A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #4 # and #5 # and the pyramid's height is #6 #. 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 · 2017-12-20T12:36:07

T S A = 74.9435.

#CH = 4 * sin (pi/3) = 3.4641#
Area of parallelogram base #= a * b1 = 5*3.4641 = color(red)(17.3205)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(6^2+ (5/2)^2)= 6.5#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4* 6.5= #color(red)(13)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(4/2)^2 )= 6.3246#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*5*6.3246 = color(red)( 15.8115)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 13)+ (2* 15.8115) = color(red)(57.623)#

Total surface area =Area of parallelogram base + Lateral surface area # = 17.3205 + 57.623 = 74.9435#

Total Surface Area # T S A = **74.9435**#enter image source here