What is the area of a regular pentagon with a side length of 10 inches and a distance from the center to a vertex of 10 inches? round your answer to the nearest tenth.

a. 242.5 in
b. 251.5 in
c. 228.5 in
d. 216.5

This equals  5 * the area of the triangle formed by the side and 2 lines from joining the center 2 the vertices.

Area of each triangle =  1/2 * base * height  = 1/2 * 10 * (sqrt(10^2-5^2) =  43.30 ins^2

area of pentagon = 5* 43.30 =  216.5 in^2