A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #7 # and #8 # 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 · 2017-12-13T09:46:02

T S A = 111.556

#CH = 7 * sin (pi/4) = 4.9497#
Area of parallelogram base #= a * b1 = 8*4.9497 = color(red)(39.5976 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(3^2+ (8/2)^2)=5#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*7* 5= #color(red)(17.5)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(3^2+(7/2)^2 )= 4.6098#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*4.6098 = color(red)( 18.4792)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 17.5)+ (2* 18.4792) = color(red)(71.8594)#

Total surface area =Area of parallelogram base + Lateral surface area # = 39.5976 + 71.8594 = 111.556#

Total Surface Area # T S A = **111.556**#enter image source here