a, Ta có:

\(\left\{{}\begin{matrix}AC\perp AB\\MN\perp AB\end{matrix}\right.\)( do MNPA là hình chữ nhật )

\(\Rightarrow AC//MN\)

\(\Rightarrow\widehat{MNB}=\widehat{PCN}\)( so le trong )

Xét \(\Delta MNB\)và \(\Delta PCN\)có:

     \(\widehat{BMN}=\widehat{NPC}=90^o\)(gt)

     \(\widehat{MNB}=\widehat{PCN}\)(chứng minh trên)

Suy ra : \(\Delta MNB\)ᔕ \(\Delta PCN\)( g.g )

\(\Rightarrow\dfrac{MN}{PC}=\dfrac{MB}{PN}\Rightarrow\dfrac{MN}{3}=\dfrac{4}{PN}\)

\(\Rightarrow MN.PN=12\)hay \(S_{ABC}=12\left(đvdt\right)\)

b, Ta có:

\(S_{ABC}=\dfrac{1}{2}AB.AC=\dfrac{1}{2}\left(4+MA\right)\left(AP+3\right)\)

          \(=\dfrac{1}{2}\left(12+MA.AP+3MA+4AP\right)=\dfrac{1}{2}\left(24+3MA+4AP\right)\)( do \(MA.AP=12\))

          \(=12+\dfrac{3MA}{2}+2AP=12+\dfrac{3MA}{2}+2.\dfrac{12}{MA}\)

    Áp dụng BDT Cô - si cho 2 số \(\dfrac{3MA}{2};\dfrac{24}{MA}\)không âm ta có:

        \(\dfrac{3MA}{2}+\dfrac{24}{MA}\ge2\sqrt{\dfrac{3MA}{2}.\dfrac{24}{MA}}=2.\sqrt{36}=12\)

Khi đó:; \(S_{ABC}\ge12+12=24\)

          Vậy \(S_{ABCmin}=24\Leftrightarrow\dfrac{3MA}{2}=\dfrac{24}{MA}\Leftrightarrow MA=4;PC=3\)

                                        \(\Leftrightarrow AB=8;AC=6\)