A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #3 # and #7 # and the pyramid's height is #4 #. 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 · 2017-12-07T05:42:05

T S A = 65.2506

#CH = 3 * sin ((3pi)/8) = 4 sin (60) = 2.7716#
Area of parallelogram base #= 7* b1 = 7*2.7716= color(red)(19.4012 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(4^2+ (7/2)^2)= 5.3151#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3* 5.3151= #color(red)(7.9727)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(3/2)^2 )= 4.272#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*4.272 = color(red)( 14.952)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 7.9727)+ (2* 14.952) = color(red)(45.8494)#

Total surface area =Area of parallelogram base + Lateral surface area # = 19.4012 + 45.8494 = 65.2506#

Total Surface Area # T S A = **65.2506**#enter image source here