A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #7 # and #1 # and the pyramid's height is #6 #. 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 · 2017-12-07T04:58:50

T S A = 55.1538

#CH = 1 * sin (pi/3) = 1 sin (60) = 0.866#
Area of parallelogram base #= 7* b1 = 7*0.866 = color(red)(6.062 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(6^2+ (7/2)^2)= 6.9462#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1* 6.9462= #color(red)(3.4731)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(1/2)^2 )= 6.0208#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*6.0208 = color(red)( 21.0727)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 3.4731)+ (2* 21.0727) = color(red)(49.0918)#

Total surface area =Area of parallelogram base + Lateral surface area # = 6.062 + 49.0918 = 55.1538#

Total Surface Area # T S A = **55.1538**#enter image source here