In a diagram,
Therefore, we need to find the volume of a truncated cone.
Let's define the quantities below,
[tex]\begin{gathered} h_t\rightarrow\text{ height of the truncate cone} \\ h\rightarrow\text{ height of the whole cone} \\ r\rightarrow\text{ radius of the cone} \\ A_b\rightarrow\text{ area of the base of the cone} \\ V\rightarrow\text{ volume} \end{gathered}[/tex]Therefore, according to the question, we can identify the relations below
a) h_t=0.30h
[tex]\begin{gathered} h_t=0.30*h \\ h=2r-2 \\ A_b=380 \end{gathered}[/tex]Thus,
[tex]\begin{gathered} \Rightarrow A_b=380 \\ and \\ A_b=\pi r^2 \\ \Rightarrow\pi r^2=380 \\ \Rightarrow r=\sqrt{\frac{380}{\pi}} \end{gathered}[/tex]Finding h_t,
[tex]\begin{gathered} r=\sqrt{\frac{380}{\pi}} \\ and \\ h_t=0.30h=0.30(2r-2)=0.6r-0.6 \\ \Rightarrow h_t=5.99885...\approx6 \\ \end{gathered}[/tex]Thus, the volume of the first case is
[tex]\begin{gathered} \Rightarrow V=\frac{1}{3}\pi r^2h=\frac{1}{3}*380*6=760 \\ \Rightarrow V=760 \end{gathered}[/tex]Notice that A_b=380, so the result of r obtained above applies in this case too.
Obtaining h_t,
[tex]h_t=0.7h=0.7(2r-2)=1.4r-1.4=13.9973\approx14[/tex]Thus,
[tex]\Rightarrow V=\frac{1}{3}*380*14=1773.3333[/tex]
