A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #4 # and #7 # and the pyramid's height is #8 #. 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 · 2017-12-06T15:00:02

T S A = 116.9005

#CH = 4 * sin (pi/3) = 4 sin (60) = 3.4641#
Area of parallelogram base #= 7* b1 = 7*3.46 = color(red)(24.2487 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(8^2+ (7/2)^2)= 8.7321#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4* 8.7321= #color(red)(17.4642)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(8^2+(4/2)^2 )= 8.2642#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*8.2462 = color(red)( 28.8617)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 17.4642)+ (2* 28.8617) = color(red)(92.6518)#

Total surface area =Area of parallelogram base + Lateral surface area # = 24.2487 + 92.6518 = 116.9005#

Total Surface Area # T S A = **116.9005**#enter image source here