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 #9 #, its base's sides have lengths of #4 #, and its base has a corner with an angle of #(5 pi)/6 #. What is the pyramid's surface area?

1 Answer
Oct 18, 2017

Total Surface Area #= **85.7944**#

Explanation:

AB = BC = CD = DA = a = 4
Height OE = h = 9
OF = a/2 = 2
# EF = sqrt(EO^2+ OF^2) = sqrt (h^2 + (a/2)^2) = sqrt(9^2+2^2)=sqrt85#

Area of #DCE = (1/2)*a*EF = (1/2)*4 *sqrt95 = 19.4936#
Lateral surface area #= 4*Delta DCE = 4*19.4936 = 77.9744#

#/_C = (5pi)/6, /_C/2 = (5pi)/12#
diagonal #AC = d_1# & diagonal #BD = d_2#
#OB = d_2/2 = BC*sin (C/2)=4*sin((5pi)/12) = 3.8637#
#OC = d_1/2 = BC cos (C/2) = 4* cos ((5pi)/12) = 1.0353#

Area of base ABCD #= (1/2)*d_1*d_2 = (1/2)(2*3.8637)(2*1.0353) = 8#

Total Surface Area #= Lateral surface area + Base area. T S A # 77.7944 + 8 = 85.7944#
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