A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #2 # and #9 # 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 · 2017-12-14T12:59:38

T S A = 82.4708

#CH = 2 * sin (pi/4) = 1.414#
Area of parallelogram base #= a * b1 = 9*1.414 = color(red)(12.726)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(6^2+ (9/2)^2)= 7.5#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*2* 7.5= #color(red)(7.5)#

#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)*9*6.0828 = color(red)( 27.3724)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 7.5)+ (2* 27.3724) = color(red)(69.7448)#

Total surface area =Area of parallelogram base + Lateral surface area # = 12.726 + 69.7448 = 82.4708#

Total Surface Area # T S A = **82.4708**#enter image source here