Geometry
posted by Bob .
Find the surface area of a right octagonal pyramid with height 2.5 yards, and its base has apothem length 1.5 yards.

The surface area is equal to the area of the base plus the area of the 8
isosceles triangles on the sides.The base consist of 8 isosceles triangles
also.
A = Ab + A(sides).
Ac = 360/N = 360/8 = 45Deg = central angle.
Base angles = (180  45) / 2 = 67.5deg
each.
tan67.5 = h/(b/2) = 1.5/(b/2),
b/2 = 1.5 / tan67.5 = 0.62,
b = 2 * 0.62 = 1.24yds = base of each
triangle.
Ab = (b*h/2)N = (1.24*1.5/2)8 = 7.44sq
yds.
Area of Sides:
Altitude = 2.5*sin67.5 = 2.31YDS.
Area of sides = (b*h/2)N =
(1.24*2.31/2)8 = 11.5 sq. yds.
Surface Area = 7.44 + 11.5 = 18.9 sq yds. 
CORRECTION!
Area of the sides:
The height(altitude) is GIVEN: 2.5 yds.
S = h/sin67.5 = 2.5 / sin67.5  2.71
yds = length of each side.
As = (b*h/2)N = (1.24*2.5/2)8 = 12.4sq yds.= Area of sides.
Surface Area = Ab + As = 7.44 + 12.4
= 19.84sq yds.