Cho hình chóp S.ABCD có đáy là h...

A

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Ta có: \(S{{M}^{2}}={{\left( 2a \right)}^{2}}-{{a}^{2}}=3{{a}^{2}}\)

\(S{{M}^{2}}=M{{N}^{2}}+S{{N}^{2}}-2MN.SN\cos {{60}^{0}}\)

\(\Leftrightarrow 3{{a}^{2}}={{\left( 2a \right)}^{2}}+S{{N}^{2}}-2.2a.SN.\frac{1}{2}\Leftrightarrow S{{N}^{2}}-2aSN+{{a}^{2}}=0\)

\(\Leftrightarrow {{\left( SN-a \right)}^{2}}=0\Leftrightarrow SN=a\)

\(SH=SN\sin {{60}^{0}}=\frac{a\sqrt{3}}{2};MP=\sqrt{{{a}^{2}}+{{a}^{2}}}=a\sqrt{2}\)

\(HN=SN\cos {{60}^{0}}=\frac{a}{2}\Rightarrow HO=a-\frac{a}{2}=\frac{a}{2}\)

Ta có: \(\frac{OM}{HM}=\frac{a}{\frac{3a}{2}}=\frac{2}{3}\) nên \(d\left( O;\left( SMP \right) \right)=\frac{2}{3}d\left( H;\left( SMP \right) \right)\)

\(PN=\sqrt{{{a}^{2}}+{{a}^{2}}}=a\sqrt{2}.\) Mà \(\frac{KH}{PN}=\frac{MH}{MN}\)

\(\Rightarrow KH=\frac{MH}{MN}.PN=\frac{\frac{3a}{2}}{2a}a\sqrt{2}=\frac{3a\sqrt{2}}{4}\)

\(\frac{1}{I{{H}^{2}}}=\frac{1}{H{{S}^{2}}}+\frac{1}{H{{K}^{2}}}=\frac{1}{{{\left( \frac{a\sqrt{3}}{2} \right)}^{2}}}+\frac{1}{{{\left( \frac{3a\sqrt{2}}{4} \right)}^{2}}}\Rightarrow IH=\frac{3a\sqrt{5}}{10}\)

\(\Rightarrow d\left( O;\left( SMP \right) \right)=\frac{2}{3}d\left( H;\left( SMP \right) \right)=\frac{2}{3}IH=\frac{2}{3}.\frac{3a\sqrt{5}}{10}=\frac{a\sqrt{5}}{5}.\)

Chọn A.