Answer:
Part 1) [tex]SA=58\ cm^2[/tex]
Part 2) [tex]SA=90\ ft^2[/tex]
Part 3) [tex]SA=168\ m^2[/tex]
Part 4) [tex]SA=96\ in^2[/tex]
Part 5) [tex]SA=660\ in^2[/tex]
Part 6) [tex]SA=96\ cm^2[/tex]
Part 7) Your friend is not correct, [tex]SA=150\ cm^2[/tex]
Part 8) [tex]SA=486\ in^3[/tex]
Part 9) [tex]SA=94\ ft^2[/tex]
Step-by-step explanation:
we know that
The surface area of a rectangular prism is equal to
[tex]SA=2B+Ph[/tex]
where
B is the area of the base of the prism
P is the perimeter of the base of the prism
h is the height of the prism
Part 1) Find the surface area of the rectangular prism
Find the area of the base B
[tex]B=(4)(1)=4\ cm^2[/tex]
Find the perimeter of the base
[tex]P=2(4+1)=10\ cm[/tex]
we have
[tex]h=5\ cm[/tex]
substitute the values in the formula of surface area
[tex]SA=2(4)+10(5)=58\ cm^2[/tex]
Part 2) Find the surface area of the rectangular prism
Find the area of the base B
[tex]B=(2)(5)=10\ ft^2[/tex]
Find the perimeter of the base
[tex]P=2(2+5)=14\ ft[/tex]
we have
[tex]h=5\ ft[/tex]
substitute the values in the formula of surface area
[tex]SA=2(10)+14(5)=90\ ft^2[/tex]
Part 3) Find the surface area of the rectangular prism
Find the area of the base B
[tex]B=(9)(6)=54\ m^2[/tex]
Find the perimeter of the base
[tex]P=2(9+6)=30\ m[/tex]
we have
[tex]h=2\ m[/tex]
substitute the values in the formula of surface area
[tex]SA=2(54)+30(2)=168\ m^2[/tex]
Part 4) Find the surface area of the triangular prism
Find the area of the triangular base B
[tex]B=(1/2)(8)(3)=12\ in^2[/tex] ---> the area of triangle
Find the perimeter of the triangular base
[tex]P=8+5+5=18\ in[/tex]
we have
[tex]h=4\ in[/tex]
substitute the values in the formula of surface area
[tex]SA=2(12)+18(4)=96\ in^2[/tex]
Part 5) Find the surface area of the triangular prism
Find the area of the triangular base B
[tex]B=(1/2)(12)(5)=30\ in^2[/tex] ---> the area of triangle
Find the perimeter of the triangular base
[tex]P=12+5+13=30\ in[/tex]
we have
[tex]h=20\ in[/tex]
substitute the values in the formula of surface area
[tex]SA=2(30)+30(20)=660\ in^2[/tex]
Part 6) Find the surface area of the triangular prism
Find the area of the triangular base B
[tex]B=(1/2)(6)(8)=24\ cm^2[/tex] ---> the area of triangle
Find the perimeter of the triangular base
[tex]P=8+6+10=24\ cm[/tex]
we have
[tex]h=2\ cm[/tex]
substitute the values in the formula of surface area
[tex]SA=2(24)+24(2)=96\ cm^2[/tex]
Part 7) we know that
The surface area of a cube is equal to the area of its six square faces
[tex]SA=6b^2[/tex]
where
b is the length side of the cube
we have
[tex]b=5\ cm[/tex]
substitute
[tex]SA=6(5^2)[/tex]
[tex]SA=150\ cm^2[/tex]
therefore
Your friend is not correct
The surface area of the cube is [tex]SA=150\ cm^2[/tex]
Part 8) we know that
The surface area of a cube is equal to the area of its six square faces
[tex]SA=6b^2[/tex]
where
b is the length side of the cube
we have
[tex]b=9\ in[/tex]
substitute
[tex]SA=6(9^2)[/tex]
[tex]SA=486\ in^3[/tex]
Part 9) Find the surface area of the rectangular prism (shipping crate)
Find the area of the base B
[tex]B=(5)(4)=20\ ft^2[/tex]
Find the perimeter of the base
[tex]P=2(5+4)=18\ ft[/tex]
we have
[tex]h=3\ ft[/tex]
substitute the values in the formula of surface area
[tex]SA=2(20)+18(3)=94\ ft^2[/tex]
