Cho hình chóp $S.ABC$ có đáy là tam giác vuông c...

B

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Ta có: $CI\cap \left( SAB \right)=B\Rightarrow \frac{d\left( C;\left( SAB \right) \right)}{d\left( I;\left( SAB \right) \right)}=\frac{CB}{IB}=2\Rightarrow d\left( C;\left( SAB \right) \right)=2d\left( I;\left( SAB \right) \right)$

$\begin{align}& IH\cap \left( SAB \right)=A\Rightarrow \frac{d\left( I;\left( SAB \right) \right)}{d\left( H;\left( SAB \right) \right)}=\frac{IA}{HA}=\frac{2}{3}\Rightarrow d\left( I;\left( SAB \right) \right)=\frac{2}{3}d\left( H;\left( SAB \right) \right) \\  & \Rightarrow d\left( C;\left( SAB \right) \right)=\frac{4}{3}d\left( H;\left( SAB \right) \right) \\ \end{align}$

Trong (ABC) kẻ $HD\bot AB\Rightarrow HD//AC$

Có: $\left. \begin{align}& AB\bot HD \\ & AB\bot SH\left( SH\bot \left( ABC \right) \right) \\ \end{align} \right\}\Rightarrow AB\bot \left( SHD \right)$

Trong (SHD) kẻ $HK\bot SD$

Có: $\left. \begin{align} & HK\bot SD \\  & HK\bot AB\left( AB\bot \left( SHD \right) \right) \\ \end{align} \right\}\Rightarrow HK\bot \left( SAB \right)\Rightarrow d\left( H;\left( SAB \right) \right)=HK$

Ta có: $\Delta ADH\sim \Delta AIB\left( g.g \right)\Rightarrow \frac{HD}{IB}=\frac{AH}{AB}$

Tam giác ABC vuông cân tại A nên $BC=AB\sqrt{2}=2a\Rightarrow AI=IB=\frac{BC}{2}=a$

$\Rightarrow AH=\frac{3}{2}AI=\frac{3}{2}a$

Suy ra $HD=\frac{IB.AH}{AB}=\frac{a.\frac{3}{2}a}{a\sqrt{2}}=\frac{3\sqrt{2}}{4}a$

Vì $SH\bot \left( ABC \right)\Rightarrow SH\bot HD\Rightarrow \Delta SHD$ vuông tại H.

$\begin{array}{l} \Rightarrow \frac{1}{{H{K^2}}} = \frac{1}{{H{S^2}}} + \frac{1}{{H{D^2}}} = \frac{7}{{8{a^2}}} + \frac{8}{{9{a^2}}} = \frac{{127}}{{72{a^2}}} \Rightarrow HK = \frac{{a\sqrt {72} }}{{\sqrt {127} }} = \frac{{6a\sqrt 2 }}{{\sqrt {127} }}.\\ \Rightarrow d\left( {C;\left( {SAB} \right)} \right) = \frac{4}{3}HK = \frac{{4.6a\sqrt 2 }}{{3\sqrt {127} }} = \frac{{8a\sqrt 2 }}{{\sqrt {127} }}. \end{array}$

Chọn B.