A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #2 # and #1 # and the pyramid's height is #6 #. 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 · 2017-12-20T12:11:55

T S A = 19.5798

#CH = 1 * sin (pi/4) = 0.707#
Area of parallelogram base #= a * b1 = 2 * 0.707 = color(red)(1.414)#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(6^2+ (2/2)^2)= 6.0828#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1* 6.0828= #color(red)(3.0414)#

#EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(1/2)^2 )= 6.0415#
Area of #Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*2*6.0415 = color(red)( 6.0415)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 3.0414)+ (2* 6.0415) = color(red)(18.1658)#

Total surface area =Area of parallelogram base + Lateral surface area # = 1.414 + 18.1658 = 19.5798#

Total Surface Area # T S A = **19.5798**#enter image source here