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Câu hỏi: Cho hình nón có bán kính đáy bằng 2 (cm), góc ở đỉnh bằng ${{60}^{o}}.$ Thể tích khối nón là
A. $V=\dfrac{8\pi \sqrt{3}}{9}\left( \text{c}{{\text{m}}^{3}} \right).$
B. $V=\dfrac{8\pi \sqrt{3}}{2}\left( \text{c}{{\text{m}}^{3}} \right).$
C. $V=8\pi \sqrt{3}\left( \text{c}{{\text{m}}^{3}} \right).$
D. $V=\dfrac{8\pi \sqrt{3}}{3}\left( \text{c}{{\text{m}}^{3}} \right).$
Ta có bán kính đáy $r=2,$ đường cao $h=\dfrac{r}{\tan {{30}^{o}}}\Rightarrow h=2\sqrt{3}.$
3999865217170
00
Vậy thể tích khối nón $V=\dfrac{1}{3}\pi {{r}^{2}}h=\dfrac{1}{3}\pi .4.2\sqrt{3}=\dfrac{8\pi \sqrt{3}}{3}\left( \text{c}{{\text{m}}^{3}} \right).$
A. $V=\dfrac{8\pi \sqrt{3}}{9}\left( \text{c}{{\text{m}}^{3}} \right).$
B. $V=\dfrac{8\pi \sqrt{3}}{2}\left( \text{c}{{\text{m}}^{3}} \right).$
C. $V=8\pi \sqrt{3}\left( \text{c}{{\text{m}}^{3}} \right).$
D. $V=\dfrac{8\pi \sqrt{3}}{3}\left( \text{c}{{\text{m}}^{3}} \right).$
3999865217170
00
Vậy thể tích khối nón $V=\dfrac{1}{3}\pi {{r}^{2}}h=\dfrac{1}{3}\pi .4.2\sqrt{3}=\dfrac{8\pi \sqrt{3}}{3}\left( \text{c}{{\text{m}}^{3}} \right).$
Đáp án D.