A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #6 # and #7 # and the pyramid's height is #2 #. If one of the base's corners has an angle of #(5pi)/6#, what is the pyramid's surface area?

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Answer 1 · 2017-12-07T05:49:20

T S A = 70.4258

#CH = 6 * sin ((5pi)/6) = 3#
Area of parallelogram base #= 7* b1 = 7*3 = color(red)(21 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(2^2+ (7/2)^2)= 4.011#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*6* 4.0311= #color(red)(12.0933)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(2^2+(6/2)^2 )= 3.6056#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*3.6056 = color(red)( 12.6196)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 12.0933)+ (2* 12.6196) = color(red)(49.4258)#

Total surface area =Area of parallelogram base + Lateral surface area # = 21 + 49.4258 = 70.4258#

Total Surface Area # T S A = **70.4258**#enter image source here