Câu hỏi 11

Ta có: \(\left\{ \matrix{M \in \left( \alpha \right) \cap \left( {ABCD} \right) \hfill \cr BC\parallel \left( \alpha \right) \hfill \cr BC \subset \left( {ABCD} \right) \hfill \cr} \right. \Rightarrow \left( \alpha \right) \cap \left( {ABCD} \right) = MN\parallel BC\,\left( {N \in AB} \right)\,\,\left( 1 \right).\)

Tương tự:

\(\eqalign{ & \left\{ \matrix{N \in \left( \alpha \right) \cap \left( {SAB} \right) \hfill \cr SA\parallel \left( \alpha \right) \hfill \cr SA \subset \left( {SAB} \right) \hfill \cr} \right. \Rightarrow \left( \alpha \right) \cap \left( {SAB} \right) = NP\parallel SA\,\left( {P \in SB} \right) \cr & \left\{ \matrix{P \in \left( \alpha \right) \cap \left( {SBC} \right) \hfill \cr BC\parallel \left( \alpha \right) \hfill \cr BC \subset \left( {SBC} \right) \hfill \cr} \right. \Rightarrow \left( \alpha \right) \cap \left( {SBC} \right) = PQ\parallel BC\,\left( {Q \in SC} \right)\,\,\left( 2 \right). \cr} \)

Từ (1) và (2) suy ra MN // PQ.

Vậy thiết diện là hình thang MNPQ.

Chọn B.