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Câu hỏi: Thể tích của khối nón có thiết diện qua trục là tam giác đều cạnh $a$ bằng
A. $\dfrac{\sqrt{3}\pi {{a}^{3}}}{48}$.
B. $\dfrac{\sqrt{3}\pi {{a}^{3}}}{24}$.
C. $\dfrac{\sqrt{3}\pi {{a}^{3}}}{8}$.
D. $\dfrac{\sqrt{3}\pi {{a}^{3}}}{12}$.
A. $\dfrac{\sqrt{3}\pi {{a}^{3}}}{48}$.
B. $\dfrac{\sqrt{3}\pi {{a}^{3}}}{24}$.
C. $\dfrac{\sqrt{3}\pi {{a}^{3}}}{8}$.
D. $\dfrac{\sqrt{3}\pi {{a}^{3}}}{12}$.
Ta có:
$\left\{ \begin{aligned}
& 2r=a \\
& l=a \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& r=\dfrac{a}{2} \\
& h=\sqrt{{{l}^{2}}-{{r}^{2}}}=\sqrt{{{a}^{2}}-\dfrac{{{a}^{2}}}{4}}=\dfrac{\sqrt{3}}{2}a \\
\end{aligned} \right.\Rightarrow V=\dfrac{\pi {{r}^{2}}h}{3}=\dfrac{1}{3}\pi {{\left( \dfrac{a}{2} \right)}^{2}}\dfrac{\sqrt{3}a}{2}=\dfrac{\sqrt{3}\pi {{a}^{2}}}{24}$.
$\left\{ \begin{aligned}
& 2r=a \\
& l=a \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& r=\dfrac{a}{2} \\
& h=\sqrt{{{l}^{2}}-{{r}^{2}}}=\sqrt{{{a}^{2}}-\dfrac{{{a}^{2}}}{4}}=\dfrac{\sqrt{3}}{2}a \\
\end{aligned} \right.\Rightarrow V=\dfrac{\pi {{r}^{2}}h}{3}=\dfrac{1}{3}\pi {{\left( \dfrac{a}{2} \right)}^{2}}\dfrac{\sqrt{3}a}{2}=\dfrac{\sqrt{3}\pi {{a}^{2}}}{24}$.
Đáp án B.