what is the area of a regular pentagon that has a radius of 10in?

Draw 5 lines. Each line joins the center and one vertex of the octagon.
The 5 lines divide the regular octagon into 5 equilateral triangles.
In each equilateral triangle, the lengths of the two equal sides are both 10 in.

The inclined angle of the two equal sides
= 360°/5
= 72°

Area of each equilateral triangle
= (1/2) × (10 in) × (10 in) × sin72° in²

The area of the regular pentagon
= 5 ×(1/2) × (10 in) × (10 in) × sin72° in²
= 238 in² (to 3 sig. fig.)

900in^2. Split it into eight triangles, find area of one, and multiply by 8.

The central angle subtended by each side of the pentagon is 72°. The radii dissect the pentagon into five isosceles triangles, each having odd angle 72° and equal sides 10 inches.

area of each triangle = 10²sin(72°)/2 = 50sin(72°)

pentagon area = 5[50sin(72°)] = 250sin(72°)

sin(72°) = √[10 - 2√(5)]/4

pentagon area = 250√[10 + 2√(5)]/4 = (125/2)√[10 + 2√(5)] in²

That is approximately 238 in².

This link can help explain sin(72°):
http://whistleralley.com/polyhedra/pentagon.htm

ooo