Question

A cylindrical container has a volume of 36 cm³. Another container made out of the same material is similar to the first container with a height that is three-fourths the height of the first container. To the nearest tenth of a cm³, what is the volume of the second container?

Answer 1

trick question ?=27

Answer 2

Formula for volume of a cylinder,
\[V=\pi r^2h\]
Now for cylinder 1,
\[V_{1}=36\]\[r_{1}=r\](let)
\[h_{1}=h\] (let)

Cylinder 2,
\[V_{2}=x\](let)\[r_{2}=r\](same radius for both cylinders)
\[h_{2}=\frac{3}{4}h\](this is given in question)
Now the next step is to make the equations,

Cylinder 1,
\[V_{1}=\pi r_{1}^2h_{1}\]\[36=\pi r^{2}h\]
Cylinder 2,
\[V_{2}=\pi r_{2}^2h_{2}\]\[x=\frac{3}{4}\pi r^{2}h\]
Now you have 2 equations and 2 variables these can be solved using elimination or substitution methods

Answer 3

A shorter method,
A cylinder's volume is directly proportional to it's height
Therefore we can write
\[V=kh\]
where k is your constant of proportionality
Cylinder 1,
\[V_{1}=kh_{1}\]
Cylinder 2,
\[V_{2}=kh_{2}\]
Divide equation 2 by equation 1,
\[\frac{V_{2}}{V_{1}}=\frac{k}{k} \times \frac{h_{2}}{h_{1}}\]\[\frac{V_{2}}{V_{1}}=\frac{h_{2}}{h_{1}}\]\[V_{2}=V_{1}.\frac{h_{2}}{h_{1}}\]
use \[h_{2}=\frac{3}{4}h_{1}\]
and\[V_{1}=36\]
In short since the radius is constant and volume is directly proportional to height, if height is 3/4th then volume would also become 3/4th