A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #3 # and #8 # and the pyramid's height is #2 #. 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 · 2017-12-12T10:21:45

T S A is 45.4164

#CH = b1 = 3 * sin (pi/6) = 1.5#
Area of parallelogram base #= a * b1 = 8*1.5 = color(red)(12 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(2^2+ (8/2)^2)= 4.4721#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3* 4.4721= #color(red)(6.7082)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(2^2+(3/2)^2 )= 2.5#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*2.5 = color(red)( 10)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 6.7082)+ (2* 10) = color(red)(33.4164)#

Total surface area =Area of parallelogram base + Lateral surface area # = 12 + 33.4164 = 45.4164#

Total Surface Area # T S A = **45.4164**#enter image source here