Đáp án:

$1)\quad V_{S.ABC}=\dfrac{3a^3\sqrt3}{8}$

$2)\quad V_{S.ABC}= \dfrac{3a^3\sqrt3}{4}$

$3)\quad V_{S.ABCD}=\dfrac{a^3\sqrt2}{3}$

Giải thích các bước giải:

Câu 1:

Ta có:

$\quad S_{ABC}=\dfrac12AB.BC.\sin\widehat{BAC}$

$\Leftrightarrow S_{ABC}=\dfrac12\cdot a\cdot 3a\cdot \sin120^\circ$

$\Leftrightarrow S_{ABC}= \dfrac{3a^2\sqrt3}{4}$

Từ $A$ kẻ $AH\perp BC$

$\Rightarrow S_{ABC}=\dfrac12AH.BC$

$\Rightarrow AH = \dfrac{2S_{ABC}}{BC}$

$\Rightarrow AH =\dfrac{2\cdot \dfrac{3a^2\sqrt3}{4}}{3a}$

$\Rightarrow AH = \dfrac{a\sqrt3}{2}$

Ta có:

$\begin{cases}AH\perp BC\quad \text{(cách dựng)}\\SA\perp BC\quad (SA\perp (ABC))\end{cases}$

$\Rightarrow BC\perp (SAH)$

$\Rightarrow BC\perp SH$

Khi đó:

$\begin{cases}(SBC)\cap (ABC)= BC\\AH\perp BC\quad \text{(cách dựng)}\\AH\subset (ABC)\\SH\perp BC\quad (cmt)\\SH\subset (SBC)\end{cases}$

$\Rightarrow \widehat{((SBC);(ABC))}=\widehat{SHA}= 60^\circ$

$\Rightarrow SA = AH.\tan\widehat{SHA}= \dfrac{a\sqrt3}{2}\cdot \tan60^\circ =\dfrac{3a}{2}$

Ta được:

$\quad V_{S.ABC}=\dfrac13S_{ABC}.SA$

$\Leftrightarrow V_{S.ABC}=\dfrac13\cdot \dfrac{3a^2\sqrt3}{4}\cdot \dfrac{3a}{2}$

$\Leftrightarrow V_{S.ABC}=\dfrac{3a^3\sqrt3}{8}$

Câu 2:

Gọi $O$ là tâm của đáy

$\Rightarrow SO\perp (ABC)$ (hình chóp đều)

$\Rightarrow \widehat{(SA;(ABC))}=\widehat{SAO}= 60^\circ$

$\Rightarrow \begin{cases}SO = SA.\sin\widehat{SAO}= 2a.\sin60^\circ = a\sqrt3\\OA = SA.\cos\widehat{SAO}= 2a.\cos60^\circ = a\end{cases}$

Ta lại có:

$\quad OA = \dfrac{AB\sqrt3}{3}$

$\Rightarrow AB = OA\sqrt3$

$\Rightarrow AB= a\sqrt3$

$\Rightarrow S_{ABC}= \dfrac{AB^2\sqrt3}{4}= \dfrac{3a^2\sqrt3}{4}$

Ta được:

$\quad V_{S.ABC}=\dfrac13S_{ABC}.SO$

$\Leftrightarrow V_{S.ABC}= \dfrac13\cdot \dfrac{3a^2\sqrt3}{4}\cdot a\sqrt3$

$\Leftrightarrow V_{S.ABC}= \dfrac{3a^3\sqrt3}{4}$

Câu 3:

Ta có:

$\begin{cases}(SAB)\perp (ABCD)\\(SAC)\perp (ABCD)\\(SAB)\cap (SAC)= SA\end{cases}$

$\Rightarrow SA\perp (ABCD)$

$\Rightarrow SA\perp BC$

Lại có: $BC\perp AB$

$\Rightarrow BC\perp (SAB)$

$\Rightarrow \widehat{(SC;(SAB))}=\widehat{BSC}= 30^\circ$

$\Rightarrow SB = \dfrac{BC}{\tan\widehat{BSC}}=\dfrac{a}{\tan30^\circ}= a\sqrt3$

Áp dụng định lý $Pythagoras$ ta được:

$\quad SB^2 = SA^2 + AB^2$

$\Rightarrow SA =\sqrt{SB^2 - AB^2}=\sqrt{3a^2 - a^2}$

$\Rightarrow SA = a\sqrt2$

Khi đó:

$\quad V_{S.ABCD}= \dfrac13S_{ABCD}.SA$

$\Leftrightarrow V_{S.ABCD}=\dfrac13\cdot a^2\cdot a\sqrt2$

$\Leftrightarrow V_{S.ABCD}=\dfrac{a^3\sqrt2}{3}$