Answer:
Part 1) The surface area of the second solid is ![82.5\ m^{2}](https://tex.z-dn.net/?f=82.5%5C%20m%5E%7B2%7D)
Part 2) The volume of the second solid is ![78\ m^{3}](https://tex.z-dn.net/?f=78%5C%20m%5E%7B3%7D)
Step-by-step explanation:
In this problem we have
![scale\ factor=\frac{1}{2}](https://tex.z-dn.net/?f=scale%5C%20factor%3D%5Cfrac%7B1%7D%7B2%7D)
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
we have
![z=\frac{1}{2}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B1%7D%7B2%7D)
![y=330\ m^{2}](https://tex.z-dn.net/?f=y%3D330%5C%20m%5E%7B2%7D)
substitute and solve for x
![x=\frac{1}{4}*330=82.5\ m^{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B1%7D%7B4%7D%2A330%3D82.5%5C%20m%5E%7B2%7D)
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
we have
![z=\frac{1}{2}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B1%7D%7B2%7D)
![y=624\ m^{3}](https://tex.z-dn.net/?f=y%3D624%5C%20m%5E%7B3%7D)
substitute and solve for x
![x=\frac{1}{8}*624=78\ m^{3}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B1%7D%7B8%7D%2A624%3D78%5C%20m%5E%7B3%7D)