Đáp án:
A
Bài làm:
Gọi $AC\cap BD=E, SE\cap MN=F, AF\cap SC=I$
$\to (AMN)\cap (SABCD)=AMIN$
Vì M, N là trung điểm SD, SB $\to F$ là trung điểm SE
Mà A,F,I thẳng hàng
$\to \dfrac{AC}{AE}.\dfrac{FE}{FS}.\dfrac{IS}{IC}=1$ (Định lý menelaus)
$\to 2.1.\dfrac{IS}{IC}=1$
$\to\dfrac{SI}{IC}=\dfrac 12$
$\to\dfrac{SI}{SI+IC}=\dfrac{1}{1+2}\to\dfrac{SI}{SC}=\dfrac 13$
Ta có :
$\dfrac{V_{SAMI}}{V_{SABC}}=\dfrac{SA}{SA}.\dfrac{SM}{SB}.\dfrac{SI}{SC}=\dfrac 16$
$\to V_{SAMI}=\dfrac 16 V_{S_{SABC}}$
$\to V_{SAMI}=\dfrac 1{12} V_{ABCD}$
$\to V_{SAMN}=2V_{SAMI}=2.\dfrac 1{12} V_{SABCD}$
$\to V_{SAMN}=\dfrac 1{6} V_{SABCD}$
$\to V_{ABCDMNI}=\dfrac{5}{6}V_{SABCD}=\dfrac 56.\dfrac 13SA.S_{ABCD}=\dfrac{5\sqrt{3}}{18}a^3$
$\to A$
