A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #4 # and #9 # and the pyramid's height is #4 #. If one of the base's corners has an angle of #(3pi)/8#, what is the pyramid's surface area?

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Answer 1 · 2017-12-07T05:08:24

T S A = 97.5919

#CH = 4 * sin ((3pi)/8) = 3.6955#
Area of parallelogram base #= 9* b1 = 9*3.6955 = color(red)(33.2595 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(4^2+ (9/2)^2)= 6.0208#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4* 6.0208= #color(red)(12.0416)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(4/2)^2 )= 4.4721#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*9*4.4721 = color(red)( 20.1246)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 12.0416)+ (2* 20.1246) = color(red)(64.3324)#

Total surface area =Area of parallelogram base + Lateral surface area # = 33.2595 + 64.3324 = 97.5919#

Total Surface Area # T S A = **97.5919**#enter image source here