A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #4 # and #5 # 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 · 2017-12-14T12:53:16

T S A = 58.5491

#CH = 4 * sin (pi/3) = 3.464#
Area of parallelogram base #= a * b1 = 5*3.464 = color(red)(17.3205)#

#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)*4* 4.717= #color(red)(9.434)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(4/2)^2 )= 4.4721#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*5*4.4721 = color(red)( 11.1803)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 9.434)+ (2* 11.1803) = color(red)(41.2286)#

Total surface area =Area of parallelogram base + Lateral surface area # = 17.3205 + 41.2286 = 58.5491#

Total Surface Area # T S A = **58.5491**#enter image source here