Question

A pyramid has a parallelogram as base and its peak is directly above its center. The sides of the base have lengths of `7and 3` and the pyramid's height is `8`. If one of the base's corners has an angle of `pi/3`, what is the pyramid's surface area?

Answer 1

Total Surface Area = 231.9635

Explanation:

`CH = 3*sin (pi/3) = (3sqrt3)/2`
Area of parallelogram base `= a* b1 = (7 * 3sqrt3)/2 = 18.1865`

`EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(8^2+ (7/2)^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`