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:
$\quad V_{S.ABCD}= \dfrac13S_{ABCD}.SA$
$\Leftrightarrow V_{S.ABCD}=\dfrac13\cdot a^2\cdot a\sqrt2$
$\Leftrightarrow V_{S.ABCD}=\dfrac{a^3\sqrt2}{3}$
