Cho hình lăng trụ ABC.A'B'C' có mặt đáy ABC là tam giác đều, độ dài cạnh

* Hướng dẫn giải

Đáp án B.

Do H là trung điểm AB nên $\frac{d\left(B;\left(ACC\text{'}A\text{'}\right)\right)}{d\left(H;\left(ACC\text{'}A\text{'}\right)\right)}=\frac{BA}{HA}=2$

 $\Rightarrow d\left(B;\left(ACC\text{'}A\text{'}\right)\right)=2dd\left(H;\left(ACC\text{'}A\text{'}\right)\right)$

Ta có $AH\text{'}\perp \left(ABC\right)$ nên $\left(\stackrel{⏜}{AA\text{'},\left(ABC\right)}\right)=\left(\stackrel{⏜}{A\text{'}A,HA}\right)=\stackrel{⏜}{A\text{'}AH}=60°$

Gọi D là trung điểm của AC thì $BD\perp AC$.

 Kẻ $HE\perp AC,\left(E\in AC\right)\to HE//BD$

Ta có $\left\{\begin{array}{l}AC\perp A\text{'}H\\ AC\perp HE\end{array}\right.\Rightarrow AC\perp \left(A\text{'}HE\right)\perp \left(ACC\text{'}A\text{'}\right)$ 

Trong $\left(A\text{'}HE\right)$ kẻ $HK\perp A\text{'}E,\left(K\in A\text{'}E\right)\Rightarrow HK\perp \left(ACC\text{'}A\text{'}\right)$

Suy ra

$d\left(H;\left(ACC\text{'}A\text{'}\right)\right)=HK\Rightarrow 2d\left(B;\left(ACC\text{'}A\text{'}\right)\right)=2HK$

Ta có $BD=\frac{2a\sqrt{3}}{2}=a\sqrt{3}\Rightarrow HE=\frac{1}{2}BD=\frac{a\sqrt{3}}{2}$

Xét tam giác vuông $A\text{'}AH$ có $AH\text{'}=AH.\tan 60°=a\sqrt{3}$

Xét tam giác vuông $A\text{'}HE$ có  $\frac{1}{H{K}^2}=\frac{1}{A\text{'}{H}^2}+\frac{1}{H{E}^2}=\frac{1}{{\left(a\sqrt{3}\right)}^2}+\frac{1}{{\left({\displaystyle \frac{a\sqrt{3}}{2}}\right)}^2}=\frac{5}{3a^2}\Rightarrow HK=\frac{a\sqrt{15}}{5}$.

Vậy $d\left(B;\left(ACC\text{'}A\text{'}\right)\right)=2HK=\frac{2a\sqrt{15}}{5}$