A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #2 # and #1 # 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 · 2017-12-07T05:28:34

T S A = 15.0888

#CH = 1 * sin ((5pi)/6) = 0.5#
Area of parallelogram base #= 2* b1 = 2*0.5 = color(red)(1 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(7^2+ (2/2)^2)= 7.0711#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1* 7.0711= #color(red)(3.5356)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(7^2+(1/2)^2 )= 7.0178#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*2*7.0178 = color(red)( 27.0178)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 7.0711)+ (2* 7.0178) = color(red)(14.0888)#

Total surface area =Area of parallelogram base + Lateral surface area # = 1 + 14.0888 = 15.0888#

Total Surface Area # T S A = **15.0888**#enter image source here