A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #8 # and #6 # and the pyramid's height is #4 #. If one of the base's corners has an angle of #pi/3#, what is the pyramid's surface area?

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Answer 1 · 2017-10-25T22:04:06

Total Surface Area # T S A = **81.5103#

#CH = 6 sin (pi/3)= 3sqrt3#
Area of parallelogram base #= 8*b1 = 24sqrt3= color(blue)(41.5692)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(4^2+ (8/2)^2)= 5.6569#
Area of #Delta AED = BEC = (1/2)*b*h_1 = (1/2)*6*5.6569=color(red)(16.9706)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(6/2)^2 )= 5#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*5 = color(red)(20)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 16.9706)+ (2* 5)= color(blue)(43.9411)#

Total surface area =Area of parallelogram base + Lateral surface area # = 41.5692 + 43.9411= color(green)(85.5103)#

Total Surface Area # T S A = **81.5103**#enter image source here