A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #2 # and #3 # 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 · 2017-12-07T05:59:05

T S A = 38.5312

#CH = 2 * sin ((5pi)/6) = 1#
Area of parallelogram base #= a* b1 = 3*1 = color(red)(3)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(7^2+ (3/2)^2)= 7.1589#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*2* 7.1589= #color(red)(7.1589)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(7^2+(2/2)^2 )= 7.0711#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*3*7.0711 = color(red)( 10.6067)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 7.1589)+ (2* 10.6067) = color(red)(35.5312)#

Total surface area =Area of parallelogram base + Lateral surface area # = 3 + 35.5312 = 38.5312#

Total Surface Area # T S A = **38.5312**#enter image source here