find the surface area of a right octagonal pyramid with

Question

height 2.5 yards, and its base has apothem length 1.5 yards

Answer 1

The pyramid consists of an octagonal base and eight isosceles triangles.

Let the centre of the base be A. Any apothem will meet a side of the octagon at the midpoint of that side.

Let that midpoint be B. Let one end of that side be C.

Then ABC is a right-angled triangle. It is one of 16 similar triangles comprising the octagon.

The angle CAB = 2pi/16 = pi/8

octagonal pyramid apothem length is AB = 1.5 yards

So CB = 1.5 tan (pi/8)

Apply half-angle formula tan (a/2) = (1-cos a)/sin a

 tan (pi/8) = [1-cos(pi/4)]/sin (pi/4)

= [1- (1/√2)/(1/√2)

=√2 -1

Substitute  tan (pi/8) = √2 -1

CB = 1.5 (√2 -1)

So base area = 16 x 1.5 x 1.5(√2 - 1) x (1/2)

= 18(√2 - 1)

= 7.456 sq. yds. (approx.)

Now, let the top of the pyramid be D.

octagonal pyramid with height 2.5 yards

BD = √[(2.5)^2+(1.5)^2]

     = √(8.5)

The total area of the sloping faces of the pyramid = 8 x √(8.5) x (1.5) x (√2  - 1)

= 6 √34 (√2  - 1)

= 14.492 sq. yds. (approx.)

If you are counting the base, total area of pyramid = 18(√2 - 1)+ 6 √34 (√2 - 1)

= (18 + 6 √34)(√2 - 1)

= 6(3 +√34 )(√2 - 1)

= 21.948 sq. yds. (approx.).