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.