Xét $ΔABC(A=90^o)$
$⇒BC^2=AB^2+AC^2$
$⇒AC=\sqrt{BC^2-AB^2}=\sqrt{5^2-3^2}=4(cm)$
$S_{ΔABC}=\dfrac{1}{2}.AH.BC=\dfrac{1}{2}.AB.AC$
$⇒AH.BC=AB.AC$
$⇒AH=\dfrac{AB.AC}{BC}=\dfrac{3.4}{5}=2,4(cm)$
Xét $ΔABH(BHA=90^o):$
$⇒BA^2=BH^2+AH^2$
$⇒BH=\sqrt{BA^2-AH^2}=\sqrt{3^2-2,4^2}=1,8(cm)$
Đáp án $B$