Ta có: $ΔSAB$ đều
$IA = IB$ $(gt)$
$\Rightarrow SI\perp AB; \, SI = \dfrac{AB\sqrt{3}}{2} = \dfrac{a\sqrt{3}}{2}$
$\begin{cases}(SAB)\cap(ABCD)=AB\\SI⊂ (SAB)\\SI\perp AB\end{cases}\Rightarrow SI\perp (ABCD)$
a) Gọi $M$ là trung điểm $CD$
⇒ $IM//BC//AD; \, IM = a$
$V_{S.ICD} = \dfrac{1}{3}.S_{ICD}.SI = \dfrac{1}{3}.\dfrac{1}{2}.IM.CD.SI = \dfrac{1}{6}.a.a.\dfrac{a\sqrt{3}}{2} = \dfrac{a^3\sqrt{3}}{12} \, (đvtt)$
b) Kẻ $DH\perp IC$
Ta có: $SI\perp (ABCD)$
$\Rightarrow SI\perp DH$
mà $DH\perp IC$ (cách dựng)
$\Rightarrow DH\perp (SIC)$
$\Rightarrow DH = d(D;(SIC))$
Ta có: $DH.IC = IM.CD = 2.S_{ICD}$
$\Rightarrow DH = \dfrac{IM.CD}{IC} = \dfrac{IM.CD}{\sqrt{BC^2 + BI^2}} = \dfrac{a^2}{\sqrt{a^2 + (\dfrac{a}{2})^2}} = \dfrac{2a\sqrt{5}}{5}$
c) Kẻ $IK\perp SM$
Ta có:
$ΔSIC = ΔSID \Rightarrow SC = SD$
$\Rightarrow ΔSCD$ cân tại $S$
$\Rightarrow SM\perp CD$
mà $CD\perp SI$ $(SI \perp (ABCD))$
$\Rightarrow CD\perp (SIM)$
$\Rightarrow CD\perp IK$
mà $IK\perp SM$ (cách dựng)
$\Rightarrow IK \perp (SCD)$
$\Rightarrow IK = d(I;(SCD))$
Áp dụng hệ thức lượng, ta được:
$\dfrac{1}{IK^2} = \dfrac{1}{SI^2} + \dfrac{1}{IM^2}$
$\Rightarrow IK^2 = \dfrac{(SI.IM)^2}{SI^2 + IM^2} = \dfrac{(\dfrac{a\sqrt{3}}{2}.a)^2}{(\dfrac{a\sqrt{3}}{2})^2 + a^2} = \dfrac{3a^2}{7}$
$\Rightarrow IK = \sqrt{\dfrac{3a^2}{7}} = \dfrac{a\sqrt{21}}{7}$
Mặt khác: $AB//CD$
$\Rightarrow AB//(SCD)$
$\Rightarrow d(A;(SCD)) = d(B;(SCD)) = d(I;(SCD))$
Vậy $d(A;(SCD)) = \dfrac{a\sqrt{21}}{7}$
