A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #8 # and #9 # 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 · 2017-12-01T13:55:26

Total surface area of the pyramid = #color(blue)(161.4237)#

#CH = 8 * sin (pi/3) = 8sin (60) = 4sqrt3 = 6.928#
Area of parallelogram base #= 9* b1 = 9*6.928 = color(red)(62.352)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(4^2+ (9/2)^2)= 6.02#
Area of #Delta AED = BEC = (1/2)*b*h_1 = (1/2)*8*6.02= color(red)(24.08)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(8/2)^2 )= 5.6569#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*9*5.6569 = color(red)(25.4558)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 24.08)+ (2* 25.4558) = color(red)(99.0717)#)#

Total surface area =Area of parallelogram base + Lateral surface area # = 62.352 + 99.0717 = 161.4237#

Total Surface Area # T S A = **161.4237**#enter image source here