Đặt tỉ số \(\dfrac{AB}{CD}=x\). Theo hình vẽ: AB > CD nên x > 1

Hạ \(PH\perp AB;PK\perp CD\)

Theo hình vẽ, ta có:

\(S_{PAB}=\dfrac{1}{2}PH.AB\Rightarrow PH=\dfrac{2S_{PAB}}{AB}=\dfrac{8}{x.CD}\)

\(S_{PCD}=\dfrac{1}{2}PK.CD\Rightarrow PK=\dfrac{2S_{PCD}}{CD}=\dfrac{4}{CD}\)

\(S_{ABCD}=\dfrac{1}{2}\left(PH+PK\right)\left(AB+CD\right)=\dfrac{1}{2}\left(\dfrac{8}{x.CD}+\dfrac{4}{CD}\right)\left(x.CD+CD\right)=\left(\dfrac{4}{x}+2\right)\left(x+1\right)\)

Mặt khác: \(S_{ABCD}=S_{PAB}+S_{PBC}+S_{PCD}+S_{PDA}=2+3+4+5=14\)

\(\Rightarrow14=\left(\dfrac{4}{x}+2\right)\left(x+1\right)\)

\(\Leftrightarrow7x=\left(2+x\right)\left(x+1\right)\)

\(\Leftrightarrow x^2-4x+2=0\)

\(\Leftrightarrow x=2\pm\sqrt{2}\)

Kết hợp với điều kiện x > 1 ta suy ra tỉ số \(\dfrac{AB}{CD}=2+\sqrt{2}\)