A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #7 # and #2 # 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 · 2017-12-07T05:18:54

T S A = 51.616

#CH = 2 * sin (pi/3) = 2 sin (60) = 1.732#
Area of parallelogram base #= 7* b1 = 7*1.732 = color(red)(12.124 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(4^2+ (7/2)^2)= 5.3151#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*2* 5.3151= #color(red)(5.3151)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(2/2)^2 )= 4.1231#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*4.1231 = color(red)( 14.4309)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 5.3151)+ (2* 14.4309) = color(red)(39.492)#

Total surface area =Area of parallelogram base + Lateral surface area # = 12.124 + 39.492 = 51.616#

Total Surface Area # T S A = **51.616**#enter image source here