Câu 1:
$V_{SABCD}=\dfrac{1}{3}.2a.a^2=\dfrac{2a^3}{3}$
Dễ dàng CM $SC\bot (AHE)$ do $AH\bot SC, AE\bot SC$
$\to SC\bot AQ\quad\forall Q\ne A\in (AHE)$
$\to Q\equiv I$
$\to A, H, K, E$ đồng phẳng
$AH\bot (SAB)\to AH\bot HK$
$AE\bot (SCD)\to AE\bot EK$
Có $\Delta SAB=\Delta SAD$ (c.g.c)
$\to AH=AE$
$\to \Delta AHK=\Delta AEK$ (ch-cgv)
$\to S_{AHK}=S_{AKE}$
$\to V_{SAHK}=V_{SKEA}=\dfrac{V_{SHKAE}}{2}$
$S_{ABC}=S_{ACD}\to V_{SABC}=V_{SACD}=\dfrac{V_{SABCD}}{2}$
$\to \dfrac{V_{SHKEA}}{V_{SABCD}}=\dfrac{V_{KSAH}}{V_{CSAB}}=\dfrac{d(K;(SAB))}{d(C;(SAB))}.\dfrac{S_{AHS}}{S_{SBA}}$
$AC=a\sqrt2\to SC=\sqrt{(2a)^2+2a}=a\sqrt6$
$\to SK=\dfrac{SA^2}{SC}=\dfrac{2a\sqrt6}{3}$
$\to \dfrac{d(K;(SAB))}{d(C;(SAB))}=\dfrac{SK}{SC}=\dfrac{2}{3}$
$SB=\sqrt{(2a)^2+a^2}=a\sqrt5$
$\to SH=\dfrac{SA^2}{SB}=\dfrac{4a}{\sqrt5}$
$\to \dfrac{S_{SHA}}{S_{SAB}}=\dfrac{SH}{SB}=\dfrac{4}{5}$
Suy ra $\dfrac{V_{SKHAE}}{V_{SABCD}}=\dfrac{2}{3}.\dfrac{4}{5}=\dfrac{8}{15}$
Vậy $V_{S.AHKE}=\dfrac{8}{15}.\dfrac{2a^3}{3}=\dfrac{16a^3}{45}$
$\Rightarrow D$
