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 #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 · 2017-12-07T06:14:19

T S A = 89.3341

#CH = 3 * sin (pi/4) = 2.1213#
Area of parallelogram base #= a * b1 = 9*2.1213 = color(red)(19.0917 )#

#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)*3* 7.5= #color(red)(11.25)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(3/2)^2 )= 6.1847#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*9*6.1847 = color(red)( 23.8712)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 11.25)+ (2* 23.8712) = color(red)(70.2424)#

Total surface area =Area of parallelogram base + Lateral surface area # = 19.0917 + 70.2424 = 89.3341#

Total Surface Area # T S A = **89.3341**#enter image source here