Find the area and perimeter of a nonagon, inscribed in a circle, with radius 4 inches
1 answer:
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 <u>perimeter</u> 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 <u>area</u> of the nonagon is
9a = 9x²/8 tan(70°) ≈ 46.281 in²
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