A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #3 # and #5 # 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 · 2017-12-07T04:49:05

T S A = 53.4781

#CH = 3 * sin (pi/4) = 2.1213#
Area of parallelogram base #= 5* b1 = 5*2.1213 = color(red)(10.6065 )#

#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)*3* 5.5902= #color(red)(8.3853)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(5^2+(3/2)^2 )= 5.2202#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*5*5.2202 = color(red)( 13.0505)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 8.3853)+ (2* 13.0505) = color(red)(42.8716)#

Total surface area =Area of parallelogram base + Lateral surface area # = 10.6065 + 42.8716 = 53.4781#

Total Surface Area # T S A = **53.4781**#enter image source here