Đáp án:
$a) \, V_{S.ABC} = \dfrac{a^3\sqrt{15}}{12}$
$b) \, V_{S.AMN} = \dfrac{a^3\sqrt{15}}{72}$
Giải thích các bước giải:
a) Ta có:
$V_{S.ABC} = \dfrac{1}{3}S_{ABC}.SA = \dfrac{1}{3}.\dfrac{a^2\sqrt3}{4}.a\sqrt5 = \dfrac{a^3\sqrt{15}}{12}$
b) Ta có:
$\dfrac{SM}{SB} = \dfrac{1}{2}$
$\dfrac{SN}{SC} = \dfrac{1}{3}$
$\Rightarrow \dfrac{V_{S.AMN}}{V_{S.ABC}} = \dfrac{SA}{SA}\cdot\dfrac{SM}{SB}\cdot\dfrac{SN}{SC} = 1.\dfrac{1}{2}.\dfrac{1}{3} = \dfrac{1}{6}$
$\Rightarrow V_{S.AMN} = \dfrac{1}{6}V_{S.ABC} = \dfrac{1}{6}.\dfrac{a^3\sqrt{15}}{12} = \dfrac{a^3\sqrt{15}}{72}$