A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #1 # and #5 # and the pyramid's height is #2 #. If one of the base's corners has an angle of #(5pi)/12#, what is the pyramid's surface area?

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Answer 1 · 2017-10-21T22:49:40

Total Surface Area # T S A = **14.4755**#

#CH = 1*sin (5pi/12) = 0.9659#
Area of parallelogram base #= a* b1 = 5 * 0.9659#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(2^2+ (5/2)^2)= 3.2016#
Area of #Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1*3.2016 = 1.6008#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(2^2+(1/2)^2 ) = 2.0616#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*5*2.0616 = 5.154#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 1.6008)+ (2*5.154 ) = 13.5096#

Total surface area =Area of parallelogram base + Lateral surface area # = 0.9659 + 13.5096 = 14.4755 #

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