Cho hình lăng trụ ABC A'B'C' có AA'=a, góc g...

D

- Hướng dẫn giải

Đáp án D.

Gọi H là trọng tâm của tam giác ABC, từ giả thiết suy ra $B\text{'}H\perp \left(ABC\right)$  .

Khi đó 

$\begin{array}{l}\left(\widehat{BB\text{'},\left(ABC\right)}\right)=\left(\widehat{BB\text{'},BH}\right)\\ =\widehat{B\text{'}BH}=60°\end{array}$

Ta có 

$\begin{array}{l}BB\text{'}=a\\ \Rightarrow BH=BB\text{'}.\cos \widehat{B\text{'}BH}\\ =a.\cos 60°=\frac{a}{2},B\text{'}H\\ =\sqrt{B\text{'}{B}^2-B{H}^2}=\frac{a\sqrt{3}}{2}\end{array}$

Gọi M là trung điểm BC, suy ra $BH=\frac{2}{3}BM$$\Rightarrow BM=\frac{3}{2}BH=\frac{3}{2}.\frac{a}{2}=\frac{3a}{4}$  .

Đặt $AC=x>0$$\Rightarrow BC=AC.\tan \widehat{BAC}$$=x.\tan 60°=x\sqrt{3}$$\Rightarrow AB=\sqrt{A{B}^2+A{C}^2}=2x$  .

Lại có 

$\begin{array}{l}BM=\sqrt{B{C}^2+C{M}^2}\\ =\sqrt{B{C}^2+\frac{A{C}^2}{4}}\\ =\sqrt{3x^2+\frac{x^2}{4}}\\ =\frac{x\sqrt{13}}{2}=\frac{3a}{4}\\ \Rightarrow x=\frac{3a}{2\sqrt{13}}\end{array}$

 $\begin{array}{l}\Rightarrow AC=\frac{3a}{2\sqrt{13}},BC=\frac{3\sqrt{3}a}{2\sqrt{13}},\\ AB=\frac{6a}{2\sqrt{13}}\\ \Rightarrow S_{\Delta ABC}=\frac{1}{2}AC.BC=\frac{9\sqrt{3}{a}^2}{104}\end{array}$

(đvdt).

Vậy $V_{A\text{'}ABC}=\frac{1}{3}B\text{'}H.S_{\Delta ABC}$$=\frac{1}{3}.\frac{a\sqrt{3}}{2}.\frac{9\sqrt{3}{a}^2}{104}=\frac{9a^3}{208}$ (đvtt).