A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #6 # and #3 # and the pyramid's height is #3 #. If one of the base's corners has an angle of #pi/4#, what is the pyramid's surface area?

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Answer 1 · 2018-01-05T03:20:54

T S A = 45.5801

#CH = 3 * sin ((pi)/4) = 2.1213#
Area of parallelogram base #= a * b1 = 6* 2.1213 = color(red)(12.7279)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(3^2+ (6/2)^2)= 4.2426#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3* 4.2426= #color(red)(6.3638)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(3^2+(3/2)^2 )= 3.3541#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*6*3.3541 = color(red)( 10.0623)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 6.3638)+ (2* 10.0623) = color(red)(32.8522)#

Total surface area =Area of parallelogram base + Lateral surface area # = 12.7279 + 32.8522 = 45.5801#

Total Surface Area # T S A = **45.5801**#enter image source here