What is the area of a regular nonagon with a perimeter of 144 inches?

Answer 1

A nonagon is a nine-sided polygon. The area of a regular polygon is:

A = (n/4)(s^2)[cos (180° /n)]/[sin (180° /n)]

So the area of the nonagon with side 16 (144/9) is:

A = (9/4)(16^2)[[cos(180°/9)]/[sin (180°/9)]]= (2.25)(256)[(cos 20°)/(sin 20°)]

A ≈ 1,582.55 ft^2 If you don't know the formula of the area of a polygon, you can find its area by multiplying by 9 the area of one of the 9 congruent isosceles triangles that are formed by connecting the center of the polygon with its vertices. But for this you need to find the altitude and the length of the side (which is the radius of the circumscribed circle) of that triangle such as:

we know the length base which is 16 ft (144/9), the angle base which is 70 (140/2), and the vertex angle which is 40 (360°/9 or 180° - 140°). By using the Law of Sines we can find the length of r. So,

r/sin 70° = 16/sin 40° multiply by sin 70° to both sides;

r = (16 sin 70°)/sin 40° sin 70 = altitude/radius

altitude = (sin 70)(radius) = (Sin 70)[(16 sin 70)/sin 40]

altitude = [16(sin 70)^2]/sin 40

Thus the area of this nonagon is:

A = 9[(1/2)(bh)] where b = 16 and h = [16(sin 70)^2]/sin 40

A = (4.5)(16) [[16(sin 70)^2]/sin 40] A ≈ 1,582.55 ft^2