# Geometry

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

• Geometry - ,

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.

• Geometry - ,

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.

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