Đáp án:

$B. \dfrac{a^3\sqrt3}{24}$

$B. \dfrac{8a^3\sqrt3}{27}$

Giải thích các bước giải:

Câu 27:

Ta có:

$SA\perp (ABC) \,(gt)$

$\Rightarrow SA \perp BC$

mà $BC\perp AB$

$\Rightarrow BC\perp (SAB)$

$\Rightarrow BC\perp SB$

Ta lại có: $(SBC) \cap (ABC) = BC$

$SB\perp BC \, (cmt)$

$AB\perp BC$

$\Rightarrow\widehat{((SBC);(ABCD))} = \widehat{SBA} = 60^o$

$\Rightarrow SA = AB.\tan60^o = a\sqrt3$

$\Rightarrow V_{SABC} = \dfrac{1}{3}S_{ABC}.SA = \dfrac{1}{3}\cdot\dfrac{1}{2}AB.BC.SA = \dfrac{1}{6}a^2.a\sqrt3 = \dfrac{a^3\sqrt3}{6}$

Mặt khác:

$M,N$ là trung điểm $SC,AC$ $(gt)$

$\Rightarrow MN= \dfrac{1}{2}SA$ (tính chất đường trung bình)

$\Rightarrow S_{CMN} = \dfrac{1}{4}S_{SAC}$

$\Rightarrow V_{B.CMN} = \dfrac{1}{4}V_{B.SAC}$

$\Leftrightarrow V_{MNBC} = \dfrac{1}{4}.\dfrac{a^3\sqrt3}{6} = \dfrac{a^3\sqrt3}{24}$

Câu 28:

Ta có:

$SA\perp (ABCD) \, (gt)$

$\Rightarrow SA\perp CB$

mà $CB\perp AB$

$\Rightarrow CB\perp (SAB)$

$\Rightarrow BC\perp SB$

Ta lại có:

$(SBC) \cap (ABCD) = BC$

$SB\perp BC \,(gt)$

$AB\perp BC$

$\Rightarrow \widehat{((SBC);(ABCD))} = \widehat{SBA}= 60^o$

$\Rightarrow AB = \dfrac{SA}{\tan60^o} = \dfrac{2a}{\sqrt3}$

Mặt khác:

$SA\perp AB$

$AB\perp AD$

$\Rightarrow AB\perp (SAD)$

$\Rightarrow AB = d(B;(SAD))$

$\Rightarrow V_{S.ABD} = \dfrac{1}{3}S_{SAD}.d(B;(SAD)) = \dfrac{1}{3}\cdot\dfrac{1}{2}SA.AD.AB = \dfrac{1}{6}.2a.2a.\dfrac{2a}{\sqrt3} = \dfrac{4a^3}{3\sqrt3}$

Bên cạnh đó:

$BG= \dfrac{2}{3}BM$ ($M$ là trung điểm $SC$)

$\Rightarrow d(G;(SAD)) = \dfrac{2}{3}d(B;(SAD))$

$\Rightarrow V_{G.SAD} = \dfrac{2}{3}V_{B.SAD} = \dfrac{2}{3}\cdot \dfrac{4a^3}{3\sqrt3} = \dfrac{8a^3}{9\sqrt3} = \dfrac{8a^3\sqrt3}{27}$