Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `8` and `6` and the pyramid's height is `9`. If one of the base's corners has an angle of `(5pi)/12`, what is the pyramid's surface area?

Answer 1

Total surface area `= 168.7614`

Explanation:

`CH = 6*sin ((5pi)/12) = 5.7956`
Area of parallelogram base `= a* b1 = 6* 5.7956=33.7736`

`EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(9^2 + 4^2) = sqrt97`
Area of `Delta AEF = BEC = (1/2)*b*h_1 = (1/2)*6*sqrt97=3sqrt97`

`EG = h_2 = sqrt(h^2+(b/2)^2 = sqrt(9^2 + 3^2 = sqrt90`
Area of `Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*sqrt90=4sqrt90`

Lateral surface area = `2* DeltaAED + 2*Delta CED`
`= 2*3sqrt97 + 2*4sqrt90 = 59.0931 + 75.8947 = 134.9878`

Total surface area =Area of parallelogram base + Lateral surface area `= 33.7736 + 134.9878 =168.7614`