A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #5 # and #3 # 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 · 2017-12-06T13:55:02

T S A = 48.5015

#CH = 3 * sin (pi/3) = 3sin (60)= 2.5981#
Area of parallelogram base #= 5* b1 = 5*2.5981 = color(red)(12.9905 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(4^2+ (5/2)^2)= 4.717#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3* 4.717 = #color(red)(7.0755)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(3/2)^2 )= 4.272#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*5*4.272 = color(red)( 10.68)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 7.0755)+ (2* 10.68) = color(red)(35.511)#

Total surface area =Area of parallelogram base + Lateral surface area # = 12.9905 + 35.511 = 48.5015#

Total Surface Area # T S A = **48.5015**#enter image source here