Answer:
16 times
Step-by-step explanation:
[tex]\textsf{Volume of a cone}=\sf \dfrac{1}{3} \pi r^2 h \quad\textsf{(where r is the radius and h is the height)}[/tex]
If only the radius is changed, the change in volume will be proportionate to the multiplicative factor squared.
[tex]\sf \implies Volume =\dfrac{1}{3} \pi (ar)^2h=\dfrac{1}{3} \pi (a^2)r^2h[/tex]
Therefore, if the cone is quadrupled (multiplied by 4), the volume of the larger cone will be 4² times greater than the volume of the smaller cone, so 16 times greater than the smaller cone.
Proof
Given:
Substituting the given values into the formula:
[tex]\sf \implies Volume =\dfrac{1}{3} \pi (6)^2(9)=108 \pi \: \:cubic\:units[/tex]
If the radius is quadrupled:
- radius = 6 × 4 = 24
- height = 9
Substituting the new given values into the formula:
[tex]\sf \implies Volume =\dfrac{1}{3} \pi (24)^2(9)=1728 \pi \: \:cubic\:units[/tex]
To find the number of times greater the volume of the large cone is than the volume of the smaller cone, divide their volumes:
[tex]\sf \implies \dfrac{V_{large}}{V_{small}}=\dfrac{1728\pi}{108\pi}=16[/tex]
So the volume of the larger cone is 16 times greater than the volume of the smaller cone.
