Câu hỏi 25

Dựng (P)

+) Qua M dựng \(PQ\parallel BD\)

+) Qua M dựng \(MT\parallel SA\)

\(\Rightarrow \left( P \right)\equiv \left( TPQ \right)\)

Tìm thiết diện.

+) Ta có: \(\left( TPQ \right)\cap \left( ABCD \right)=PQ\)

+) \(\left\{ \begin{align}  & Q\in \left( TPQ \right),\,Q\in \left( SAB \right) \\ & TM\parallel SA \\\end{align} \right.\)

\(\Rightarrow \left( TPQ \right)\cap \left( SAB \right)=QE\parallel TM\parallel SA\)

+) Tương tự \(\left( TPQ \right)\cap \left( SAD \right)=PE\parallel TM\parallel SA\)

\(\Rightarrow\) Thiết diện là ngũ giác PQETF

Tính diện tích thiết diện

\(\begin{align}  & {{S}_{TD}}=2{{S}_{MQET}} \\ & +)\,MQ\parallel OB\Rightarrow \frac{MQ}{OB}=\frac{AM}{AO}\Rightarrow MQ=AM=x \\ & +)\,TM\parallel SA\Rightarrow \frac{TM}{SA}=\frac{CM}{CA}\Rightarrow TM=\frac{\left( a\sqrt{2}-x \right)a}{a\sqrt{2}}=\frac{a\sqrt{2}-x}{\sqrt{2}} \\ & +)\,QE\parallel SA\Rightarrow \frac{QE}{SA}=\frac{BQ}{BA}=\frac{OM}{OA}\Rightarrow QE=\frac{a\sqrt{2}-2x}{\sqrt{2}} \\ & +)\,{{S}_{TD}}=\left( MT+QE \right).MQ=\left( \frac{a\sqrt{2}-x}{\sqrt{2}}+\frac{a\sqrt{2}-2x}{\sqrt{2}} \right).x=\frac{\left( 2a\sqrt{2}-3x \right).x}{\sqrt{2}} \\\end{align}\)