Answer:
Part 1) The volume of the paperweight is [tex]576\ cm^{3}[/tex]
Part 2) The total surface area of the paperweight is [tex]400\ cm^{2}[/tex]
Step-by-step explanation:
Part 1) what is the volume of the paperweight?
we know that
The volume of the paperweight is equal to the volume of the square pyramid plus the volume of the cube
step 1
Find the volume of the pyramid
The volume of the pyramid is equal to
[tex]V=\frac{1}{3}BH[/tex]
where
B is the area of the square base
H is the height of the pyramid
[tex]B=8^{2}=64\ cm^{2}[/tex]
[tex]H=3\ cm[/tex]
substitute
[tex]V=\frac{1}{3}(64)(3)=64\ cm^{3}[/tex]
step 2
Find the volume of the cube
The volume of the cube is equal to
[tex]V=b^{3}[/tex]
[tex]V=8^{3}=512\ cm^{3}[/tex]
step 3
Find the volume of the paperweight
[tex]64\ cm^{3}+512\ cm^{3}=576\ cm^{3}[/tex]
Part 2) what is the total surface area of the paperweight?
we know that
The total surface area of the paperweight is equal to the surface area of 5 faces of the cube plus the lateral area of the pyramid
step 1
Find the surface area of 5 faces of the cube
[tex]SA=5b^{2}[/tex]
[tex]SA=5(8^{2})=320\ cm^{2}[/tex]
step 2
Find the lateral area of the pyramid
[tex]LA=4[\frac{1}{2}bh][/tex]
[tex]LA=4[\frac{1}{2}(8)(5)]=80\ cm^{2}[/tex]
step 3
Find the total surface area of the paperweight
[tex]320\ cm^{2}+80\ cm^{2}=400\ cm^{2}[/tex]
