A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #6 # and #4 # and the pyramid's height is #7 #. 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 · 2017-12-20T12:26:07

T S A = 116.4108

#CH = 4 * sin ((5pi)/6) = 8#
Area of parallelogram base #= a * b1 = 6*8 = color(red)(48)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(7^2+ (6/2)^2)= 7.6158#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4* 7.6158= #color(red)(15.2316)#

#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)*6*6.3246 = color(red)( 18.9738)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 15.2316)+ (2* 18.9738) = color(red)(68.4108)#

Total surface area =Area of parallelogram base + Lateral surface area # = 48 + 68.4108 = 116.4108#

Total Surface Area # T S A = **116.4108**#enter image source here