Truncated cone 6

Calculate the volume of the truncated cone whose bases consist of an inscribed circle and a circle circumscribed to the opposite sides of the cube with the edge length a=1.

Correct answer:

V =  0.4469

Step-by-step explanation:

$$\begin{gathered} a=1 \\ h=a=1 \\ {r}_1=a/2=1/2=\frac{1}{2}=0.5 \\ {r}_2=\sqrt{2}/2\cdot a=\sqrt{2}/2\cdot 1\doteq 0.7071 \\ V=\frac{\pi \cdot h}{3}\cdot (r_1^2+r_1\cdot r_2\cdot r_2^2)=\frac{3.1416\cdot 1}{3}\cdot (0.5^2+0.5\cdot 0.7071\cdot 0.7071^2)\doteq 0.4469 \\ \text{ Verifying Solution: } \\ V=\frac{\pi \cdot a}{3}\cdot (a^2/4+a^2/4\cdot \sqrt{2}+a^2/2) \\ V=\frac{\pi \cdot a^3}{3}\cdot (1/4+\sqrt{2}/4+1/2) \\ V=(\frac{\sqrt{2}}{12}+\frac{1}{4})\cdot \pi \cdot a^3 \end{gathered}$$

Try another example