A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 8 8 and 6 6 and the pyramid's height is 4 4. If one of the base's corners has an angle of pi/3π3, what is the pyramid's surface area?

1 Answer
sankarankalyanam
Oct 25, 2017

Total Surface Area T S A = **81.5103TSA=∗81.5103

Explanation:

CH = 6 sin (pi/3)= 3sqrt3CH=6sin(π3)=3√3
Area of parallelogram base = 8*b1 = 24sqrt3= color(blue)(41.5692)=8⋅b1=24√3=41.5692

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(4^2+ (8/2)^2)= 5.6569EF=h1=√h2+(a2)2=√42+(82)2=5.6569
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*6*5.6569=color(red)(16.9706)

EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(6/2)^2 )= 5
Area of Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*5 = color(red)(20)

Lateral surface area = 2* DeltaAED + 2*Delta CED
=( 2 * 16.9706)+ (2* 5)= color(blue)(43.9411)

Total surface area =Area of parallelogram base + Lateral surface area = 41.5692 + 43.9411= color(green)(85.5103)

Total Surface Area # T S A = **81.5103**#enter image source here

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