A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #3 # and #9 # and the pyramid's height is #3 #. 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 · 2017-12-02T13:58:32

Total surface area of the pyramid = #color(blue)(65.5039)#

#CH = 3 * sin (pi/4) = 3sin (45) = 3/sqrt2 = 2.1213#
Area of parallelogram base #= 9* b1 = 9*2.1213 = color(red)(19.0919 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(3^2+ (9/2)^2)= 5.4083#
Area of #Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3*5.4083 = color(red)(8.1125)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(3^2+(3/2)^2 )= 3.3541#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*9*3.3541 = color(red)( 15.0935)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 8.1125)+ (2* 15.0935) = color(red)(46.412)#

Total surface area =Area of parallelogram base + Lateral surface area # = 19.0919 + 46.412 = 65.5039#

Total Surface Area # T S A = **65.5039**#enter image source here