Lời giải:
Hạ đường cao $AH$ xuống cạnh $BC$ ($H\in BC$)
Khi đó:
\(S_{ABC}=\frac{AH.BC}{2};S_{ABD}=\frac{AH.BD}{2}\)
\(\Rightarrow \frac{S_{ABC}}{S_{ABD}}=\frac{BC}{BD}=\frac{BC}{BC-CD}=\frac{24}{24-4}=\frac{6}{5}\)
\(\Rightarrow S_{ABC}=\frac{6}{5}.S_{ABD}=\frac{6}{5}.17=20,4\) (cm vuông)
b) \(\frac{S_{ABC}}{S_{ABD}}=\frac{6}{5}=1,2=120:100=120(\text{%})\)