A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #3 # 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 · 2017-12-06T14:26:43

T S A = 52.1571

#CH = 3 * sin (pi/4) = 3 sin (45) = 3/sqrt2 = 2.1213#
Area of parallelogram base #= 7* b1 = 7*2.1213 = color(red)(14.8491 )#

#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)*3* 4.6098= #color(red)(6.9147)#

#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)*7*3.3541 = color(red)( 11.7393)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 6.9147)+ (2* 11.7393) = color(red)(37.308)#

Total surface area =Area of parallelogram base + Lateral surface area # = 14.8491 + 37.308 = 52.1571#

Total Surface Area # T S A = **52.1571**#enter image source here