Cho hình chóp SABC có đáy là tam giác đều cạnh a, SA vuông góc

* Hướng dẫn giải

Chọn A

$$\begin{gathered} \mathrm{Gọi} \mathrm{M} \mathrm{là} \mathrm{trung} \mathrm{điểm} \mathrm{của} \mathrm{BC} \mathrm{khi} \mathrm{đó} \mathrm{AM}\perp \mathrm{BC} \\ \mathrm{Ta} \mathrm{có} \left\{\begin{array}{l}\mathrm{AM}\perp \mathrm{BC}\\ \mathrm{SA}\perp \mathrm{BC} ( \mathrm{do} \mathrm{SA} \perp (\mathrm{ABC}\left)\right)\end{array}\right. \\ \Rightarrow \mathrm{BC}\perp \left(\mathrm{SAM}\right) \\ \Rightarrow \measuredangle (\left(\mathrm{SBC}),\left(\mathrm{ABC}\right)\right)=\measuredangle \mathrm{SMA}={60}^{\circ } \\ \mathrm{Khi} \mathrm{đó} \mathrm{SA}=\mathrm{AM}.\tan {60}^{\circ }=\frac{\mathrm{a}\sqrt{3}}{2}.\sqrt{3}=\frac{3\mathrm{a}}{2} \\ {\mathrm{S}}_{△\mathrm{ABC}}=\frac{{\mathrm{a}}^2\sqrt{3}}{4} \\ \Rightarrow {\mathrm{V}}_{\mathrm{SABC}}=\frac{1}{3}.\mathrm{SA}.{\mathrm{S}}_{△\mathrm{ABC}}=\frac{1}{3}.\frac{3\mathrm{a}}{2}.\frac{{\mathrm{a}}^2\sqrt{3}}{4}=\frac{{\mathrm{a}}^3\sqrt{3}}{8} \end{gathered}$$