A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #7 # and #1 # 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 · 2017-12-23T12:35:04

T S A = 46.2275

#CH = 1 * sin ((pi)/4) = 0.7071#
Area of parallelogram base #= a * b1 = 7*0.7071 = color(red)(4.9497)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(5^2+ (7/2)^2)= 6.1033#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1* 6.1033= #color(red)(3.0517)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(5^2+(1/2)^2 )= 5.0249#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*5.0249 = color(red)( 17.5872)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 3.0517)+ (2* 17.5872) = color(red)(41.2778)#

Total surface area =Area of parallelogram base + Lateral surface area # = 4.9497 + 41.2778 = 46.2275#

Total Surface Area # T S A = **46.2275**#enter image source here