What is the area of a regular heptagon (7 sided polygon) with a radius of 4 feet?

Question

What is the area of a regular heptagon (7 sided polygon) with a radius of 4 feet?

Answer 1

draw the heptagon

draw one of the seven triangles from the center of the circle

draw the height, h

the angle at the apex, at the center of the circle, is bisected by the height

call this angle  θ

now take the cos of the angle and the sin of the angle

cos θ=h/x where h is the height and x is half of the length of one side

sin θ=x/r wher r is the radius

A=(1/2)(b)(h)

A=(1/2)(2x)(h)

A=xh

x=rsin θ and h=rcos θ

A=(rsin θ)(rcos θ)

A=r²sin θcos θ

from the trig identity sin2 θ=2sin θcos θ we get sin θcos θ=(1/2)(sin2θ)

A=r²(1/2)(sin2θ)

2 θ=360/7

2 θ=51.428571

A=(1/2)(r²)(sin51.428571º)

A=(1/2)(4²)(0.7818314778)

A=(1/2)(16)(0.7818314778)

A=8*0.7818314778

A=6.254651816 square feet for the area of the triangle

there are 7 triangles so multiply this answer by 7 to get the area of the heptagon

A=7*6.254651816

A=43.78256271 square feet for the area of the heptagon (round off as needed)