarrow\dfrac{1}{144}=\dfrac{1}{\dfrac{9}{16}AC^2}+\dfrac{1}{AC^2}\\ \Leftrightarrow\dfrac{1}{144}=\dfrac{16+9}{9AC^2}\Leftrightarrow9AC^2=144\cdot25=3600\\ \Leftrightarrow AC^2=400\Leftrightarrow AC=20\left(cm\right)\Leftrightarrow AB=\dfrac{3}{4}\cdot20=15\left(cm\right)\\ \Leftrightarrow BC=\sqrt{AB^2+AC^2}=25\left(cm\right)\left(pytago\right)\)

Áp dụng HTL tam giác

\(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=\dfrac{15^2}{25}=9\left(cm\right)\\CH=\dfrac{AC^2}{BC}=\dfrac{20^2}{25}=16\left(cm\right)\end{matrix}\right.\)