A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #2 # and #9 # and the pyramid's height is #4 #. If one of the base's corners has an angle of #pi/3#, what is the pyramid's surface area?

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Answer 1 · 2017-12-12T11:22:19

T S A = 64.7376

#CH = 2 * sin (pi/3) = 1.732#
Area of parallelogram base #= a * b1 = 9*1.732 = color(red)(15.588 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(4^2+ (9/2)^2)= 6.0208#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*2* 6.0208= #color(red)(6.0208)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(2/2)^2 )= 4.1231#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*9*4.1231 = color(red)( 18.554)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 6.0208)+ (2* 18.554) = color(red)(49.1496)#

Total surface area =Area of parallelogram base + Lateral surface area # = 15.588 + 49.1496 = 64.7376#

Total Surface Area # T S A = **64.7376**#enter image source here