A pyramid has a parallelogram as base and its peak is directly above its center. The sides of the base have lengths of 7and 37and3 and the pyramid's height is 88. If one of the base's corners has an angle of pi/3π3, what is the pyramid's surface area?

1 Answer
sankarankalyanam
Oct 20, 2017

Total Surface Area = 231.9635

Explanation:

CH = 3*sin (pi/3) = (3sqrt3)/2CH=3⋅sin(π3)=3√32
Area of parallelogram base = a* b1 = (7 * 3sqrt3)/2 = 18.1865=a⋅b1=7⋅3√32=18.1865

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(8^2+ (7/2)^2)= 76.25EF=h1=√h2+(a2)2=√82+(72)2=76.25
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3*76.25=114.375

EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(8^2+(3/2)^2 ) = 66.25
Area of Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*3*66.25= 99.375

Lateral surface area = 2* DeltaAED + 2*Delta CED
= 114.375 + 99.375 = 213.75

Total surface area =Area of parallelogram base + Lateral surface area = 18.1865 + 213.75 = 231.9365

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