\(AD=\dfrac{1}{3}AC\)
=>\(S_{ABD}=\dfrac{1}{3}\cdot S_{ACB}=\dfrac{1}{3}\cdot20,25=6,75\left(dm^2\right)=675\left(cm^2\right)\)
Ta có: \(S_{ABD}+S_{DBC}=S_{ABC}\)
=>\(S_{BDC}=2025-675=1350\left(cm^2\right)\)
Xét ΔDBC có DK là đường cao
nên DK*BC=2*SBDC
=>\(DK\cdot50=2\cdot1350=2700\)
=>DK=54(cm)