A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #5 # and #3 # and the pyramid's height is #9 #. 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 · 2017-12-20T12:44:54

T S A = 88.132

#CH = 3 * sin ((5pi)/12) = 2.8978#
Area of parallelogram base #= a * b1 = 5*2.8978 = color(red)(14.489)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(9^2+ (5/2)^2)= 9.3408#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3* 9.3408= #color(red)(14.0112)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(9^2+(3/2)^2 )= 9.1241#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*5*9.1241 = color(red)( 22.8103)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 14.0112)+ (2* 22.8103) = color(red)(73.643)#

Total surface area =Area of parallelogram base + Lateral surface area # = 14.489 + 73.643 = 88.132#

Total Surface Area # T S A = **88.132**#enter image source here