Cho lăng trụ tam giác ABCA'B'C' có đáy ABC là tam...

A $\dpi{100} V_{ABCA'B'C'}= a^{3}$ ; $\dpi{100} d_{B'\rightarrow (A'ACC')}$ = $\dpi{100} \frac{2\sqrt{39}a}{13}$

B $\dpi{100} V_{ABCA'B'C'}$ = $\dpi{100} 2a^{3}$ ; $\dpi{100} d_{B'\rightarrow (A'ACC')}$ = $\dpi{100} \frac{2\sqrt{39}a}{13}$

C $\dpi{100} V_{ABCA'B'C'}$ = $\dpi{100} 2a^{3}$ ; $\dpi{100} d_{B'\rightarrow (A'ACC')}$ = $\dpi{100} \frac{\sqrt{39}a}{13}$

D $\dpi{100} V_{ABCA'B'C'}= a^{3}$ ; $\dpi{100} d_{B'\rightarrow (A'ACC')}$ = $\dpi{100} \frac{\sqrt{39}a}{13}$

Đáp án

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Có: B = $\dpi{100} S_{ABC}=\frac{1}{2}.AB.AC$ = $\dpi{100} \frac{a^{2}\sqrt{3}}{2}$ (0,5đ)

Có: AG = $\dpi{100} \frac{2}{3}.AN$ = $\dpi{100} \frac{2}{3}.\frac{1}{2}.BC= \frac{2a}{3}$ (0,5đ)

Xét tam giác vuông A'GA có:

tan 60 = $\dpi{100} \frac{A'G}{AG}$

=> A'G = AG. tan 60 = $\dpi{100} \frac{2a\sqrt{3}}{3}=h$ (0,5đ)

=> $\dpi{100} V_{ABCA'B'C'}= a^{3}$ (0,5đ)

Có: $\dpi{100} \frac{d_{B\rightarrow (A'ACC')}}{d_{G\rightarrow (A'ACC')}}=\frac{MB}{MG}= 3$

=> $\dpi{100} d_{B\rightarrow (A'ACC')}= 3d_{G\rightarrow (A'ACC')}$ (0,5đ)

Kẻ GK ⊥ AC.

=> GK // AB

Kẻ GI ⊥ A'K

=> $\dpi{100} d_{G\rightarrow (A'ACC')}= GI$ (0,5đ)

Xét tam giác ABM có GK //AB

$\dpi{100} \frac{GK}{AB}=\frac{MG}{MB}=\frac{1}{3}$

=> $\dpi{100} GK = \frac{1}{3}.AB=\frac{a}{3}$ (0,5đ)

Có: $\dpi{100} A'G = \frac{2\sqrt{3}a}{3}$

=> $\dpi{100} \frac{1}{GI^{2}}=\frac{1}{A'G^{2}}+\frac{1}{GK^{2}}$

=> $\dpi{100} GI= \frac{2a}{\sqrt{39}}$ (0,5đ)

=> $\dpi{100} d_{G\rightarrow (A'ACC')}=GI$ = $\dpi{100} \frac{2\sqrt{39}a}{39}$ (0,5đ)

=> $\dpi{100} d_{B\rightarrow (A'ACC')}=3d_{G\rightarrow (A'ACC')} =$ $\dpi{100} \frac{2\sqrt{39}a}{13}$ (0,5đ)

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