\(\dfrac{QH}{HR}=\dfrac{1}{2}\Rightarrow HR=2QH\)
Xét tam giác PQR vuông tại P có PH là đường cao ta có:
\(PH^2=HR\cdot QH\)
\(\Rightarrow4^2=2QH\cdot QH\)
\(\Rightarrow16=2QH^2\)
\(\Rightarrow QH^2=8\)
\(\Rightarrow QH=2\sqrt{2}\left(cm\right)\)
\(\Rightarrow HR=2\cdot2\sqrt{2}=4\sqrt{2}\left(cm\right)\)
\(\Rightarrow QR=4\sqrt{2}+2\sqrt{2}=6\sqrt{2}\left(cm\right)\)