A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #2 # and #7 # and the pyramid's height is #6 #. If one of the base's corners has an angle of #pi/4 #, what is the pyramid's surface area?

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Answer 1 · 2017-12-06T14:14:34

T S A = 66.37

#CH = 2 * sin (pi/4) = 2 sin (45) = 1.414#
Area of parallelogram base #= 7* b1 = 7*1.414 = color(red)(9.898 )#

#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)*2* 6.9462 = #color(red)(6.9462)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(2/2)^2 )= 6.0828#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*6.0828 = color(red)( 21.2898)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 6.9462)+ (2* 21.2898) = color(red)(56.472)#

Total surface area =Area of parallelogram base + Lateral surface area # = 9.898 + 56.472 = 66.37#

Total Surface Area # T S A = **66.37**#enter image source here