A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #3 # 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 · 2017-12-12T10:35:28

T S A is 22.4158

#CH = 1 * sin (pi/4) = 0.707#
Area of parallelogram base #= a * b1 = 3*0.707 = color(red)(2.121 )#

#EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(5^2+ (3/2)^2)= 5.22#
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1* 5.22= #color(red)(2.61)#

#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)*3*5.0249 = color(red)( 7.5374)#

Lateral surface area = #2* DeltaAED + 2*Delta CED#
#=( 2 * 2.61)+ (2* 7.5374) = color(red)(20.2948)#

Total surface area =Area of parallelogram base + Lateral surface area # = 2.121 + 20.2948 = 22.4158#

Total Surface Area # T S A = **22.4158**#enter image source here