A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #2 # and #7 # 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 · 2017-12-26T13:14:42

T S A = 41.2552

#CH = 2 * sin ((pi)/4) = 1.4142#
Area of parallelogram base #= a * b1 = 7*1.4142 = color(red)(9.8994)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(3^2+ (7/2)^2)= 4.6098#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*2* 4.6098= #color(red)(4.6098)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(3^2+(2/2)^2 )= 3.1623#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*3.1623 = color(red)( 11.0681)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 4.6098)+ (2* 11.0681) = color(red)(31.3558)#

Total surface area =Area of parallelogram base + Lateral surface area # = 9.8994 + 31.3558 = 41.2552#

Total Surface Area # T S A = **41.2552**#enter image source here