A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is #3 #, its base has sides of length #2 #, and its base has a corner with an angle of #(3 pi)/8 #. What is the pyramid's surface area?

1 Answer
Feb 11, 2017

A = surface area #= 46.4 # #units^2#

Explanation:

The surface area of a pyramid is

#A = "base area" xx 1/2 xx "base perimeter" xx "slant height"#

The base is the rhombus:

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Area of a rhombus #= 1/2 pq#

where #p = BD and q = AC#

The internal angles of a quadrilateral #= 360^@ = 2pi = 32/16pi#

The diagonals of a rhombus are perpendicular.

The diagonals bisect the angles. This means if one corner angle is #3/8pi# then the half angle is #1/2 3/8pi = 3/16pi#

Use trigonometry to find the diagonal lengths:

#BE = DE = 2 sin 3/16 pi = 1.11114#; #BD = 2.22228#

#AE = CE = 2cos 3/16 pi = 1.66294#; #AC = 3.32588#

The base area is #1/2 (2.22228)(3.32588) = 3.6955# #units^2#

Base perimeter = #4(2) = 8#

To find the slant height use Pythagorean Theorem.

http://www.mathcaptain.com/geometry/lateral-area.html

slant height = #sqrt (a^2 + height^2)#
where #a# = the length of the altitude, the perpendicular line from a side of the rhombus to the center of the rhombus.

#a = BE sin 5/16pi = .9239#.
Slant height = sqrt (.9239^2 + 3^2) = 3.139.

The Pyramid's surface area is

#A = 3.6955 (1/2) (8)(3.139) = 46.4# #units^2#