Answer: two times
Step-by-step explanation:
Given
height of cone A and B [tex]h_a=h_b=5\ in.[/tex]
The volume of cone A [tex]V_a=20.9\ in.^3[/tex]
The volume of cone B is 4 times of cone A
the volume of a cone is [tex]\frac{1}{3}\pi r^2h[/tex]
The volume of cone A
[tex]V_a=\dfrac{1}{3}\pi r_a^2h_a=20.9\quad \ldots(i)[/tex]
The volume of cone B
[tex]V_b=\dfrac{1}{3}\pi r_b^2h_b=4\times 20.9\quad \ldots(ii)[/tex]
divide (i) and (ii) we get
[tex]\Rightarrow \dfrac{r_a^2}{r_b^2}=\dfrac{1}{4}\\\Rightarrow \dfrac{r_a}{r_b}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\\\Rightarrow \dfrac{d_a}{d_b}=\dfrac{1}{2}\\\Rightarrow d_b=2d_a[/tex]
thus, diameter of cone B is twice the diameter of A
