Answer:
Part 1) Sphere The surface area is equal to [tex]SA=196\pi\ m^{2}[/tex] and the volume is equal to [tex]V=\frac{1,372}{3}\pi\ m^{3}[/tex]
Part 2) Cone The surface area is equal to [tex]SA=(16+4\sqrt{65})\pi\ units^{2}[/tex] and the volume is equal to [tex]V=\frac{112}{3}\pi\ units^{3}[/tex]
Part 3) Triangular Prism The surface area is equal to [tex]SA=51.57\ mm^{2}[/tex] and the volume is equal to [tex]V=17.388\ mm^{3}[/tex]
Step-by-step explanation:
Part 1) The figure is a sphere
a) Find the surface area
The surface area of the sphere is equal to
[tex]SA=4\pi r^{2}[/tex]
we have
[tex]r=14/2=7\ m[/tex] ----> the radius is half the diameter
substitute
[tex]SA=4\pi (7)^{2}[/tex]
[tex]SA=196\pi\ m^{2}[/tex]
b) Find the volume
The volume of the sphere is equal to
[tex]V=\frac{4}{3}\pi r^{3}[/tex]
we have
[tex]r=14/2=7\ m[/tex] ----> the radius is half the diameter
substitute
[tex]V=\frac{4}{3}\pi (7)^{3}[/tex]
[tex]V=\frac{1,372}{3}\pi\ m^{3}[/tex]
Part 2) The figure is a cone
a) Find the surface area
The surface area of a cone is equal to
[tex]SA=\pi r^{2} +\pi rl[/tex]
we have
[tex]r=4\ units[/tex]
[tex]h=7\ units[/tex]
Applying Pythagoras Theorem find the value of l (slant height)
[tex]l^{2}=r^{2} +h^{2}[/tex]
substitute the values
[tex]l^{2}=4^{2} +7^{2}[/tex]
[tex]l^{2}=65[/tex]
[tex]l=\sqrt{65}\ units[/tex]
so
[tex]SA=\pi (4)^{2} +\pi (4)(\sqrt{65})[/tex]
[tex]SA=16\pi +4\sqrt{65}\pi[/tex]
[tex]SA=(16+4\sqrt{65})\pi\ units^{2}[/tex]
b) Find the volume
The volume of a cone is equal to
[tex]V=\frac{1}{3}\pi r^{2}h[/tex]
we have
[tex]r=4\ units[/tex]
[tex]h=7\ units[/tex]
substitute
[tex]V=\frac{1}{3}\pi (4)^{2}(7)[/tex]
[tex]V=\frac{112}{3}\pi\ units^{3}[/tex]
Part 3) The figure is a triangular prism
a) The surface area of the triangular prism is equal to
[tex]SA=2B+PL[/tex]
where
B is the area of the triangular base
P is the perimeter of the triangular base
L is the length of the prism
Find the area of the base B
[tex]B=\frac{1}{2} (2.7)(2.3)=3.105\ mm^{2}[/tex]
Find the perimeter of the base P
[tex]P=2.7*3=8.1\ mm[/tex]
we have
[tex]L=5.6\ mm[/tex]
substitute the values
[tex]SA=2(3.105)+(8.1)(5.6)=51.57\ mm^{2}[/tex]
b) Find the volume
The volume of the triangular prism is equal to
[tex]V=BL[/tex]
where
B is the area of the triangular base
L is the length of the prism
we have
[tex]B=3.105\ mm^{2}[/tex]
[tex]L=5.6\ mm[/tex]
substitute
[tex]V=(3.105)(5.6)=17.388\ mm^{3}[/tex]
