Đáp án:

$\text{Câu 1: }7a^3$

$\text{Câu 2: } 20a^3\sqrt{3}$

$\text{Câu 3: } \dfrac{4a\sqrt{6}}{9}$

$\text{Câu 4: } \dfrac{3a\sqrt{3}}{8}$

Giải thích các bước giải:

Câu 1:

$V_{S.ABCD} = \dfrac{1}{3}.S_{ABCD}.SA $

$= \dfrac{1}{3}.\dfrac{1}{2}.(AD+BC).AB.SA = \dfrac{1}{6}.(4a+3a).2a.3a = 7a^3$ (đvtt)

Câu 2:

Ta có: $SA\perp (ABCD)$

$\Rightarrow \widehat{(SC;(ABCD))} = \widehat{SCA} = 60^o$

$\Rightarrow SA = AC\sqrt{3} = 5a\sqrt{3}$

Áp dụng định lý Pytago, ta được:

$AC^2 = AB^2 +BC^2$

$\Rightarrow BC^2 = AC^2 - AB^2 = 25a^2 - 9a^2 = 16a^2$

$\Rightarrow BC = 4a$

$\Rightarrow V_{S.ABCD} = \dfrac{1}{3}.S_{ABCD}.SA=\dfrac{1}{3}.AB.BC.SA=\dfrac{1}{3}.3a.4a.5a\sqrt{3} = 20a^3\sqrt{3}$

Câu 3:

Ta có: $ΔSBC$ đều, $SB = SC = BC = a$

$\Rightarrow S_{SBC} = \dfrac{a^2\sqrt{3}}{4}$

Ta lại có: $V_{S.ABC} = V_{A.SBC} = \dfrac{1}{3}.S_{SBC}.d(A;(SBC))$

$\Leftrightarrow \dfrac{a^3\sqrt{2}}{36} = \dfrac{1}{3}.\dfrac{a^2\sqrt{3}}{4}.d(A;(SBC))$

$\Leftrightarrow d(A;(SBC)) = \dfrac{3.\dfrac{a^3\sqrt{2}}{36}}{\dfrac{a^2\sqrt{3}}{4}} = \dfrac{4a\sqrt{6}}{9}$

Câu 4:

Ta có: $(ACD)\subset (ABCD)$

$\Rightarrow d(S;(ACD)) = d(S;(ABCD))$

Ta lại có: $V_{S.ABCD} = \dfrac{1}{3}.S_{ABCD}.d(S;(ABCD))$

$\Leftrightarrow \dfrac{a^3\sqrt{3}}{8} = \dfrac{1}{3}a^2.d(S;(ABCD))$

$\Leftrightarrow d(S;(ABCD)) = \dfrac{3a\sqrt{3}}{8}$