What is the area of a regular 16-gon with a radius of 15 feet? Round your answer to the nearest tenth.

Split up the 16-gon into 16 congruent isosceles triangles. Cut each of these isosceles triangles in half to get a right triangle.

The radius of the 16-gon corresponds to the hypotenuse of any of these 32 right triangles. As you probably know, the area of a triangle is half the product of its base and height. For a right triangle, the base and height can be taken to be the two legs. So you need to find their lengths.

Recall that the exterior angles of any convex polygon sum to 360º in measure. For a regular polygon, these angles are all congruent. There are 16 of them in this case, so they each have measure (360/16)º = 22.5º.

Interior angles are supplementary to exterior angles, which means each interior angle of the 16-gon has measure (180 - 22.5)º = 157.5º.

When we split up the 16-gon into isosceles triangles, the "base angles" are equal in measure to half the measure of the interior angles, since each hypotenuse bisects the interior angle. So each base angle has measure 78.75º.

In each right triangle, we then have base b and height h such that

sin(78.75º) = h/15

cos(78.75º) = b/15

Solve for h and b to get

b = 15 cos(78.75º)

h = 15 sin(78.75º)

Then the area of each right triangle is

1/2 b h = 225/2 sin(78.75º) cos(78.75º) = 225/4 sin(157.5º)

and so the area of the 16-gon is

32(1/2 b h) = 1800 sin(157.5º) ≈ 688.8 sq. ft.

lol

step-by-step explanation:

the answer

b. they are parallel and will never intersect.