Đáp án:

$1)\quad D.\ \dfrac{16a^3}{45}$

$2)\quad D.\ \dfrac{1}{12}$

Giải thích các bước giải:

Câu 1:

Áp dụng hệ thức lượng trong $\triangle SAB$ vuông tại $A$ đường cao $AH$ ta được:

$\quad SA^2 = SH.SB$

$\Rightarrow \dfrac{SA^2}{SB^2}=\dfrac{SH}{SB}$

$\Rightarrow \dfrac{SH}{SB}=\dfrac{4a^2}{4a^2+ a^2}=\dfrac45$

Tương tự ta được:

$\dfrac{SK}{SC}= \dfrac{SA^2}{SC^2}=\dfrac{4a^2}{4a^2 + 2a^2}=\dfrac23$

$\dfrac{SE}{SD}=\dfrac{SA^2}{SD^2}=\dfrac{4a^2}{4a^2 + a^2}=\dfrac45$

Ta có:

$+)\quad \dfrac{V_{S.AHK}}{V_{S.ABC}}=\dfrac{SH}{SK}\cdot \dfrac{SK}{SC}=\dfrac45\cdot \dfrac23 = \dfrac{8}{15}$

$\Rightarrow V_{S.AHK}=\dfrac{8}{15}V_{S.ABC}=\dfrac{4}{15}V_{S.ABCD}$

$+)\quad \dfrac{V_{S.AKE}}{V_{S.ACD}}=\dfrac{SE}{SD}\cdot \dfrac{SK}{SC}=\dfrac45\cdot \dfrac23 = \dfrac{8}{15}$

$\Rightarrow V_{S.AKE}=\dfrac{8}{15}V_{S.ACD}=\dfrac{4}{15}V_{S.ABCD}$

Do đó:

$V_{SAHKE}= V_{S.AHK} + V_{S.AKE}= \dfrac{8}{15}V_{S.ABCD}$

$\Rightarrow V_{SAHKE}= \dfrac{8}{15}\cdot \dfrac{1}{3}\cdot a^2\cdot 2a = \dfrac{16a^3}{45}$

Câu 2:

Ta có:

$\dfrac{V_{S.MNP}}{V_{S.ABC}}=\dfrac{SM}{SA}\cdot \dfrac{SN}{SB}\cdot \dfrac{SP}{SC}$

$\Leftrightarrow \dfrac{V_{S.MNP}}{V_{S.ABC}}= \dfrac12\cdot\dfrac12\cdot \dfrac13 = \dfrac{1}{12}$

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 $V_{S.AHKE}=\dfrac{8}{15}.\dfrac{2a^3}{3}=\dfrac{16a^3}{45}$

$\Rightarrow D$