Equilateral cone

A cup has the shape of a right circular cone whose slant height equals the diameter of its base (i.e. the axial cross-section is an equilateral triangle). It is supposed to hold 0.2 litres of liquid when filled to 1 cm below the rim. Calculate its diameter.

Final Answer:

d =  9.5906 cm

Step-by-step explanation:

$$\begin{gathered} s=d=2r \\ V=0.2\text{l}\to {\text{cm}}^3=0.2\cdot 1000{\text{cm}}^3=200{\text{cm}}^3 \\ V=\frac{1}{3}\pi r^{2}h \\ {s}^2=r^2+h^2 \\ (2r{)}^2=r^2+h^2 \\ V=\frac{1}{3}\pi r^2\sqrt{s^2-r^2} \\ \frac{3V}{\pi }=r^2\sqrt{4r^2-r^2} \\ \frac{3V}{\pi }=r^2\sqrt{3r^2} \\ \frac{3V}{\pi }=r^3\sqrt{3} \\ Q=\frac{3\cdot V}{\pi \cdot \sqrt{3}}=\frac{3\cdot 200}{3.1416\cdot \sqrt{3}}\doteq 110.2658 \\ r=\sqrt[3]{Q}=\sqrt[3]{110.2658}\doteq 4.7953\text{cm} \\ d=2\cdot r=2\cdot 4.7953\doteq 9.5906\text{cm} \\ \text{ Verifying Solution: } \\ s=2\cdot r=2\cdot 4.7953\doteq 9.5906\text{cm} \\ h=\sqrt{s^2-r^2}=\sqrt{9.5906^2-4.7953^2}\doteq 8.3057\text{cm} \\ {V}_2=\frac{1}{3}\cdot \pi \cdot r^2\cdot h=\frac{1}{3}\cdot 3.1416\cdot 4.7953^2\cdot 8.3057=200{\text{cm}}^3 \end{gathered}$$

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