Triangular 6610

The curved part of the rotating cylinder is four times larger than the area of its base. Determine the volume of the regular triangular prism inscribed in the cylinder. The radius of the bottom of the cylinder is 10 cm.

Correct answer:

V =  2598.0762 cm³

Step-by-step explanation:

$$\begin{gathered} r=10\text{cm} \\ {S}_1=\pi \cdot r^2=3.1416\cdot 10^2\doteq 314.1593{\text{cm}}^2 \\ {S}_2=4\cdot S_1=4\cdot 314.1593\doteq 1256.6371{\text{cm}}^2 \\ {S}_2=2\pi \cdot r\cdot h \\ h=\frac{S_2}{2\pi \cdot r}=\frac{1256.6371}{2\cdot 3.1416\cdot 10}=20\text{cm} \\ r=\frac{\sqrt{3}}{3}a \\ a=r/\frac{\sqrt{3}}{3}=10/\frac{\sqrt{3}}{3}=10\sqrt{3}\text{cm}\doteq 17.3205\text{cm} \\ {S}_3=\frac{\sqrt{3}}{4}\cdot a^2=\frac{\sqrt{3}}{4}\cdot 17.3205^2=75\sqrt{3}{\text{cm}}^2\doteq 129.9038{\text{cm}}^2 \\ V=S_3\cdot h=129.9038\cdot 20=1500\sqrt{3}=2598.0762{\text{cm}}^3 \end{gathered}$$

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