Answer:
Part 1) The surface area of the second solid is [tex]82.5\ m^{2}[/tex]
Part 2) The volume of the second solid is [tex]78\ m^{3}[/tex]
Step-by-step explanation:
In this problem we have
[tex]scale\ factor=\frac{1}{2}[/tex]
Part 1)
we know that
The ratio of the surface areas of two similar solids is equal to the scale factor squared
Let
x------> the surface area of the second solid (reduced solid)
y------> the surface area of the first solid (original solid)
z-----> the scale factor
[tex]z^{2}=\frac{x}{y}[/tex]
we have
[tex]z=\frac{1}{2}[/tex]
[tex]y=330\ m^{2}[/tex]
substitute and solve for x
[tex](\frac{1}{2})^{2}=\frac{x}{330}[/tex]
[tex]x=\frac{1}{4}*330=82.5\ m^{2}[/tex]
Part 2)
we know that
The ratio of the volumes of two similar solids is equal to the scale factor elevated to the cube
Let
x------> the volume of the second solid (reduced solid)
y------> the volume of the first solid (original solid)
z-----> the scale factor
[tex]z^{3}=\frac{x}{y}[/tex]
we have
[tex]z=\frac{1}{2}[/tex]
[tex]y=624\ m^{3}[/tex]
substitute and solve for x
[tex](\frac{1}{2})^{3}=\frac{x}{624}[/tex]
[tex]x=\frac{1}{8}*624=78\ m^{3}[/tex]
