Xét ΔHIK có DE//KI
nên \(\dfrac{x}{10}=\dfrac{\sqrt{3}}{5}\)
hay \(x=2\sqrt{3}\left(cm\right)\)
Ta có DE//KI áp dụng định lí Ta-let đảo có: \(\dfrac{DH}{DK}=\dfrac{HE}{EI}\Leftrightarrow\dfrac{\sqrt{3}}{5}=\dfrac{x}{10}\Rightarrow x=2\sqrt{3}\)
ta có: DE // KI
\(\Leftrightarrow\dfrac{HD}{KD}=\dfrac{HE}{IE}\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{5}=\dfrac{x}{10}\)
\(\Leftrightarrow5x=10\sqrt{3}\)
\(\Leftrightarrow x=2\sqrt{3}\)
Xét tam giác HKI có DE//KI
áp dụng đl ta-lét có:
\(\dfrac{x}{EI}=\dfrac{HD}{DK}=\dfrac{\sqrt{3}}{5}\)
\(\Leftrightarrow\dfrac{x}{10}=\dfrac{\sqrt{3}}{5}\Rightarrow x=\dfrac{\sqrt{3}}{5}.10=2\sqrt{3}\)
