Tam giác ABC vuông tại A
\(\Rightarrow\)\(AB^2+AC^2=BC^2\)(định lý Pi-ta-go)
\(\Leftrightarrow\)\(AB^2+AC^2=24^2=576\)(1)
Thay \(AB=\frac{2}{3}AC\) vào (1), ta được:
\(\frac{4}{9}AC^2+AC^2=576\)\(\Leftrightarrow\)\(\frac{13}{9}AC^2=576\)
\(\Rightarrow\)\(AC^2=576:\frac{13}{9}\)\(=\frac{5184}{13}\)
\(\Rightarrow\)\(AC=\frac{72\sqrt{13}}{13}\left(cm\right)\)\(\Rightarrow\)\(AB=\frac{2}{3}.\frac{72\sqrt{13}}{13}=\frac{48\sqrt{13}}{13}\left(cm\right)\)