3110 B-CR
Volume of Cylinders, Cones, and Spheres
Pre-Test

1. The base diameter and the height of a cone are both equal to [tex]$x$[/tex] units.
Which expression represents the volume of the cone, in cubic units?

A. [tex]\pi x^2[/tex]
B. [tex]2 \pi x^3[/tex]
C. [tex]\frac{1}{3} \pi x^2[/tex]
D. [tex]\frac{1}{12} \pi x^3[/tex]



Answer :

Let's begin by understanding the problem: we need to find an expression that represents the volume of a cone given that both the base diameter and the height of the cone are equal to [tex]\( x \)[/tex] units. For a cone, the volume [tex]\( V \)[/tex] is calculated using the formula:

[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]

where:
- [tex]\( r \)[/tex] is the radius of the base of the cone
- [tex]\( h \)[/tex] is the height of the cone

Given that the diameter of the base of the cone is [tex]\( x \)[/tex] units, we can find the radius [tex]\( r \)[/tex] by dividing the diameter by 2:

[tex]\[ r = \frac{x}{2} \][/tex]

The height of the cone [tex]\( h \)[/tex] is also given as [tex]\( x \)[/tex] units.

Now we substitute [tex]\( r \)[/tex] and [tex]\( h \)[/tex] into the volume formula:

[tex]\[ V = \frac{1}{3} \pi \left( \frac{x}{2} \right)^2 x \][/tex]

To simplify this, we need to first square the radius [tex]\( \frac{x}{2} \)[/tex]:

[tex]\[ \left( \frac{x}{2} \right)^2 = \frac{x^2}{4} \][/tex]

Now, let's substitute this into our volume equation:

[tex]\[ V = \frac{1}{3} \pi \left( \frac{x^2}{4} \right) x \][/tex]

Next, we multiply [tex]\( \frac{x^2}{4} \)[/tex] by [tex]\( x \)[/tex]:

[tex]\[ V = \frac{1}{3} \pi \frac{x^2 \cdot x}{4} = \frac{1}{3} \pi \frac{x^3}{4} = \frac{1}{12} \pi x^3 \][/tex]

Thus, the expression that represents the volume of the cone in cubic units is:

[tex]\[ \boxed{\frac{1}{12} \pi x^3} \][/tex]