Đáp án: $C$
Giải thích các bước giải:
Gọi $AC\cap BD=O$
Ta có $ABCD$ là hình thang có đáy lớn là $AB, AB=2CD\to AB//CD$
$\to\dfrac{OA}{OC}=\dfrac{OB}{OD}=\dfrac{AB}{CD}=\dfrac21$
$\to \dfrac{AO}{AO+OC}=\dfrac{2}{2+1}\to\dfrac{AO}{AC}=\dfrac23$
$\to\dfrac{AC}{AO}=\dfrac32$
Gọi $SO\cap AN=E$
Qua $E$ kẻ đường thẳng $MP$ song song $BD, M\in SD, P\in SB$
Ta có $A,E,N$ thẳng hàng
$\to \dfrac{AC}{AO}\cdot\dfrac{EO}{ES}\cdot\dfrac{NS}{NO}=1$
$\to \dfrac{3}{2}\cdot\dfrac{EO}{ES}\cdot1=1$ vì $N$ là trung điểm $SC$
$\to \dfrac{EO}{ES}=\dfrac23$
$\to \dfrac{EO+ES}{ES}=\dfrac{2+3}{3}$
$\to\dfrac{SO}{SE}=\dfrac53$
$\to\dfrac{SE}{SO}=\dfrac35$
Mà $MP//BD$
$\to\dfrac{SM}{SD}=\dfrac{SP}{SB}=\dfrac{SE}{SO}=\dfrac35$
Ta có:
$AB=2CD\to S_{ABC}=2S_{ACD}\to S_{ABC}=\dfrac23S_{ABCD}, S_{ACD}=\dfrac13S_{ABCD}$
$\to V_{SACD}=\dfrac13V_{SABCD}, V_{SABC}=\dfrac23V_{SABCD}$
Ta có:
$\dfrac{V_{SAMN}}{V_{SACD}}=\dfrac{SM}{SD}.\dfrac{SN}{SC}.\dfrac{SA}{SA}$
$\dfrac{V_{SAMN}}{V_{SACD}}=\dfrac{2}{3}.\dfrac{1}{2}.1=\dfrac13$
$\to S_{AMN}=\dfrac13S_{SACD}=\dfrac19V_{SABCD}$
Lại có:
$\dfrac{V_{SANP}}{V_{SABC}}=\dfrac{SA}{SA}.\dfrac{SN}{SC}.\dfrac{SP}{SB}=1.\dfrac12.\dfrac23=\dfrac13$
$\to V_{SANP}=\dfrac13V_{SABC}=\dfrac29V_{SABCD}$
$\to V_{SAMNP}=V_{SAMN}+V_{SANP}=\dfrac13V_{SABCD}$
$\to C$
