Đáp án:

$a) \,\,V_{S.BCD} = \dfrac{a^3\sqrt2}{12}$

$V_{S.AMN} = \dfrac{a^3\sqrt2}{48}$

$b)\,\,\widehat{((SCD);(ABCD))} \approx 55^o$

$c)\,\,d(SA;CD) = \dfrac{a\sqrt6}{3}$

Giải thích các bước giải:

a) Ta có: $ABCD$ là hình vuông cạnh $a$

$\Rightarrow AC = BD = a\sqrt2$

$\Rightarrow S_{ABCD} = a^2$

Gọi $O = AC\cap BD$

$\Rightarrow OA = OB =OC = OD = \dfrac{\sqrt2}{2}$

$\Rightarrow SO\perp (ABCD)$

Áp dụng định lý Pytago, ta được:

$SA^2 = SO^2 + OA^2$

$\Rightarrow SO = \sqrt{SA^2 - OA^2} = \sqrt{a^2 - \dfrac{a^2}{2}} = \dfrac{a\sqrt2}{2}$

Ta có:

$S_{BCD} = S_{ABD} = \dfrac{1}{2}S_{ABCD}$

$\Rightarrow S_{BCD} = S_{ABD} = \dfrac{a^2}{2}$

Do đó:

$V_{S.BCD} = \dfrac{1}{3}S_{BCD}.SO = \dfrac{1}{3}\cdot\dfrac{a^2}{2}\cdot\dfrac{a\sqrt2}{2} = \dfrac{a^3\sqrt2}{12}$

Ta có:

$SM = MB = \dfrac{1}{2}SB$

$SN = ND = \dfrac{1}{2}SD$

$\Rightarrow MN = \dfrac{1}{2}BD$ (tính chất đường trung bình)

$\Rightarrow S_{AMN} = \dfrac{1}{4}S_{SBD}$

$\Rightarrow V_{A.SMN} = \dfrac{1}{4}V_{A.SBD} = \dfrac{1}{4}V_{S.ABD} = \dfrac{1}{4}V_{S.BCD}$

$\Rightarrow V_{A.SMN} = \dfrac{1}{4}\cdot\dfrac{a^3\sqrt2}{12} = \dfrac{a^3\sqrt2}{48}$

b) Gọi $H$ là trung điểm $CD$

$\Rightarrow OH\perp CD$

$\Rightarrow OH = \dfrac{1}{2}CD = \dfrac{a}{2}$

Do $SC = SD = a$

$HC = HD = \dfrac{1}{2}CD$

$\Rightarrow SH\perp CD$

Ta có:

$\begin{cases}(ABCD)\cap (SCD)= CD\\OH\perp CD, \,OH\subset (ABCD)\\SH\perp CD,\,SH\subset (SCD)\end{cases}$

$\Rightarrow \widehat{((SCD);(ABCD))} = \widehat{SHO}$

$\Rightarrow \tan\widehat{SHO} = \dfrac{SO}{OH} = \dfrac{\dfrac{a\sqrt2}{2}}{\dfrac{a}{2}} = \sqrt2$

$\Rightarrow \widehat{SHO} \approx 55^o$

c) Ta có:

$CD//AB$

$\Rightarrow CD//(SAD)$

$\Rightarrow d(CD;(SAD)) = d(CD;SA)$ $(SA\subset (SAD))$

$\Rightarrow d(H;(SAD)) = d(CD;SA)$

Ta lại có:

$AB\perp SO\quad (SO\perp (ABCD))$

$AB\perp OH \quad (OH\perp CD)$

$\Rightarrow AB\perp (SHO)$

Kéo dài $OH$ cắt $AD$ tại $K$

$\Rightarrow AB\perp (SKH)$

Kẻ $HI\perp SK$

$\Rightarrow AB\perp HI$

$\Rightarrow HI\perp (SAD)$

$\Rightarrow HI = d(H;(SAD))$

Áp dụng định lý Pytago, ta được:

$SA^2 = AK^2 + SK^2$

$\Rightarrow SK = \sqrt{SA^2 - AK^2} = \sqrt{a^2 - \dfrac{a^2}{4}} = \dfrac{a\sqrt3}{2}$

Xét $ΔSHK$ có:

$SO.HK = HI.SK= 2S_{AHK}$

$\Rightarrow HI = \dfrac{SO.HK}{SK} = \dfrac{\dfrac{a\sqrt2}{2}\cdot a}{\dfrac{a\sqrt3}{2}} = \dfrac{a\sqrt6}{3}$