Question

Đã trả lời bởi chuyên gia

Answer 1

Gọi V là thể tích khối chóp S.ABCD.

Chia khối chóp S.CDMN làm 2 khối chóp: S.CDM và S.CMN

Ta có: \({V_{S.ABC}} = {V_{S.ACD}} = \frac{1}{2}{V_{S.ABCD}}\)

\(\frac{{{V_{S.CDM}}}}{{{V_{S.CDA}}}} = \frac{{SM}}{{SA}} = \frac{1}{2}\)

\( \Rightarrow {V_{S.CDM}} = \frac{1}{2}{V_{S.CDA}} = \frac{1}{2} \cdot \frac{1}{2}{V_{S.ABCD}} = \frac{1}{4}{V_{S.ABCD}}\)

\(\frac{{{V_{S.CMN}}}}{{{V_{S.CAB}}}} = \frac{{SM}}{{SA}} \cdot \frac{{SN}}{{SB}} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\)

\( \Rightarrow {V_{S.CMN}} = \frac{1}{4}{V_{S.CAB}} = \frac{1}{4} \cdot \frac{1}{2}{V_{S.ABCD}} = \frac{1}{8}{V_{S.ABCD}}\)

Do đó: \({V_{S.CDMN}} = {V_{S.CDM}} + {V_{S.CMN}} = \frac{1}{4}{V_{S.ABCD}} + \frac{1}{8}{V_{S.ABCD}} = \frac{3}{8}{V_{S.ABCD}}\)

\( \Rightarrow \frac{{{V_{S.CDMN}}}}{{{V_{S.CDAB}}}} = \frac{3}{8}\).

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Answer 2