A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #7 # and #2 # and the pyramid's height is #8 #. 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 · 2017-12-23T12:42:02

T S A = 86.835

#CH = 2 * sin ((3pi)/8) = 1.8478#
Area of parallelogram base #= a * b1 = 7*1.8478 = color(red)(12.9346)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(8^2+ (7/2)^2)= 8.7321#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*2* 8.7321= #color(red)(8.7321)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(8^2+(2/2)^2 )= 8.0623#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*8.0623 = color(red)( 28.2181)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 8.7321)+ (2* 28.2181) = color(red)(73.643)#

Total surface area =Area of parallelogram base + Lateral surface area # = 12.9346 + 73.9004 = 86.835#

Total Surface Area # T S A = **86.835**#enter image source here