Area of polygon

Question

Find the area of a regular polygon with 6 sides that has a perimeter of 42 inches and an another of 5 inches

Answer 1

A = ns^2/4tan(180/n)

n= 6 sides

s = length of one side = 42/6 = 7 inches

A= 6(7)^2 divided by 4tan30 = 6(49)/4(1/sqr3) = 3(49)/(2/sqr3) =3(49)(sqr3)/2 = 147sqr3/2 = about 73.5sqr3 in^2 = about 127.3 for a hexagon with perimeter 42 inches

Another method is

A=3s^2sqr3/2 = 3(7)^2(sqr3)/2 =73.5sqr3 = 127.3 = area of 6 equilateral triangles that make up the hexagon

or

divide the hexagon into a rectangle in the middle with 2 triangles on right and left. the interior angles are 120 each, so the triangles have 120-30-30 angles. divide those two triangles into two right triangles, with 4 right triangles 30-60-90 angles and hypotenuses = 7 inches. shorter side = 7/2 = 3 1/2 inches. longer side = 3.5(sqr3). Area of the right triangles are bh/2 = 3.5(3.5)sqr3(1/2) = 6.125sqr3

multiply by 4 = 24.5sqr3. add the area of the rectangle which is 7 by 7sqr3 = 49sqr3 They sum to 73.5sqr3 = about 127.3 in^2

For a perimeter of 5 inches

s= 5/6

n=6

A= 6(5/6)^2/4tan(180/6) = (25/6)/2sqr3 = 25sqr3/12= about 6.5 in^2