Question

the radius of the cone is 1.75 inches, and it’s height is 3.5 inches. If the diameter of the bubble gum ball is 0.5 inches, what is the closest approximation of the volume of the cone that can be filled with flavored ice?

Answer 1

Step-by-step explanation:

The volume of a cone is:

`V_(c)=(1)/(3)\pi r^2 h \\ \\ \\ Where: \\ \\ r:\text{Radius of the cone} \\ \\ h:\text{height of the cone}`

Here:

`r=1.75in \\ \\ h=3.5in`

So, substituting into the formula:

`V_(c)=(1)/(3)\pi (1.75)^2(3.5)\approx 11.22in`

The gum ball is spherical. so the volume of a sphere is:

`V_(e)=(4)/(3)\pi r^3 \\ \\ \\ Where: \\ \\ r:\text{Radius of the gum ball} \\ \\ \\ \text{We know that the diameter of this ball is} \ d=0.5in \ \text{so the radius is:} \\ \\ r=0.5/2=0.25in`

Therefore:

`V_(e)=(4)/(3)\pi (0.25)^3\approx 0.065`

Finally, the volume (V) of the cone that can be filled with flavored ice is:

`V=V_(c)-V_(e)=11.22-0.065 \\ \\ \boxed{V\approx 11.15in^3}`