A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #8 # and #7 # 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 · 2017-10-25T14:23:55

Total Surface Area #T S A = **70.4414**#

#CH = 7 * sin (pi/3)= (7sqrt3)/2#
Area of parallelogram base #= 8 * b1 = (8 * 7sqrt3)/2 = color(red)(48.4974)#

#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)*7*5.6569 = color (red)(5.6569)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(7/2)^2 )= 5.3151#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*5.3151 = color (red)(5.3151)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 5.6569)+ (2* 5.3151)= color(blue)21.944#

Total surface area =Area of parallelogram base + Lateral surface area # = 48.4974 + 21.944= color (green)(70.4414)#

Total Surface Area # T S A = **70.4414**#enter image source here