Đáp án:

$V_{S.ADNM} = \dfrac{a^3\sqrt6}{16}$ 

Giải thích các bước giải:

Ta có:

$\begin{cases}BD\perp AC\\SA\perp BD\quad (SA\perp (ABCD))\end{cases}$
$\Rightarrow BD\perp (SAC)$

$\Rightarrow BD\perp SO$

Khi đó:

$\begin{cases}(SBD)\cap (ABCD) = BD\\AO\perp BD\\AO\subset (ABCD)\\SO\perp BD\quad (cmt)\\SO\subset (SBD)\end{cases}$

$\Rightarrow \widehat{((SBD);(ABCD))} = \widehat{SOA} = 60^\circ$

$\Rightarrow SA = OA.\tan60^\circ = \dfrac{a\sqrt2}{2}\cdot \sqrt3 = \dfrac{a\sqrt6}{2}$

$\Rightarrow V_{S.ABCD} = \dfrac13S_{ABCD}.SA = \dfrac13\cdot a^2\cdot \dfrac{a\sqrt6}{2} = \dfrac{a^3\sqrt6}{6}$

Ta được:

$\quad V_{S.ADNM} = V_{S.AMN} + V_{S.AND}$

$\Leftrightarrow V_{S.ADNM} = \dfrac14V_{S.ABC} + \dfrac12V_{S.ACD}$

$\Leftrightarrow V_{S.ADNM} = \dfrac18V_{S.ABCD} + \dfrac14V_{S.ABCD}$

$\Leftrightarrow V_{S.ADNM} = \dfrac{3}{8}V_{S.ABCD}$

$\Leftrightarrow V_{S.ADNM} = \dfrac38\cdot \dfrac{a^3\sqrt6}{6}$

$\Leftrightarrow V_{S.ADNM} = \dfrac{a^3\sqrt6}{16}$