Đáp án:
$V_{S.A'B'C}=\dfrac{a^3\sqrt3}{12}$
Giải thích các bước giải:
Ta có:
$S_{ABC}=\dfrac12AB.AC.\sin\widehat{AK}$
$\Rightarrow S_{ABC}=\dfrac12.a.2a.\sin60^o = \dfrac{a^2\sqrt3}{2}$
Do đó:
$V_{S.ABC}=\dfrac13S_{ABC}.SA =\dfrac13\cdot\dfrac{a^2\sqrt3}{2}\cdot3a =\dfrac{a^3\sqrt3}{2}$
Ta lại có:
$\dfrac{V_{S.A'B'C}}{V_{S.ABC}}=\dfrac{SA'}{SA}\cdot\dfrac{SB'}{SB}\cdot\dfrac{SC}{SC}=\dfrac12\cdot\dfrac13\cdot1 =\dfrac16$
$\Rightarrow V_{S.A'B'C}=\dfrac16V_{S.ABC}=\dfrac16\cdot\dfrac{a^3\sqrt3}{2}=\dfrac{a^3\sqrt3}{12}$