Question

[thuy]

ΔABC vuông tại A, AB<AC, đường cao AH dài 4cm, BC=10cm. Tính \(\frac{AB}{AC}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

Xét ΔABC vuông tại A có 

\(\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{AH^2}\)

\(\Leftrightarrow\dfrac{1}{100-AC^2}+\dfrac{1}{AC^2}=\dfrac{1}{16}\)

\(\Leftrightarrow\dfrac{AC^2+100-AC^2}{AC^2\left(100-AC^2\right)}=\dfrac{1}{16}\)

\(\Leftrightarrow100AC^2-AC^4=1600\)

\(\Leftrightarrow AC^4-100AC^2+1600=0\)

\(\Leftrightarrow AC^4-80AC^2-20AC^2+1600=0\)

\(\Leftrightarrow\left(AC^2-80\right)\left(AC^2-20\right)=0\)

=>\(AC=2\sqrt{5}\left(cm\right)\)

=>\(AB=4\sqrt{5}\left(cm\right)\)

=>AB/AC=2