A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #2 # and #8 # 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?

Question

1119 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-13T09:55:58

T S A = 42.8561

#CH = 2 * sin (pi/4) = 1.414#
Area of parallelogram base #= a * b1 = 8*1.414 = color(red)(11.3137)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(3^2+ (4/2)^2)= 3.6056#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*2* 3.6056= #color(red)(3.6056)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(3^2+(1/2)^2 )= 3.0414#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*3.0414 = color(red)( 12.1656)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 3.6056)+ (2* 12.1656) = color(red)(31.5424)#

Total surface area =Area of parallelogram base + Lateral surface area # = 11.3137 + 31.5424 = 42.8561#

Total Surface Area # T S A = **42.8561**#enter image source here