Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `4` and `8` and the pyramid's height is `7`. If one of the base's corners has an angle of `(5pi)/6`, what is the pyramid's surface area?

Answer 1

Total surface area `= 74.2409`

Explanation:

`CH = 4*sin ((5pi)/6) = 2`
Area of parallelogram base `= a* b1 = 8*2=16`

`EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(7^2+4^2)= sqrt65`
Area of `Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4*sqrt65=2sqrt65`

`EG = h_2 = sqrt(h^2+(b/2)^2 = sqrt(7^2+2^2) = sqrt53`
Area of `Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*sqrt53=4sqrt53`

Lateral surface area = `2* DeltaAED + 2*Delta CED`
`= 2*2sqrt65 + 2*4sqrt53= 32.249 + 58.2409`

Total surface area =Area of parallelogram base + Lateral surface area `= 16 + 58.2409 = 74.2409`