A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #4 # and #7 # 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 · 2017-12-26T12:52:06

T S A = 91.8558

#CH = 4 * sin ((pi)/4) = 2.8284#
Area of parallelogram base #= a * b1 = 7*2.8284 = color(red)(19.7988)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(6^2+ (7/2)^2)= 6.9462#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4* 6.9462= #color(red)(13.8924)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(4/2)^2 )= 6.3246#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*6.3246 = color(red)( 22.1361)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 13.8924)+ (2* 22.1361) = color(red)(72.057)#

Total surface area =Area of parallelogram base + Lateral surface area # = 19.7988 + 72.057 = 91.8558#

Total Surface Area # T S A = **91.8558**#enter image source here