Cho hình chóp S.ABCD có đáy là hình vuông cạnh a t...

C

- Hướng dẫn giải

Ta có:

\(\frac{{HI}}{{AD}} = \frac{{CH}}{{CA}} = \frac{3}{4} \Rightarrow HI = \frac{{3a}}{4}\)

\(\tan {60^0} = \frac{{SH}}{{HI}} \Rightarrow SH = \frac{{3\sqrt 3 }}{4}a\)

\(SI = \sqrt {S{H^2} + H{I^2}} = \sqrt {{{\left( {\frac{{3\sqrt 3 a}}{4}} \right)}^2} + {{\left( {\frac{{3a}}{4}} \right)}^2}} = \frac{3}{2}a\)

\(d\left( {G,\left( {SCD} \right)} \right) = \frac{3}{2}d\left( {J,\left( {SCD} \right)} \right) = \frac{2}{3}d\left( {K,\left( {SCD} \right)} \right) = \frac{2}{3}.\frac{4}{3}.d\left( {H,\left( {SCD} \right)} \right)\)

\( = \frac{8}{9}d\left( {H,\left( {SCD} \right)} \right) = \frac{8}{9}HL = \frac{8}{9}.\frac{{SH.HI}}{{SI}} = \frac{8}{9}\frac{{\frac{{3\sqrt 3 }}{4}a.\frac{{3a}}{4}}}{{\frac{{3a}}{2}}} = \frac{{\sqrt 3 }}{3}a\)