Given: Base area of cone i.e., [tex]A = \pi r^2=27.4 \ {in}^2[/tex]
and Volume of cone i.e., [tex]V = \frac{1}{3} \pi r^2 h = 54.8 \ in^3[/tex]
Since, [tex] \pi r^2=27.4 [/tex]
Substituting above in: [tex] \frac{1}{3} \pi r^2 h = 54.8 [/tex], we get:
[tex] \frac{1}{3} (27.4) h = 54.8[/tex]
i.e., [tex]h = 54.8 (\frac{3}{27.4})[/tex]
i.e., [tex]h = 2(3)[/tex]
i.e., [tex]h = 6 \ in[/tex]
