Answer:
Part 1) Triangle: [tex]P=24\sqrt{3}\ in[/tex]  and  [tex]A=48\sqrt{3}\ in^2[/tex]
Part 2) Square: [tex]P=40\sqrt{2}\ ft[/tex]  and  [tex]A=200\ ft^2[/tex]
Part 3) Hexagon 1: [tex]P=90\ cm[/tex]  and  [tex]A=\frac{675\sqrt{3}}{2})\ cm^2[/tex]
Part 4) Hexagon 2: [tex]P=540\ m[/tex]  and  [tex]A=12,150\sqrt{3}\ m^2[/tex]
Step-by-step explanation:
Part 1) we have an equilateral triangle
we know that
[tex]A=\frac{1}{2}ap[/tex]
where
a is the apothem
P is the perimeter
step 1
Find the length side of the equilateral triangle
Let
b ----> the length side of triangle
we know that
The measure of each interior angle is 60 degrees
so
[tex]tan(30^o)=\frac{4}{b/2}[/tex] ---> by TOA (opposite side divided by the adjacent side)
[tex]tan(30^o)=\frac{1}{\sqrt{3}}[/tex]
equate
[tex]\frac{4}{b/2}=\frac{1}{\sqrt{3}}[/tex]
[tex]\frac{8}{b}=\frac{1}{\sqrt{3}}[/tex]
[tex]b=8\sqrt{3}\ in[/tex]
step 2
Find the perimeter
[tex]P=3b[/tex]
[tex]P=3(8\sqrt{3})=24\sqrt{3}\ in[/tex]
step 3
Find the area
[tex]A=\frac{1}{2}(4)(24\sqrt{3})=48\sqrt{3}\ in^2[/tex]
Part 2) we have a square
we know that
[tex]A=\frac{1}{2}ap[/tex]
where
a is the apothem
P is the perimeter
step 1
Find the length side of the square
Let
b ----> the length side of the square
we know that
The diagonal is half the radius of the square
Applying the Pythagorean Theorem
[tex]D^2=b^2+b^2[/tex]
[tex]D^2=2b^2[/tex]
we have
[tex]D=2a=2(10)=20\ ft[/tex]
[tex]20^2=2b^2[/tex]
[tex]b^2=200\\b=\sqrt{200}\ ft[/tex]
simplify
[tex]b=10\sqrt{2}\ ft[/tex]
step 2
Find the perimeter
[tex]P=4b=4(10\sqrt{2})=40\sqrt{2}\ ft[/tex]
step 3
Find the area
[tex]A=\frac{1}{2}ap[/tex]
The apothem is half the length side of the square
[tex]a=5\sqrt{2}\ ft[/tex]
[tex]A=\frac{1}{2}(5\sqrt{2})(40\sqrt{2})[/tex]
[tex]A=200\ ft^2[/tex]
Part 3) we have a regular hexagon
we know that
[tex]A=\frac{1}{2}ap[/tex]
where
a is the apothem
P is the perimeter
step 1
Find the length side of the hexagon
we know that
The length side of a regular hexagon is equal to the radius
Let
b ----> the length side of the hexagon
we have
[tex]b=30/2=15\ cm[/tex] ----> the radius is half the diameter
step 2
Find the perimeter
[tex]P=6b=6(15)=90\ cm[/tex]
step 3
Find the area
[tex]A=\frac{1}{2}ap[/tex]
The apothem is the height of an equilateral triangle
Applying the Pythagorean Theorem
[tex]b^2=(b/2)^2+a^2[/tex]
[tex]15^2=(15/2)^2+a^2[/tex]
[tex]a^2=225-(225/4)[/tex]
[tex]a^2=675/4\\a=\frac{\sqrt{675}}{2}\ cm[/tex]
simplify
[tex]a=\frac{15\sqrt{3}}{2}\ cm[/tex]
substitute in the formula of area
[tex]A=\frac{1}{2}(\frac{15\sqrt{3}}{2})(90)[/tex]
[tex]A=\frac{675\sqrt{3}}{2})\ cm^2[/tex]
Part 4) we have a regular hexagon
we know that
[tex]A=\frac{1}{2}ap[/tex]
where
a is the apothem
P is the perimeter
step 1
Let
b ---> the length side of the hexagon
we have
[tex]b=90\ m[/tex]
Find the perimeter
[tex]P=6b=6(90)=540\ m[/tex]
step 2
Find the area
[tex]A=\frac{1}{2}ap[/tex]
The apothem is the height of an equilateral triangle
Applying the Pythagorean Theorem
[tex]b^2=(b/2)^2+a^2[/tex]
[tex]90^2=(90/2)^2+a^2[/tex]
[tex]a^2=8,100-2,025[/tex]
[tex]a^2=6,075\\\\a=\sqrt{6,075}\ m[/tex]
simplify
[tex]a=45\sqrt{3}\ m[/tex]
substitute in the formula of area
[tex]A=\frac{1}{2}(45\sqrt{3})(540)[/tex]
[tex]A=12,150\sqrt{3}\ m^2[/tex]
