A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #1 # and #4 # 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 · 2017-12-25T12:10:07

T S A = 13.0748

#CH = b1 = 1 * sin ((pi)/6) = 0.5#
Area of parallelogram base #= a * b1 = 4*0.5 = color(red)(2)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(2^2+ (4/2)^2)= 2.8284#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1* 2.8284= #color(red)(1.4142)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(2^2+(1/2)^2 )= 2.0616#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*4*2.0616 = color(red)( 4.1232)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 1.4142)+ (2* 4.1232) = color(red)(11.0748)#

Total surface area =Area of parallelogram base + Lateral surface area # = 2 + 11.0748 = 13.0748#

Total Surface Area # T S A = **13.0748**#enter image source here