Q:

A spherical balloon was inflated such that it had a diameter of 24 centimeters. Additional helium was then added to the balloon until it had a diameter of 36 centimeters. Approximately what volume of additional helium was added to the balloon? 904 cm³ 7235 cm³ 17,194 cm³ 24,429 cm³

Accepted Solution

A:
To solve this we are going to use the formula for the volume of a sphere: [tex]V= \frac{4}{3} \pi r^3[/tex]
where 
[tex]r[/tex] is the radius of the sphere 

Remember that the radius of a sphere is half its diameter; since the first radius of our sphere is 24 cm, [tex]r= \frac{24}{2} =12[/tex]. Lets replace that in our formula: 
[tex]V= \frac{4}{3} \pi r^3[/tex]
[tex]V= \frac{4}{3} \pi (12)^3[/tex]
[tex]V=7238.23 cm^3[/tex]

Now, the second diameter of our sphere is 36, so its radius will be: [tex]r= \frac{36}{2} =18[/tex]. Lets replace that value in our formula one more time:
[tex]V= \frac{4}{3} \pi r^3[/tex]
[tex]V= \frac{4}{3} \pi (18)^3[/tex]
[tex]V=24429.02[/tex]

To find the volume of the additional helium, we are going to subtract the volumes:
Volume of helium=[tex]24429.02cm^3-7238.23cm^3=17190.79cm^3[/tex]

We can conclude that the volume of additional helium in the balloon is approximately 17,194 cm³.