A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #6 # 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 · 2017-12-25T12:25:01

T S A = 52.7226

#CH = 1 * sin ((pi)/6) = 0.5#
Area of parallelogram base #= a * b1 = 6*0.5 = color(red)(3)#

#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)*1* 7.6158= #color(red)(3.8079)#

#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)*6*7.0178 = color(red)( 21.0534)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 3.8079)+ (2* 21.0534) = color(red)(49.7226)#

Total surface area =Area of parallelogram base + Lateral surface area # = 3 + 49.7226 = 52.7226#

Total Surface Area # T S A = **52.7226**#enter image source here