A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #4 # and #9 # and the pyramid's height is #2 #. If one of the base's corners has an angle of #(3pi)/8#, what is the pyramid's surface area?

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Answer 1 · 2017-12-06T13:26:20

T S A = 67.7804

#CH = 4 * sin (pi/4) = 4 sin (45) = 4/sqrt2 = 2.8284#
Area of parallelogram base #= 9* b1 = 9*2.8284 = color(red)(22.6272 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(2^2+ (9/2)^2)= 4.9244#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4* 4.9244 = #color(red)(9.8488)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(2^2+(4/2)^2 )= 2.8284#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*9*2.8284 = color(red)( 12.7278)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 9.8488)+ (2* 12.7278) = color(red)(45.1532)#

Total surface area =Area of parallelogram base + Lateral surface area # = 22.6272 + 45.1532 = 67.7804#

Total Surface Area # T S A = **67.7804**#enter image source here