A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #8 # and #1 # and the pyramid's height is #7 #. 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 · 2017-12-06T13:40:58

T S A = 68.2048

#CH = 8 * sin ((5pi)/6) = 4#
Area of parallelogram base #= 1* b1 = 1*4 = color(red)(4 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(7^2+ (8/2)^2)= 8.0623#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1* 8.0623 = #color(red)(4.0312)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(7^2+(1/2)^2 )= 7.0178#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*7.0178 = color(red)( 28.0712)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 4.0312)+ (2* 28.0712) = color(red)(64.2048)#

Total surface area =Area of parallelogram base + Lateral surface area # = 4 + 64.2048 = 68.2048#

Total Surface Area # T S A = **68.2048**#enter image source here