Question

Answer 1

Áp dụng HTL trong tam giác ABC vg tại A có đg cao AH:

\(BH.HC=AH^2\)

\(\Rightarrow xy=6^2=36\Rightarrow y=\dfrac{36}{x}\)

\(\left\{{}\begin{matrix}AB^2=BH.BC\\AC^2=HC.BC\end{matrix}\right.\)\(\Rightarrow\dfrac{AB^2}{AC^2}=\dfrac{BH.BC}{HC.BC}=\dfrac{BH}{HC}\)

\(\Rightarrow\dfrac{x}{y}=\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\)

\(\Rightarrow x:\dfrac{36}{x}=\dfrac{9}{16}\Rightarrow\dfrac{x^2}{36}=\dfrac{9}{16}\Rightarrow x^2=\dfrac{81}{4}\Rightarrow x=4,5\left(cm\right)\)

\(\Rightarrow y=\dfrac{36}{x}=\dfrac{36}{4,5}=8\left(cm\right)\)