Question

Đã trả lời bởi chuyên gia

Answer 1

Chọn D.

Đặt

\{\begin{cases} V_{1} = V_{S A B I K N} \\ V_{2} = V_{N B C D I K} \end{cases} \to \frac{V_{1}}{V_{2}} = ?

*

V_{S . A B C D} = \frac{1}{3} . \frac{a \sqrt{6}}{2} a^{2} = \frac{\sqrt{6}}{6} a^{3}

*

V_{N . B M C} = \frac{1}{3} . N H . S_{Δ B M C} = \frac{1}{3} . \frac{S O}{2} . S_{Δ B M C} = \frac{1}{3} \frac{a \sqrt{6}}{4} . \frac{1}{2} . a .2 a = \frac{\sqrt{6}}{12} a^{3}

* Nhận thấy K là trọng tâm của tam giác SMC

\to \frac{M K}{M N} = \frac{2}{3}

*

\frac{V_{M . D I K}}{V_{M . C B N}} = \frac{M D}{M C} . \frac{M I}{M B} . \frac{M K}{M N} = \frac{1}{2} . \frac{1}{2} . \frac{2}{3} = \frac{1}{6}

\begin{array}{l} \to V_{2} = V_{M . C B N} - V_{M . D I K} = \frac{5}{6} V_{M . C B N} = \frac{5}{6} . \frac{\sqrt{6}}{12} a^{3} = \frac{5 \sqrt{6}}{72} a^{3} \\ \to V_{1} = V_{S . A B C D} - V_{2} = \frac{\sqrt{6}}{6} a^{3} - \frac{5 \sqrt{6}}{72} a^{3} = \frac{7 \sqrt{6}}{72} a^{3} \to \frac{V_{1}}{V_{2}} = \frac{\frac{7 \sqrt{6}}{72} a^{3}}{\frac{5 \sqrt{6}}{72} a^{3}} = \frac{7}{5} \end{array}

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Answer 2

Chọn D.

Đặt

\{\begin{cases} V_{1} = V_{S A B I K N} \\ V_{2} = V_{N B C D I K} \end{cases} \to \frac{V_{1}}{V_{2}} = ?

*

V_{S . A B C D} = \frac{1}{3} . \frac{a \sqrt{6}}{2} a^{2} = \frac{\sqrt{6}}{6} a^{3}

*

V_{N . B M C} = \frac{1}{3} . N H . S_{Δ B M C} = \frac{1}{3} . \frac{S O}{2} . S_{Δ B M C} = \frac{1}{3} \frac{a \sqrt{6}}{4} . \frac{1}{2} . a .2 a = \frac{\sqrt{6}}{12} a^{3}

* Nhận thấy K là trọng tâm của tam giác SMC

\to \frac{M K}{M N} = \frac{2}{3}

*

\frac{V_{M . D I K}}{V_{M . C B N}} = \frac{M D}{M C} . \frac{M I}{M B} . \frac{M K}{M N} = \frac{1}{2} . \frac{1}{2} . \frac{2}{3} = \frac{1}{6}

\begin{array}{l} \to V_{2} = V_{M . C B N} - V_{M . D I K} = \frac{5}{6} V_{M . C B N} = \frac{5}{6} . \frac{\sqrt{6}}{12} a^{3} = \frac{5 \sqrt{6}}{72} a^{3} \\ \to V_{1} = V_{S . A B C D} - V_{2} = \frac{\sqrt{6}}{6} a^{3} - \frac{5 \sqrt{6}}{72} a^{3} = \frac{7 \sqrt{6}}{72} a^{3} \to \frac{V_{1}}{V_{2}} = \frac{\frac{7 \sqrt{6}}{72} a^{3}}{\frac{5 \sqrt{6}}{72} a^{3}} = \frac{7}{5} \end{array}