Lời giải:
Vì $SA\perp (ABCD)$ nên 

$60^0= \angle (SC, (ABCD))=\angle (SC, AC)=\widehat{SCA}$

Ta có:

$AC=\sqrt{a^2+(2a)^2}=\sqrt{5}a$

$\frac{SA}{AC}=\tan \widehat{SCA}=\tan 60^0=\sqrt{3}$

$\Rightarrow SA=\sqrt{15}a$
$V_{S.ABCD}=\frac{1}{3}.SA.S_{ABCD}$

$=\frac{1}{3}.\sqrt{15}a.a.2a=\frac{2\sqrt{15}}{3}a^3$

Ta có: \(S_{ABCD}=\dfrac{\left(BC+AD\right).AB}{2}=\dfrac{3}{2}a^2\)

a, \(h=SA=AB.tan60^o=a\sqrt{3}\)

\(\Rightarrow V=\dfrac{1}{3}.S_{ABCD}.h=\dfrac{1}{3}.\dfrac{3}{2}a^2.a\sqrt{3}=\dfrac{\sqrt{3}}{2}a^3\)

b, \(h=SA=AD.tan45^o=2a\)

\(\Rightarrow V=\dfrac{1}{3}.S_{ABCD}.h=\dfrac{1}{3}.\dfrac{3}{2}a^2.2a=a^3\)

c, Dễ chứng minh được SC vuông góc với CD tại C \(\Rightarrow\widehat{SCA}=30^o\)

\(\Rightarrow h=SA=AC.tan30^o=AD.sin45^o.tan30^o=\dfrac{\sqrt{6}}{3}a\)

\(\Rightarrow V=\dfrac{1}{3}.S_{ABCD}.h=\dfrac{1}{3}.\dfrac{3}{2}a^2.\dfrac{\sqrt{6}}{3}a=\dfrac{\sqrt{6}}{6}a^3\)

a/ Kẻ \(CE//BD\Rightarrow BD//\left(SCE\right)\Rightarrow d\left(SC,BD\right)=d\left(BD,\left(SCE\right)\right)=d\left(B,\left(SCE\right)\right)\)

\(AB\cap\left(SCE\right)=\left\{E\right\}\Rightarrow\dfrac{d\left(B,\left(SCE\right)\right)}{d\left(A,\left(SCE\right)\right)}=\dfrac{EB}{EA}=\dfrac{1}{2}\)

\(\widehat{CAE}=\dfrac{1}{2}\widehat{DAB};\widehat{AEC}=\widehat{BDC}=\dfrac{1}{2}\widehat{ADC};\widehat{DAB}+\widehat{ADC}=180^0\Rightarrow\widehat{CAE}+\widehat{AEC}=90^0\Rightarrow\widehat{ACE}=90^0\)

\(\Rightarrow AC\perp EC\)

\(\left\{{}\begin{matrix}SA\perp CE\\AC\perp CE\end{matrix}\right.\Rightarrow CE\perp\left(SAC\right)\Rightarrow\left(SCE\right)\perp\left(SAC\right)\)

Kẻ \(AH\perp SC\Rightarrow AH\perp\left(SCE\right)\Rightarrow d\left(A,\left(SCE\right)\right)=AH=\dfrac{SA.AC}{\sqrt{SA^2+AC^2}}=..\)

\(\Rightarrow d\left(SC,BD\right)=d\left(B,\left(SCE\right)\right)=\dfrac{AH}{2}=...\)

b/ \(AD//BC\Rightarrow AD//\left(SBC\right)\Rightarrow d\left(SC,AD\right)=d\left(AD,\left(SBC\right)\right)=d\left(A,\left(SBC\right)\right)\)

Kẻ \(AK\perp BC\Rightarrow\left\{{}\begin{matrix}SA\perp BC\\AK\perp BC\end{matrix}\right.\Rightarrow\left(SBC\right)\perp\left(SAK\right)\)

Kẻ \(AM\perp SK\Rightarrow AM\perp\left(SBC\right)\Rightarrow d\left(A,\left(SBC\right)\right)=AM=\dfrac{SA.AK}{\sqrt{SA^2+AK^2}}=...=d\left(SC,AD\right)\)

a: Xét ΔBAC có BA=BC và góc ABC=60 độ

nên ΔABC đều

=>\(S_{ABC}=\dfrac{a^2\sqrt{3}}{4}\)

=>\(S_{ABCD}=\dfrac{a^2\sqrt{3}}{2}\)

Chọn đáp án A

+ Ta có

nên K là trọng tâm của tam giác BCD

+ Ta dễ dàng chứng minh được SH  ⊥ (BKH) ⇒ SB, (BKH) = SBH

Theo đề có:

\(\left\{{}\begin{matrix}CD\perp AD\\CD\perp SA\end{matrix}\right.\)

=> \(CD\perp\left(SAD\right)\)

<=> \(d\left(C,\left(SAD\right)\right)=CD=a\)

`HaNa♬`