Question

Find the area and perimeter of a nonagon, inscribed in a circle, with radius 4 inches​

Answer 1

If it's a regular nonagon, each exterior angle has measure 360°/9 = 40°, so the supplementary interior angles each have measure 180° - 40° = 140°.

We can split up the nonagon into 9 congruent isosceles triangles whose longer side coincides with the radius of the circle. (The language in the question is a bit ambiguous; I think it's clearler to say the nonagon is itself circumscribed by the circle, so that each of its vertices lie on the circle. See attached image - not mine, but it's in public domain.)

As the image suggests, each triangle will have interior angles measuring 70°, 70°, and 40°.

We can use trigonometry to find the side length of the nonagon, i.e. the length of the shortest side of each triangle - I'll call it x. For example, using the law of cosines,

x² = 4² + 4² - 2 • 4 • 4 cos(40°)   ⇒   x = 4√2 √(1 - cos(40°))

Then the perimeter of the nonagon is

9x = 36√2 √(1 - cos(40°)) in ≈ 24.623 in

The area of a triangle is 1/2 the product of its base and height. The base will be b = x, but to get the height we cut the isosceles triangle in half to produce a right triangle. We use trig again to determine its height h:

tan(70°) = h/(b/2)   ⇒   h = 1/2 x tan(70°)

so each triangle contributes an area of

a = 1/2 bh = 1/4 x² tan(70°) ≈ 5.142 in²

and so the area of the nonagon is

9a = 9x²/8 tan(70°) ≈ 46.281 in²