Đáp án:

$A.\ V = \dfrac{a^3\sqrt6}{16}$

Giải thích các bước giải:

Gọi $O$ là tâm của đáy

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

Do $\triangle SAB=\triangle SAD\ (c.g.c)$

$\Rightarrow SB = SD$

Lại có: $OB = OD$

nên $SO\perp BD$

Ta có:

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

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

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

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

Mặt khác:

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

$\Leftrightarrow V_{S.ADNM} = \dfrac{SM}{SB}\cdot \dfrac{SN}{SC}\cdot V_{S.ABC} + \dfrac{SN}{SC}\cdot V_{S.ACD}$

$\Leftrightarrow V_{S.ADNM} = \dfrac12\cdot \dfrac12V_{S.ABC} + \dfrac12V_{S.ACD}$

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

$\Leftrightarrow V_{S.ADNM} = \dfrac14\cdot \dfrac12V_{S.ABCD} + \dfrac12\cdot \dfrac12V_{S.ABCD}$

$\Leftrightarrow V_{S.ADNM} = \dfrac38V_{S.ABCD} = \dfrac38\cdot \dfrac{a^3\sqrt6}{6}$

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