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

* Hướng dẫn giải

Đáp án D

Gọi N là trung điểm của BC

Ta có $AB//MN\Rightarrow d\left(AB;SM\right)=d\left(A;\left(SMN\right)\right)$

$SA=AC\tan 60°=5a\sqrt{3}$

$SM=\sqrt{{\left(5a\sqrt{3}\right)}^2+{\left(\frac{5a}{2}\right)}^2}=\frac{5a\sqrt{13}}{2}$

$S{N}^2=S{B}^2+B{N}^2=S{A}^2+A{B}^2+{\left(\frac{BC}{2}\right)}^2={\left(5a\sqrt{3}\right)}^2+{\left(3a\right)}^2+{\left(2a\right)}^2=88a^2$

$\Rightarrow SN=2a\sqrt{22}$

$MN=\frac{AB}{2}=\frac{3a}{2}$

Ta có:

$S{M}^2=N{S}^2+N{M}^2-2NS.NM.cos\widehat{MNS}\iff {\left(\frac{5a\sqrt{13}}{2}\right)}^{22}=88a^2+{\left(\frac{3a}{2}\right)}^2-2.2a.\sqrt{22}.\frac{3a}{2}cos\widehat{MNS}$

$cos\widehat{MNS}=\frac{3}{2\sqrt{22}}\Rightarrow \sin \widehat{MNS}=\sqrt{\frac{79}{88}}$

$S_{SMN}=\frac{1}{2}NM.NS.sin\stackrel{⏜}{MNS}=\frac{1}{2}.\frac{3a}{2}.2a\sqrt{22}.\sqrt{\frac{79}{88}}=\frac{3a^2\sqrt{79}}{4}$

$S_{AMN}=\frac{1}{4}{S}_{ABC}=\frac{1}{4}.\frac{1}{2}.3a.4a=\frac{3a^2}{2};V_{S.AMN}=\frac{1}{3}SA.S_{AMN}=\frac{1}{3}.5a\sqrt{3}.\frac{3a^2}{2}=\frac{5a^3\sqrt{3}}{2}$

 $d\left(A;\left(SMN\right)\right)=\frac{3V_{S.AMN}}{S_{SMN}}=\frac{3.\frac{5a^3\sqrt{3}}{2}}{\frac{3a^2\sqrt{79}}{4}}=\frac{10a\sqrt{3}}{\sqrt{79}}$