A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #6 # and #5 # and the pyramid's height is #6 #. If one of the base's corners has an angle of #(5pi)/6#, what is the pyramid's surface area?

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Answer 1 · 2017-12-12T11:14:04

T S A = 87.541

#CH = 5 * sin (pi/6) = 2.5#
Area of parallelogram base #= a * b1 = 6*2.5 = color(red)(15 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(6^2+ (6/2)^2)= 6.7082#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*5* 6.7082= #color(red)(16.7705)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(5/2)^2 )= 6.5#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*6*6.5 = color(red)( 19.5)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 16.7705)+ (2* 19.5) = color(red)(72.541)#

Total surface area =Area of parallelogram base + Lateral surface area # = 15 + 72.541 = 87.541#

Total Surface Area # T S A = **87.541**#enter image source here