Answer:
The volume of the reduced globe is [tex]V=523.3\ in^{3}[/tex]
Step-by-step explanation:
step 1
we know that
The volume of a sphere is equal to
[tex]V=\frac{4}{3}\pi r^{3}[/tex]
we have
[tex]r=20/2=10\ in[/tex] -----> the radius is half the diameter
substitute
The volume of the original globe is
[tex]V=\frac{4}{3}(3.14)(10)^{3}[/tex]
[tex]V=4,186.7\ in^{3}[/tex]
step 2
If the dimensions of the globe were reduced by half,
then
The new radio is equal to
[tex]r=10/2=5\ in[/tex]
The volume of the reduced globe is
[tex]V=\frac{4}{3}(3.14)(5)^{3}[/tex]
[tex]V=523.3\ in^{3}[/tex]
therefore
The volume of the reduced globe is [tex]V=523.3\ in^{3}[/tex]
Compare the volumes
[tex]4,186.7/523.3=8[/tex]
Remember that
If two solids are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube
The scale factor is 1/2
so
[tex]\frac{1}{2}^{3}=\frac{1}{8}[/tex]
therefore
The original globe is eight times the volume of the reduced globe
