Bài toán thể tích kết hợp với bài toán tính khoảng cách giữa hai đường thẳng chéo nhau

A $\dpi{100} V_{ABCA'B'C'}=$ $\dpi{100} \frac{3a^{3}}{2}$ ; $\dpi{100} d_{AA'\rightarrow B'C}$ $\dpi{100} = \frac{a\sqrt{3}}{2}$

B $\dpi{100} V_{ABCA'B'C'}=$ $\dpi{100} \frac{3a^{3}}{4}$ ; $\dpi{100} d_{AA'\rightarrow B'C}$ = $\dpi{100} \frac{3a\sqrt{3}}{2}$

C $\dpi{100} V_{ABCA'B'C'}=\frac{3a^{3}}{4}$ ; $\dpi{100} d_{AA'\rightarrow B'C}$ $\dpi{100} = \frac{a\sqrt{3}}{2}$

D $\dpi{100} V_{ABCA'B'C'}=$ $\dpi{100} \frac{3a^{3}}{2}$ ; $\dpi{100} d_{AA'\rightarrow B'C}$ = $\dpi{100} \frac{3a\sqrt{3}}{2}$

A $V_{S.ABD}= \frac{\sqrt{3}a^{3}}{12}$ ; $d_{DE\rightarrow SC}= \frac{\sqrt{5}a}{5}$

B $V_{S.ABD}= \frac{\sqrt{3}a^{3}}{12}$ ; $d_{DE\rightarrow SC}= \frac{2\sqrt{5}a}{5}$

C $V_{S.ABD}= \frac{2\sqrt{3}a^{3}}{12}$ ; $d_{DE\rightarrow SC}= \frac{\sqrt{5}a}{5}$

D $V_{S.ABD}= \frac{2\sqrt{3}a^{3}}{12}$ ; $d_{DE\rightarrow SC}= \frac{2\sqrt{5}a}{5}$

A $\dpi{100} V_{ABCA'B'C'}=$ $\dpi{100} \frac{3a^{3}}{2}$ ; $\dpi{100} d_{AA'\rightarrow B'C}$ $\dpi{100} = \frac{a\sqrt{3}}{2}$

B $\dpi{100} V_{ABCA'B'C'}=$ $\dpi{100} \frac{3a^{3}}{4}$ ; $\dpi{100} d_{AA'\rightarrow B'C}$ = $\dpi{100} \frac{3a\sqrt{3}}{2}$

C $\dpi{100} V_{ABCA'B'C'}=\frac{3a^{3}}{4}$ ; $\dpi{100} d_{AA'\rightarrow B'C}$ $\dpi{100} = \frac{a\sqrt{3}}{2}$

D $\dpi{100} V_{ABCA'B'C'}=$ $\dpi{100} \frac{3a^{3}}{2}$ ; $\dpi{100} d_{AA'\rightarrow B'C}$ = $\dpi{100} \frac{3a\sqrt{3}}{2}$

A $V_{S.ABD}= \frac{\sqrt{3}a^{3}}{12}$ ; $d_{DE\rightarrow SC}= \frac{\sqrt{5}a}{5}$

B $V_{S.ABD}= \frac{\sqrt{3}a^{3}}{12}$ ; $d_{DE\rightarrow SC}= \frac{2\sqrt{5}a}{5}$

C $V_{S.ABD}= \frac{2\sqrt{3}a^{3}}{12}$ ; $d_{DE\rightarrow SC}= \frac{\sqrt{5}a}{5}$

D $V_{S.ABD}= \frac{2\sqrt{3}a^{3}}{12}$ ; $d_{DE\rightarrow SC}= \frac{2\sqrt{5}a}{5}$

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Toán học - Lớp 12