Geometry help please

Question

2. If a heptagon has an area of 105 in2, what is the measure of one side?

Each 105/7=15 sq in

360 / 7 

2sa + 51.4 = 180

sa = 64.3

A=(1/2)bh

tan(26) = 0.48157385

tan(25.71425)=(0.5b)/h

h=0.5b/0.48157385

15=(1/2)(b)[(0.5b)/0.48157385)

30=b(0.5b)/(0.48157385)

30=b^2/0.9631476

30*0.9631476=b^2

28.894428=b^2

b=√28.894428

b=5.375 inches

I do not know what to do after this to solve for the measure of one side.

Answer 1

Here is a basic formula that you can use to find the side length given the area and vice-versa.

A=s^2n/4tanθ where s=side length, n=# of sides, θ=1/2 * central angle of one of the triangles

360/7=51.428571, 51.428571/2=25.7142857

A=s^2*7/4tan25.71428571°

105=7s^2/4*0.481574625

105=7s^2/1.9262985

202.2613425=7s^2

28.8944775=s^2

take the square root of each side

s=5.37535836 inches

it looks like basically what you did put into a formula that you can memorize and always use for these types of problems; works for all regular polygons given area or side length

Answer 2

You already got it. Through your work, you found the length of the base of a triangle formed by two adjacent vertices of the heptagon and the centroid of that heptagon. That triangle's base is a side of the heptagon.