A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #3 # and #2 # and the pyramid's height is #5 #. 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 · 2017-10-25T21:44:14

Total Surface Area # T S A = 33.2695

#CH = 2 * sin (pi/4)= sqrt2#
Area of parallelogram base #= 3* b1 = 3sqrt2= color(red)(4.2426)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(5^2+ (5/2)^2)= 5.5902#
Area of #Delta AED = BEC = (1/2)*b*h_1 = (1/2)*2*5.5902= color(red)(5.5902)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(5^2+(2/2)^2 )= 5.099#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*3*5.099= color(red)(8.9233)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 5.5902)+ (2* 8.9233)= color(red)(29.0269)#

Total surface area =Area of parallelogram base + Lateral surface area # = 4.2426+ 29.0269 = 33.2695#

Total Surface Area # T S A = **33.2695**#enter image source here