We are asked to find how the volume of the given cone would change if the height is doubled.
Therefore, the radius of the bigger cone is r=4 cm and the height is h=18 cm. Now, let's find the volume of this cone.
\begin{gathered}
V_\text{big}=\dfrac{1}{3}\pi (4)^2(18)=96\pi
\end{gathered}
Since the ratio between the volume of the big cone and the old cone is 96Ï€/48Ï€=2, the volume of the cone doubles if we double its height.
b
As in Part A, we are given the cone of the radius of r=4 centimeters, and the height of h=9 centimeters. From Part A we know that the volume of the given cone is V_\text{cone}=48\pi cubic centimeters. We are asked to find how the volume of the given cone would change if the radius of the base is doubled.
Therefore, the radius of the bigger cone is r=8 cm, and the height is h=9 cm. Now, let's find the volume of this cone.
\begin{gathered}
V_\text{big}=\dfrac{1}{3}\pi (8)^2(9)=192\pi
\end{gathered}
Since the ratio between the volume of the big cone and the old cone is 192Ï€/48Ï€=4, the volume of the cone quadruples if we double its radius.
c
As in Part A, we are given the cone of the radius of r=4 centimeters, and the height of h=9 centimeters, which has the volume of V_\text{cone}=48\pi cubic centimeters. We are asked to find how the volume of the given cone would change if the height of the cone is doubled, and the radius of the base is doubled.
Therefore, the radius of the bigger cone is r=8 cm, and the height is h=18 cm. Now, let's find the volume of this cone.
\begin{gathered}
V_\text{big}=\dfrac{1}{3}\pi (8)^2(18)=384\pi
\end{gathered}
Since the ratio between the volume of the big cone and the old cone is 384Ï€/48Ï€=8, the volume of the cone octuples if we double its radius and its height.
